Game of NEET 3.0 ⚔ Chemistry | Thermodynamics Complete Chapter| NEET 2026 | Wassim Bhatt Sir

Hey guys, good evening. Good evening, good evening and welcome back again to your academy English channel. I hope all of you doing great, having a good time. So my dear students, let me know quickly in the chats if all of you can hear me. If I'm perfectly audible, visible to every one of you, let me know quickly in the chats. Everyone, let me know quickly in the chats if all Of you can hear me. If I am perfectly audible, visible to every one of you. Good evening people. Good evening. Good evening and welcome back. Good

evening and welcome back. All right, I had to delay the session a bit. Yeah, because there were some technical glitchs. So, it took some time to fix them. That's the reason the session got delayed a bit. All right. So, I believe I'm perfectly Audible, visible to everyone. Yeah. Right. I believe I'm perfectly audible visible to everyone. So my dear students as you all must be knowing today we are starting one of the amazing chapters of your chemistry right and you know the name of the chapter is thermodynamics. So it's going to be a marathon of

the chapter thermodynamics wherein I shall be completing entire thermodynamics Chapter in this particular session itself. Yeah, I shall be completing the entire thermodynamics chapter in this particular session itself. So are you guys ready to be with me till the end? I just want to know that first. I just want to know that first. Are you guys ready to be with me till the end? I don't know how long it's going to take. It can take 10 hours, 11 hours, 12 hours, 13 hours, whatever hours. I have No idea. Let's see. Yeah. And I know a

lot of students have been asking me about the inorganic organic as well. So I'll be taking some of the organic inorganic chapters as well in this particular series. Don't worry. Don't worry. That's already planned. That is already planned. Don't worry about that. Yeah, perfect. So, as you know, I keep on Starting the session with one simple question. What is that question? Which part of the country you guys are watching me right now from? Which state you guys are watching me from? I just want to know that first. I just want to know that first. Which

part of the country you guys are watching me from right now? Okay. Tamil Nadu, Karnataka, Hana, Hyderabad, Maharashtra, [snorts] Talangaram, Andhra Pradesh, Rajbag. Okay. All right. All right. Kashmir. Perfect. Perfect. Perfect. So should we get going then? Should we start? Should we start people all right? All right. Okay. Perfect. So without wasting lot of our time, let's get going then. Let's get started with this amazing chapter that is thermodynamics which majority of the Students they consider it tough as well right but I'll be starting from the basics I'll be starting from the scratch right you

won't be feeling it tough at all yeah okay perfect and at the same time the session PDF of this particular session I shall be sharing on my telegram if you are not the part of my telegram channel yet be that part of the telegram channel as soon as possible which is wasimin but Chemistry official W A S S I M P H A T Chemistry official is the name of the telegram channel my dear students on which I keep on sharing on which I keep on sharing the session PDFs handwritten session PDFs okay all right

perfect then so let's get going let's get started with the chapter thermodynamics my dear students this thermodynamics chapter I shall be starting From the basic terminologies. There are some basic terminologies which you need to know in order to understand this particular chapter in a proper and effective way. And what are those basic terminologies? Let's get familiar with those basic terminologies first. Right? So first of all, as I've already mentioned the term here that is system. Okay. Let's first of all read the definition of system and then I'll make you understand what the system exactly Is

all about. My dear students, if you look at the definition of the system carefully, look at it. The part of the universe which is under thermodynamic investigation. Surrounding the part of the universe except system boundary anything which separates system from the surroundings. Now what does all this mean? Try to understand very carefully what I'll be saying. My dear students, imagine that I have Taken a container over here. Imagine that this is a container and in this container for example, I am keeping H2O liquid. I am keeping H2O liquid in this particular container. I'm keeping water

inside this container. Okay. Let's say I want to increase the temperature of this water. For example, I want to increase the temperature of this water. Let's assume I want to increase the temperature of the water by 1° centigrade. For example, I want to increase the temperature of this water by how much? By 1° centigrade. And for example, I have to do an investigation wherein I have to check how much amount of heat how much amount of heat is required. For example, I have to do an investigation about this water in the container. And what is

the investigation? I have to check how much amount of heat is to be supplied to this water to increase its Temperature by 1°. So first of all, I have taken a container over here. In this container, I have taken water. Now I want to increase the temperature of the water by 1°ree and I have to do one investigation about this particular water which is in the container. What is the investigation? I have to check how much heat has to be supplied to this water to increase its temperature by 1°ree. Okay, this is the investigation. So

my dear Students, can I say this water in the container right now is under thermodynamic investigation. This water in the container right now is under thermodynamic investigation. And let me tell you that part of the universe, that part of the universe which will be under thermodynamic investigation. What do you call that as? You call that as the system. So I would say right now this water is under investigation. So water I'll be calling as my system. Okay? For Example, this is the marker. This is the marker. Okay? I want to do some thermodynamic investigation related

to this marker. I want to do some thermodynamic investigation related to this particular marker. So I would say this marker is right now under my investigation. Okay. So this marker will become my system. So that part of the universe that part of the universe which will be under thermodynamic investigation you call that part of the Universe as a system. Point number one. Point number two my dear students. Whatever will be there in the universe except system. Whatever will be there in the universe except system that is something which you call as the surrounding. For example,

this water is right now under investigation. So water is my system. Now apart from this water, whatever is there in the universe, apart from this water, whatever is there in the universe, that is what I'll be Calling as surrounding. That's what I'll be calling as surrounding. Okay? And let me tell you anything anything that separates system from the surroundings. Anything that separates system from the surroundings is what I'll be calling as the boundary. So as you can see these walls of the container they are separating system from the surroundings. So these are what I'll be

calling as this is what I'll be calling as the boundary. So I believe All the three terminologies are clear to you. System surrounding boundary. System is the part of the universe which is under thermodynamic investigation. Apart from the system, whatever is there in the universe, that's what you call as surrounding. And anything that is going to separate your system from the surroundings, you'll be calling that as the boundary. Am I clear to everyone? Let me know once in the chat quickly. Am I clear to everyone people? [snorts] Everybody. Everybody. People. Yeah. Great. Now comes the

types of the boundaries. Now comes the types of the boundaries. So how many different types of boundaries do we have? Okay. How many different types of walls do we have? My dear students, there is one type of the boundary which is called as real Boundary. Another type is called as imaginary boundary. Another type is called as diiaothermic boundary. And another type is called as adiabatic boundary. Now what are these boundaries? How do we classify them? Understand properly. See real boundary is the one that can be seen by the eyes. A real boundary is the one

that can be seen by the ice. Okay. Imaginary boundary cannot be seen by the eyes. Imaginary boundary is the one that Cannot be seen by the eyes. What is meant by that? First of all, see my dear students, do you see these walls of the container? Yes, I can see this. I can see these walls. I can see this wall of the container right which is separating system from the surrounding and that wall that boundary which you can see by your eyes that yes it is separating system from the surroundings you'll be calling this particular

boundary as the real boundary. So in short real boundary Is nothing but it is such type of the boundary that can be seen by the eyes. Number one. Number two imaginary. What is an imaginary boundary? in order to make you understand what is an imaginary boundary. Hey guys, for example, this is one container which I'm taking over here and I'm keeping the container open on one side. I'm keeping the container open on one side. The container is open on one side. Okay? Here in this particular container for example, I'm keeping Water. I'm keeping water inside

this particular container. Right? And imagine that this water in the container it is under investigation. This water is under investigation. So this water becomes my water. This water becomes my system. Okay. So this water in the container is my what? Is my system. Correct? Now my dear students outside this container what do you see? I would say outside the container there will be Atmosphere. That means there will be atmospheric gases outside the container. There will be atmospheric gases outside the container. Now since the container is open on one side, I would say these atmospheric gases

would have entered this part of the container as well. These atmospheric gases would have entered this part of the container as well because the container is open on one side. Okay? First of all, this water in the container is my system, right? Anything apart from the system is what I call a surrounding. So this is my surrounding. This is my surrounding. So these atmospheric gases I'm calling my surrounding here. Okay? Now people try to understand very carefully. If I select this part of the wall of the container, if I select this part of the wall

of the container, do you see this part of the wall of the container? This part, the shaded part, the shaded part, do you see this shaded part, this wall Of the container is separating system from the surroundings? Yes. So I'll be calling it as the real boundary. Do you see this part of the container? Absolutely, I can see it. This wall of the container is separating system from the surroundings. So it is my real boundary. Do you see this wall of the container? Yes, it is separating system from the surroundings. So, it is my real

boundary. It is my real boundary. Now, my dear students, try to understand Carefully. Try to understand very carefully. See, first of all, this is the top layer of water in the container. This is the top layer of water in the container. And above the top layer of water, above the top layer of water, what do we have? We have got atmospheric gases. Above the top layer of water, there there is atmospheric gas basically, right? Can I say between the top layer of water? Between the top layer of water and Between the top layer of water

and these atmospheric gases. Can I say between the top layer of water and the atmospheric gases there will be a fine boundary below which there will be water above which there will be atmospheric gases. Will you be able to see that boundary by your eyes? No. So I would say there is one imaginary boundary. There is one imaginary boundary below which there is water above which there are atmospheric gases. And that boundary here I'll be Calling as the imaginary boundary. Yes. That boundary here I'll be calling as the imaginary boundary. Perfect. Now comes a dithermic

boundary. What is a dithermic boundary? Simple again. Diiothermic boundary is the one that allows that allows the passage of heat that allows the passage of heat that allows the passage of heat between system and surroundings Between system and surroundings. So that particular wall that particular boundary through which heat can pass that particular wall that particular boundary through which heat can pass either from system to surroundings or from surroundings to system you'll be calling that particular wall you'll be calling that particular boundary as the diiaothermic boundary. Okay through which heat can pass. Yeah. Now my dear

students what would be adabatic Boundary? Adabatic boundary is the one through which adabatic boundary is the one through which heat does not pass. Adabatic boundary is the one through which heat does not pass. If heat is not passing through a wall, right, you'll be calling that particular wall as the adabatic wall. Adabatic boundary. So adabatic boundary it does not allow the passage of heat either from system to Surroundings or from surrounding to system. Right? I believe I'm clear with all these three terminologies. Perfect. [snorts] Right. My dear students, can you see this bottle? Okay. This

bottle. What is the best thing about this particular bottle? If you put hot water in this bottle, it takes time to cold. It takes time to cool down. If you put hot water in this particular bottle, it takes time to cool Down. Okay. What does that mean? What does that mean? What does that mean? I would say its walls, its walls are going what? They are trying to block the passage of heat between the system and surroundings, between water and the surroundings. They're trying to block the passage of heat, right? They're trying to block the

passage of heat between the water and the surroundings. And that particular boundary, that Particular wall, that particular wall which does not allow the passage of heat between the system and surroundings, you'll be calling that wall, that boundary as the adabatic boundary. Okay? And let me tell you, let me tell you, there is no boundary. There is no wall existing in the universe which is 100% adabatic. Okay, this is not your 100% adabatic wall. This is not your 100% adabatic wall. Because if this was 100% adabatic, then the hot water in this Bottle would not cool

down. Okay. So there is no wall as such which is 100% adabatic. I believe I'm clever with this. Now moving on. Moving on to something important. What is that? That is types of the system. That is types of the system. What are the different types of the systems which we have? My dear students, we have three types of the systems which we have to study. Number one, number one is called as open system. Number one is called as open system. Number two is called as the closed system. And number three is called as the isolated

system open system, closed system and the isoloplated system. So what is an open system? What is a closed system and what is an isolated system? Let's talk about them one by one. Okay, these are again Very simple simple things. Try to understand carefully what I'll be saying. My dear students, open system is basically the one. Just give me a second. Just give me a second. Just just give me a second. Okay. Open system is the one that allows the passage of energy As well as the matter as well as the matter. that allows the passage

of energy as well as the matter with the surroundings with the surroundings. Number one. Number two, what is a closed system? It allows the passage of energy comma not the matter. Nor the matter with the surroundings with the surroundings. And what is an isolated system? It neither allows the exchange of energy. It neither allows the exchange of energy nor the matter with the surroundings. It neither allows the exchange of energy nor the matter with the surroundings. Now what is meant by all this? Let's try to understand. See guys, just to make you understand. For example,

I'm taking An open container here and in this open container, I'm keeping water. Let's say this water in the container is under investigation. If the water in the container is under investigation, I'll be calling this water as my system. Number one. Number two, assume that these walls are diiaothermic. Assume that these walls are diiaothermic. If the walls are dithermic, that means through these walls heat can pass. through these walls Heat can pass right now what will happen this water is under investigation and outside the water what do we have we have got atmosphere so basically

in short I would say outside this container what do we have we have got surroundings we have got surroundings outside the container okay now my dear students since walls are dithermic so through walls heat can pass for example imagine that I'm supplying heat to this water. Imagine that I'm supplying heat to this water. I'm supplying heat to this water which is there in the container. When I'll be supplying heat to this water, what will happen? This water will try to evaporate. And when the water evaporates, these vapors will start going into the surroundings. These vapors

will start going into the surroundings. So, can I say between the system and surroundings, exchange of matter is taking place. Exchange of Matter. Vapors are going to the surroundings. That means exchange of matter is taking place and at the same time at the same time do you see exchange of energy also taking place? Absolutely. Exchange of energy as well as matter both are happening. So that particular system which allows the exchange of energy as well as matter with the surroundings. That particular system which allows the exchange of energy as well as matter with the Surroundings.

You'll be calling that particular system as a open system. Right? Now what is a closed system? Do one thing. Close this from the top. If you close this from the top, will there be exchange of matter between the system and surroundings? No. There won't be any exchange of matter with the surroundings. And that particular system which allows the exchange of energy between the system and surroundings, not the matter, you call that particular System as the closed system. And similarly that particular system which neither allows the exchange of energy nor the matter with the surroundings you'll

be calling that particular system as the isolated system which neither allows the exchange of energy nor the matter with the surroundings. You'll be calling that as the isolated system. Okay. Perfect. And my dear students let me tell you one more important thing. There is only one example of a perfectly isolated system. If I talk about a perfectly isolated system, there is only one example of the perfectly isolated system. Otherwise, there is no perfectly isolated system. For example, if you think you have kept water in this thermos flask, you have kept water in this thermos flask.

Right? Now, it is closed. So, exchange of Matter is not happening between system and surroundings. Right? Since exchange of energy is also I would say exchange of energy is not happening between the system and surroundings. Right? I would say 100% exchange of energy right I I would say these walls of the container they are not blocking the heat up to 100%. But still somehat heat is being exchanged between water and the surroundings. Okay. So this is not the Example of the perfectly you can call it as the perfectly is you can call it as the

isolated system to some extent right but it's not perfectly isolated it's not perfectly isolated but what is the example of the perfectly isolated system there's only one example of the perfectly isolated system what is that that is if you take universe if you take entire universe under thermodynamic investigation if you take entire universe under thermodynamic Investigation my dear students if you take entire entire universe. If you take entire universe, let's say this is your entire universe. Let's say this is your entire universe. Imagine the universe is inside the circle. Entire universe is inside the circle.

So outside this universe, what is there? Nothing. Outside this universe, what is there? Nothing. Everything. What is there that is inside this universe? Universe everything. Everything is inside the Circle. If you take this entire universe under investigation, so this universe is becoming your system. If you take this entire universe under investigation. So this universe is becoming my system. Now outside this universe there is nothing guys is there anything apart from the universe? Universe is everything right? I'm considering entire universe under investigation. So entire universe is becoming a system. So if there are no Surroundings will

there be any energy exchange? Will there be any matter exchange? So this universe if taken under investigation right it becomes your perfectly isolated system. So there is only one perfectly isolated system right? What is that? That is your universe. If you take it under thermodynamic investigation I believe this is clear to everyone. If yes let me know quickly in the chats. Let me know quickly in the chats if this Is clear. Everyone. Everyone. People. >> Yeah. All right, people. Perfect. Just give me a second. Just give me a second and we shall continue. Just give

me a second. All right, sir. You remind me of that Austrian painter. Austrian painter. Yeah. Is that? Do I remind you of an Austrian painter? All right, Bill. Let's move on. Let's Move on. Let's move on. So I believe the types of the system is absolutely clear to you. Let's move on to one more amazing term, one more important term. What is that? That is extensive and intensive properties. What are extensive properties and what are intensive properties? Try to understand carefully guys what these are because questions are asked from this particular topic. Right? And understand

what kind of questions will be asked. So before Before making you understand what these extensive properties are all about, let me write the definition. I would say these are the properties of the system. These are the properties of the system whose value chains whose value chains on changing whose value change on changing the size of the system Whose value change on changing the size of the system or the amount of substance present in the system or the amount of substance present in the system. These are the properties of the system whose value change whose value

change on changing the size of the system or the amount of substance present in the system. What does that mean? We'll understand that in some time. But before That you need to know what are the example of these extensive properties. You first of all need to know the examples of these extensive properties. And what are the examples? I'm writing the examples here. These are important questions are asked from the examples. Okay. have an eye on these examples properly. So first of all mass volume mass volume moles Heat capacity heat capacity internal energy internal energy enthalpy

enthalpy Entropy Entropy Gives free energy gives free energy. These are all the examples of what? These are all the examples of extensive properties. Right? So these are the properties of the System whose value change on changing the size of the system or the amount of substance present in the system. What do we call these properties as extensive properties. And here only I'll write otherwise intensive. Here only I'll write otherwise intensive. Otherwise intensive. So these are basically the examples of extensive properties. And let me write the examples of intensive properties as Well. Examples of intensive properties.

What are the examples of intensive properties? First of all, I'll be starting with temperature. We have got pressure, temperature, comma, pressure. You have got specific heat capacity, specific heat capacity. We have got molar heat capacity. We have got molar heat capacity. We have got concentration. Concentration, marity etc. Right? You have got pH, boiling point, boiling point, emf of the cell. These are all the examples of intensive properties. So let's try to understand the meaning of these uh intensive and extensive properties. Now let's try to understand the meaning of them. Let's try to understand the meaning

of them. See guys, imagine that. For example, this is a Block. It is an iron block. For example, imagine this is an this is the iron block which I'm taking over here. For example, this iron block is right now under thermodynamic investigation. For example, this iron block, this iron block is right now under thermodynamic investigation. So this complete iron block is my system right now. Let's say its mass is M. Let's say it volume is V. Let's say it temperature is T. This entire iron block is right now under This entire iron block is right

now under thermodynamic investigation. So this entire iron block I'll be calling as my system. So this entire block is my system whose mass is M, volume is V, temperature is T. Okay. Now for example, what I'm trying to do, let's assume that I'm dividing this iron block into two equal parts. Let's say I divided this iron block into two equal parts. And now I'm taking this smaller part into consideration. Now I'm taking the Smaller part into investigation. Since I divided this iron block into two equal parts and I'm taking this smaller part of the iron

block into investigation. So now the smaller part will be my system. Earlier earlier this entire iron block was my system. Now the smaller block now the smaller part of the block is my system. Okay. Is my system. Understand? Earlier this entire iron block was under investigation. So entire Iron block was my system earlier. Now I divided this iron block into two parts. And now I'm taking the smaller part. So this smaller part is right now under investigation. So this smaller part will be my system. Now tell me what will be the mass of this smaller

part? Is it going to be m or mx2? It's going to be mx2. What is going to be the volume of the smaller part v or vx2? It's going to be vx2. What is going to be the temperature of this block? The Temperature of the smaller block is going to be T only. It's simple. For example, this is the marker. Right? If you divide this marker into two equal parts, what will happen to the temperature of the smaller part? Whatever is the temperature of this complete marker, same is going to be the temperature of the

smaller part of the marker as well. Right? Now, my dear students, do you see earlier this entire block was under investigation? So, Entire block was earlier my system. Now, do you see the size of system getting changed? I'm changing the size of the system. I'm changing the size of the system. Right? Since I'm changing the size of the system, what is happening to mass? Mass is changing from m to mx2. What is happening to volume? Volume is changing from V to V by2. Those properties those properties whose value change those properties whose Value change on

changing the size of the system. those properties whose value change on changing the size of the system you'll be calling those properties as the extensive properties. So I would say on changing the size of the system, mass is changing, volume is changing. So mass and volume I'll be calling as extensive properties here. But is the temperature of the system changing? On changing the size of the system, temperature of the system does Not change. So I'll be calling this as the intensive property. So temperature is your intensive property. As simple as that. Right? So those properties

of the system whose value change on changing the size of the system you call those properties as extensive properties and those properties whose value does not change on changing the size of the system you call them as the intensive properties. Yes, someone is saying what about the Density? Density is also the example of intensive property. You can write it here also. Density is the example of intensive property. Okay, these these are the examples of extensive these are the examples of extensive properties and these are the examples of intensive properties. Correct? Now my dear students I

believe you understood the meaning of what is extensive and what is intensive. So in short size dependent Amount of substance dependent is your is your extensive property. Size independent amount of substance independent will be your intensive property. Size dependent means extensive. Size independent means intensive. Simple. Now there are some important features important parameters about this extensive and intensive properties. I would say some important some important features. Some important features of Extensive and intensive properties. Some important features of extensive and intensive properties. Let's try to understand what are these important features of extensive and intensive properties.

Try to understand people what I'll be saying. Point number one. Point number one. What is point number one? Point number one is division of extensive properties. Division of extensive properties. Division of extensive properties makes the new property makes the new property as intensive. Division of extensive properties makes the new property as intensive. What does that mean? What does that mean? What does that mean? See, let me make you understand this particular point. For example, M stands for mass. Mass is an example of extensive property. V stands for volume. Volume is the example of extensive property.

If I divide the two mass by volume becomes my density and density as I already told it is the example of intensive property. So whenever you divide to extensive properties whenever you divide to extensive properties the new property that arises the new property that comes into existence that is going to be always your intensive property. Right? For example N stands for moles Moles extensive. V stands for volume. Volume extensive moles per unit volume is what you call as concentration. So concentration becomes your intensive property. Okay. Again one important point. So whenever you divide two extensive

properties the new property that arises is what you call as the intensive property. Okay. Point number two my dear students. Point number two what is that? Extensive properties Are Extensive properties are additive in nature extensive properties are additive in nature. Whereas intensive properties Whereas intensive properties are non-additive. Extensive properties are additive in nature whereas intensive properties are non-additive. What does that mean? Try to understand this also. For example, Let's say this is a container and this container is completely filled with water. We have got one more container. Even this particular container is completely filled with

water. Let's say mass of this water is M. Volume is V. Temperature is T. Mass of this water is M. Volume is V. Temperature is T. So how many properties I defined here? I defined three properties. Mass, volume and temperature. Mass, volume and temperature. Out of these three Properties, these two are extensive and this is intensive. Mass and volume is extensive. Temperature is intensive. Right? Now for example, if I mix them up, if I mix them up in a bigger sized container, if I mix them up in a bigger sized container, so all this water

and all this water will finally reach this container will finally reach this container. Now tell me what is the mass of water in this container? I would say the mass of Water in this container will be m + m 2 m. Volume of water in this container will be v + v that is 2 v. But temperature of water over here will be still T only. For example, I've got one glass of water. One more glass of water. Temperature of water in this glass is 25°. Temperature of water in this glass is also 25°. Now

mix them up in a bigger sized container. When you mix them up in the bigger sized container, all that Water goes into that bigger size container. What will be the temperature of that water also? Temperature of that water also will be how much? It will be 25° centiggrade only. As simple as that. This is common sense. Now my dear students, tell me one thing. How many extensive properties did we have? We had two extensive properties, mass and volume. And there was one intensive property that was temperature. Did you see m + m makes it 2

m. V + v makes it 2 V. But t + t did not make it as 2t. T + t did not make it as 2t. So I would say these extensive properties extensive properties which is mass and volume here. These extensive properties you can directly add or subtract but intensive properties you cannot directly add or subtract. Intensive properties you could not directly add or subtract. This is one simple point which you have to remember. Extensive properties they're additive in nature whereas intensive Properties are non-additive in nature. Yeah. Am I clear with this everyone? Am

I clever with this everyone? Everyone in the chats wants people quick. perfect guys. Now there is one more point which I would want to tell you here. What is that? I'm writing it. Remember whenever an extensive property Whenever an extensive property is expressed, Whenever an extensive property is expressed, per unit mass per unit mole or per unit volume per unit volume. The new property becomes intensive. The new property becomes intensive. What does that mean? It's again a simple statement. Remember it carefully. See guys, for example, The C C is what I call as heat capacity

which we have to study in some time. Heat capacity. Heat capacity is the example of what? It is the example of extensive property. It is example of extensive property. Now if I do one thing, if I write heat capacity per unit mass, heat capacity per unit mass, this heat capacity per unit mass is called as the specific heat capacity, it is called as specific heat capacity. And this specific heat capacity, it is Intensive in nature. It is intensive in nature. For example, heat capacity which is the extensive property. If you express it per unit mole,

if you express it per unit mole, it becomes more read capacity. It becomes more read capacity. heat capacity was extensive expressed per unit mole. It's called as molar heat capacity is called as molar heat capacity which is the example of intensive property. So basically in short remember just one thing remember Just one thing whenever an extensive property you are going to express per unit mass per unit mole or even per unit volume the new property that arises the new property that comes into existence that is going to be your intensive property. For example, you can

take one more. Let's say entropy. Entropy is the example of extensive property. Entropy is the example of extensive property. Express it per unit mole. It becomes molar entropy. It becomes molar entropy. And this molar entropy will be your intensive property. So remember whenever an extensive property is expressed per unit mass per unit mole or even per unit volume the new property that comes into existence is what I'll be calling as the intensive property let me know once in the chats if all these things till now are clear to every one of you quickly in the

chats people everybody quickly in the chats I'm looking at the Chats only I am looking at your chats only everyone Everyone my dear students understood. Understood. All right view. Amazing. Just give me a second and we shall continue. Okay, perfectly done. Let's move on now. Let's move on. I believe your extensive intensive property is clear. I believe your extensive properties are Clear. Intensive properties are clear. Let's move on to one more point. What is that? You call them as the state parameters. Have you heard state parameters? or sometimes you call them as the state variables

as well. State parameters or state variables. What are these state parameters? What are these state variables? I would say these are the parameters. These are the parameters That defines that defines the state of the system. These are the parameters that defines the state of the system. For example, pressure, volume, temperature, moles. These are the parameters which actually defines what? They define the state of the system. What is meant by that? You'll understand it in some time. But Before that I want to write one more point. If the state of the system, If the state of

the system is to be changed, If the state of the system is to be changed, the values of state parameters, the values of the state parameters needs to be changed. The values of the state parameters needs To be changed. Third point on changing on changing minimum on changing minimum one state parameter on changing minimum one state parameter state of the system automatically means state of the system automatically changes. What is meant by this? Try to understand properly what is this all about. See guys, for example, Imagine that here I have taken a container in this

container. For example, I'm keeping an ideal gas. Let's say this is the ideal gas present in the container. First of all, from now onwards, from now onwards, from now onwards, whenever I use the term system, from now onwards, whenever I use the term system, understand I'm talking about that particular ideal gas which is there in the container. Whenever from now onwards I use the term system, understand I'm talking about that particular ideal gas which is kept inside the container. Which is kept inside the container. Ideal gas will be my system from now onwards. In our

syllabus we have to discuss basically thermodynamics of ideal gas. So whenever from now onwards I will be using the term system. Understand? I will be talking about the ideal gas. Understand? I'll be talking About the ideal gas. Now my dear students first of all since I have taken the ideal gas in the container let's say when the ideal gas was there in the container. Let's say its pressure was P1 its volume was V_sub1 its temperature was T1. Let's say its pressure was P1, volume was B1, temperature was T1. And I would say this ideal gas

is for example present at its initial state. This ideal Gas is present at its initial state. Okay. So I would say these are basically the parameters. These are basically the variables which are defining the state of the system. These are the parameters which are telling you that the system is present at its initial state. These are the parameters. These are the parameters. These are the parameters which are defining the state of the system. These are the parameters which are telling you these are the parameters Which are telling you that the system is present at its

initial state. Now if I do one thing, if I do one thing, if I change the pressure of the gas from P1 to P2, if I change the volume of the gas from V_sub_1 to V_sub_2, if I change the temperature of the gas from T1 to TS2. Now these are the parameters which are telling you that now the gas is not at its initial state. The gas is at its final state. The gas is at its final state. Okay. So Can I say can I say one simple thing? In order to change the state of

the system, in order to change the state of the system, the value of state parameters needs to be changed. Since I change the pressure of the gas from P1 to P2, volume from V_sub1 to P2, temperature from P1 to T2. Whenever you change the value of the state parameters, the state of the system will change. The state of the system will change and minimum minimum if you change even one state Parameter since over here in this particular case I changed the value of all the three state parameters. But even if I change one state parameter,

the state of the system automatically change. Okay? The state of the system automatically change. Remember that. So what are state parameters? These are the parameters. These are the variables which define the state of the system. Okay? Whenever you want to change the State of the system, whenever you want to send the system from initial state to final state, you will have to change the value of state parameters. Minimum if I change one state parameter the state of the system automatically change. Got it. Perfect. Now comes one more important term. What is that? That is a

state function. One more important parameter. What is that? That is a state function. Now what is a state function? What is a state function? Let me tell you my dear students. These are the thermodynamic functions. These are the thermodynamic functions whose value depends whose value depends only on the initial state and final state of the system. These are the thermodynamic parameters. These are the thermodynamic parameters whose value depends on the initial state And final state of the system only. The second point which I'm writing over here, let me tell you these are path independent. These

are path independent. And what are the examples of these state functions? Let me write some seven examples which you are going to remember. Which you are going to remember. Pressure volume, temperature, internal energy, enthalpy, entropy, gives free energy. These are the seven state functions which you are going to remember. My dear students, these state functions, these are those thermodynamic functions whose value depends only on the initial state and final state of the system. Their value only depends on the initial state and final state of the system. How the state of the system are changed? How

the state of the system are changed? They have got nothing to do with that. Their value only depends on the initial state Of the system and the final state of the system. How the state of the system has been changed. They have got nothing to do with that. And these functions are what we call as the state functions. These are what we call as the state functions. Okay. One important point about these state functions. My dear students, for example, I'm writing a term A here. A. Let's say this A is a state function. Let's say

this A is a state function. This A can be anything. P V T U H S SG anything. This A is for example a state function. Now people if I multiply this A with D, it becomes DA. It becomes DA. This DA I'll be calling as this DA I'll be calling as the small change in A. This DA I'll be calling as the small change in A. the small change in a if I integrate this da, It becomes delta and delta I'll be calling as the total change in a. This delta I'll be calling as

total change in a can take an example as well to have a clear understanding better understanding of this. For example, I'm writing U over here. U stands for internal energy. U stands for internal energy which we'll be discussing in some time. So U stands for internal energy. Internal energy is a state function. If I multiply this U with D, it becomes DU. Du I'll be calling as the small change in internal energy. I'll be calling it as the small change in internal energy. If I integrate this du, it becomes deltaU. DeltaU I'll be calling as

the total change, the large change in the value of internal energy. Okay, perfect. I believe this is also clear to everyone. Now comes after state function, now comes a term that's what you call as the path function. So from the name itself, from the name itself, you can easily write its definition. Yeah, you can easily pen down its definition. So tell me, tell me people. What will be the path functions? Let me first of all tell you path functions. They also they also depend on they also depend on initial state and Final state of the

system. But at the same time they are not path independent. They are not path independent. They are path dependent functions. They are not path independent. They are part dependent functions. Okay, they're part dependent dependent functions. And there are two examples which you have to remember here. Examples of path functions are heat and work. Heat and work. These are the examples [clears throat] of what? These are the examples of path functions. Heat and work. They depend on the path followed to change the state of the system. See when you will be changing the state of the

system when you will be changing the state of the system right there would be some path which you would be following to change the state of the system and the value of heat and work the value of path functions the value of path path functions it depends on how you are Changing the state of the system it depends on how you are changing the state of the system now there is a point which we have to remember here what is that for example I'm writing a term A here. Imagine this A is a path function.

Imagine that A is a path function. Okay. If I multiply this A with D, it becomes DA. DA I won't be calling as small change in A. I will be calling it as small amount of A. I will be calling it as small amount of A. And if I integrate this DA, it becomes capital A. And this capital A I'll be calling as this was small amount of A and now it becomes total amount of A. Now it becomes total amount of A. For example, you can take an example as well to have the better

understanding. You can take the example as well. For example, I'm writing heat. Heat is represented by Q. Heat is Represented by Q. And heat is the example of what? Heat is the example of path function. If I multiply this Q with D, it becomes DQ. DQ I'll be calling as small amount of heat. I won't be calling it as small change in heat. I'll be calling it as the small amount of heat because small change in heat does not make any significance. So this dq is called as small amount of heat. If I integrate it,

small amount becomes total Amount which I represent with Q not with delta Q. This I'll be calling as the total amount of heat. If I'm clear, let me know once in the chat still here. Everybody guys, everybody in the chats quick. Everybody in the chats quick. Yeah. Everybody in the chats >> [snorts] >> I think you're done. Yes, I'll be digging the solution chapter as well. We have got engineer Vasim in the house. Well, engineer Vasim is one of my amazing friends who is in the chats. You can see, right? One of the great physics

teachers as well. All right people, let's move on. Let's move on. Let's move on. Let's move on to something called as thermodynamic process. remote dynamic process. Thermodynamic process well right now I'm just going to give you the introduction of thermodynamic process. Right now I'll just give you the introduction of thermodynamic process. And in this chapter we have to discuss the thermodynamic processes in very much depth. Okay. But right now let me first of all give you the introduction of thermodynamic process and in the detailed way we are going to study them in some time after

2 to 3 hours but Right now just the introduction. Yeah right now just the introduction. So what is the thermodynamic process? I would say the thermodynamic process it is nothing but the path followed. It is nothing but the path followed to change. The state of the system path followed to change the state of the system. What does that mean? What does that mean? What does that mean? Try to understand people. The path followed to Change the state. What does that mean? Try to understand very carefully what I'll be saying. Imagine that this is the container

which I'm taking and imagine in this container imagine in this container we have got an ideal gas let's say we have got an ideal gas present in the container for example this ideal gas is right now under thermodynamic investigation now if this ideal gas right now is under Thermodynamic investigation so this ideal gas is my system right now okay my dear students Assume that the pressure of this ideal gas right now is P1. Volume of this ideal gas is V_sub_1. Temperature of this ideal gas for example is P1. Okay, I'm assuming the ideal gas is

kept at its initial state. Right? Now, the ideal gas is kept at its initial state. Okay, now as I've told you already, in order to change the state of the system, if you want to Change the state of the system, what you have to do? You have to change the value of the state parameters. You have to change the value of the state parameters. If you minimum change one state parameter, the state of the system is automatically going to change. Let me do one thing. Let me do one thing, my dear students. Let me do

one thing. Let me do one thing. Let me change the temperature of the gas from T1 to TS2. Let me change the volume of the gas from V_sub1 to V_sub_2. But let me keep the pressure of the gas still P1. Let me keep the pressure of the gas still P1. Now do you see I changed two state parameters. Volume and temperature I changed right. I changed two state parameters and you know minimum if one state parameter is changed the state of the system automatically changed. So I would say now the ideal gas is present at

its final state. I would say now the ideal gas is present at its final state. My dear students, do you understand one thing? I changed the state of the system. I changed the state of the system, but I did not let the pressure of the gas to change. Did you realize that? I changed the state of the system, but I made sure that the pressure of the system should not change. I made sure that pressure of the gas Does not change. Should not change. No doubt I changed the state of the system but I made

sure that the pressure of the gas should not change. Okay. So in short I can say something like this. I changed the state of the system by keeping the pressure of the system constant. I changed the state of the system by keeping the pressure of the system constant. I could have changed I could have Changed the state of the system by keeping the volume of the system constant as well. I could have changed the state of the system by keeping the temperature of the system constant as well. Yeah. So I would say I have got

many paths available. I have got many parts available by means of which I can change the state of the system. I can change the state of the system by keeping the pressure constant. I can change the State of the system by keeping the volume constant. I can change the state of the system by keeping the temperature constant. So I've got many parts available. I've got many parts available by means of which I can change the state of the system. Now out of all these paths that path which I will be following, that path which I

will be following to change the state of the system, that is something I'll be calling as the thermodynamic process. What is a thermodynamic process? It is the path followed to change the state of the system. And that thermodynamic process in which pressure of the system is kept constant, you call that thermodynamic process as isobaric process. And that thermodynamic process in which volume of the system is kept constant, you call that as the isocoric process. And that thermodynamic system that thermodynamic process in which the Temperature of the system is kept constant you call that as the

isothermal process. Similarly there are other processes which we'll be discussing. Right? Like your adabatic process, polyropic processes, cyclic process etc etc. Okay. So what is a thermodynamic process here? It is the path followed. It is the path followed to change what? It is a path followed to change the state of the system. That's all. It's a path followed to change the state of the system. That's all. Yeah. Perfect guys. All clear with this. Everyone in the chats once. I'm looking at your chats only. I am looking at your chats only. Perfect guys. Now let's try

to understand these processes a bit more in detail comparatively. First of all, I'm writing the heading Here as isobaric process. I'm writing the heading here as isobaric process. So tell me first of all when can I say when can I say that the gas is undergoing isobaric process? When can I say that the gas is undergoing isobaric process? I will say when the state of the system change when you will be changing the state of the system but you will make sure during the change in the state during the change in the state pressure of

the gas should not Change you'll make sure that during the change in the state the pressure of the system should not change at that point of time I can say that the gas is undergoing isobaric process so when when the state of the system change but you are going to make sure that pressure of the system does not change you are going to make sure that the pressure of the system does not change at that point of time I'll be saying that the gas is undergoing which process isobaric Process so I would say in isobaric

process pressure of the system is constant if pressure is constant delta P change in pressure will be zero change in pressure will be zero Yeah. Now we will at the same time since you have basically taken an ideal gas in the container whose pressure you are not changing whose pressure you are keeping constant as per your ideal gas equation is concerned. PV is equal to NRT. Since Right now you are keeping the pressure of the gas constant number of moles are already constant. R is already constant. So volume and temperature they are going to be

directly proportional. So whenever you will be having an ideal gas kept at constant pressure remember at that point of time the ideal gas is going to follow this equation V is equal to K * T where K is a constant okay V is equal to K * T where K is a constant or from this equation you can make you can Make one more you can say V_sub1 by V_sub_2 has to be equal to T1 divid by TS2 but these equations are only valid when you have got ideal gas in the container and that ideal

gas is kept at constant pressure. That ideal gas is kept at constant pressure. Okay, that ideal gas is kept at constant pressure. Perfect. Now my dear students, you can show it graphically as well. You can Show it graphically as well. See how. For example, I want to plot a graph between pressure and volume. I want to plot a graph between pressure and volume for an ideal gas which is undergoing isobaric process. Isobaric means pressure of the gas should not change. Pressure of the gas should remain constant. So change the volume, increase the volume, decrease the

volume, pressure should not change. Right? So this graph is what you call as Isobar. For example, you are plotting a graph between pressure and temperature for an ideal gas which is undergoing isobaric process. So change the temperature of the gas but pressure of the gas should not change constant this also what you call as isobar right or if you want to plot a graph between volume and temperature absolutely you can do that as well since you know v is equal to kt v is equal to k * t volume You are plotting along y ais

temperature in kelvin you are plotting along x ais this becomes my m y= mx means a straight line y= mx means a straight line like this. Okay, this is the initial state of the gas. This is the final state of the gas. At the initial state of the gas, whatever would be the value of v_sub_1 by t_sub_1 at the final state of the gas, the value of v_sub_2 by t2 should be the same. Then only if v_sub_1 by t1 At the initial state will be equal to v_sub2 by ts2 at the final state, then only

you'll be calling it as the isobar. Then only you'll be calling it as the isobar. I believe this particular point is absolutely clear to everyone. Yes. Now my dear students talk about one more process. I'm just giving you the overview of the things. I'm not teaching these processes right now in detail. I'll be teaching them in detail in some time. Yeah. Guys, everything's clear. Quickly tell All right. So let's move on to one more term that is your isocoric process. So as far as the name is concerned you can easily define the isocoric process. So

tell me one thing when should I say when should I say that the gas is undergoing isocoric process tell me that when should I say that the gas is Undergoing isocoric process whenever you will be changing the state of the gas whenever you'll be changing the state of the gas but you will make sure during the change in state volume of the gas does not change you'll make sure that volume of the gas remains is constant at that point of time you can say the gas is undergoing which process isocoric process right right people so

if volume is constant I Can say delta v change in volume that has to be zero okay now since you have taken an ideal gas in the container whose state you are changing by keeping its volume constant since you are keeping volume of the ideal gas constant n and r they already constant. So I would say pressure and temperature they'll be directly proportional right now. If you change the proportionality sign it's going to be P is equal to KT. So I would say this P is equal to KT. It Is valid for an ideal gas

that will be undergoing isocoric process that will be undergoing an isocoric process. T is equal to KT or you can say P1 divided by P2 be has to be equal T1 divided by TS2. All these equations are valid only All these equations are valid only if the volume of the gas is kept constant. If the volume of the ideal gas is kept constant. Okay. Now people I believe you can easily make the graphs as well. For example, you have to plot a Graph you have to plot a graph for example between between pressure and volume.

Let's say you are plotting a graph between pressure and volume. For an ideal gas which is undergoing isocoric process or an ideal gas which will be undergoing isocoric process. See since I have to show that volume of the system is constant. I have to show that volume of the system is constant. So change the pressure increase the pressure decrease The pressure volume of the gas should not change. Then only I'll be calling this isocore. Okay. Similarly, for example, I'm plotting a graph between volume and temperature. Volume and temperature. Change the temperature. Increase the temperature, decrease

the temperature, but volume of the gas should not change. Then only you'll be calling this iso core. Okay? For example, I want to plot a graph between pressure and temperature As well. I want to plot a graph between pressure and temperature as well. So, it becomes y= mx format. y= mx means straight line like this. So at initial state whatever should be the value of P1 by T1 same has to be the value of P2 by T2 at the final state then only you can say this is the iso core wherein the volume of the

gas has kept con has been kept constant right now similarly for example The third process which we have to discuss in detail after some time what is that that is your isoothermal process what is an isoothermal process what happens in in an isothermal process. Let me tell you that whenever you will be changing the state of the gas by making sure by making sure temperature of the gas should remain constant whenever you'll be changing the state of the gas but you will make sure that the temperature of the gas does not change. Right? Whenever you'll

be changing the state of the gas, whenever you'll be changing the state of the system, but you'll make sure the temperature of the gas, temperature of the system should not change at that point of time, you can say the gas is undergoing isothermal process. And in case of isothermal processes, since temperature is constant, so I would say delta t value has to be zero. Delta t value has to be zero. Right? Now my dear students like this this is the equation for an ideal gas undergoing isobaric process. This is an equation for an ideal gas

undergoing isocoric process. Similarly there is one equation for an ideal gas which will be undergoing isothermal process and that equation is like this PV is equal to K. And this equation is valid only for the ideal gas which under goes isoothermal process. And from this Particular equation you can say P1 V_sub_1 will be equal to P2 V2. So all these equations all these equations this PV is equal to K P1 V1 is equal to P2 V2 they are valid only for an ideal gas which will be undergoing isoothermal process. Yeah. Now my dear students, if

I want to plot a grab between pressure and volume, if I want to plot a graph between pressure and volume for an ideal gas Which under goes isothermal process, then I will be using this equation. This equation says PV is equal to K or I can say P is equal to K by V. What do you think? Pressure and volume are they directly proportional or inversely? Pressure and volume they are inversely proportional to each other. And if pressure and volume are inversely proportional that means the graph between pressure and volume has to be like this

rectangular hyperola. Whenever you are supposed to plot a graph between two variables which are inversely proportional to each other the nature of the graph is going to be rectangular hyperola and you call this particular graph as the iso. You call this particular graph as the isotherm. The remaining two graphs you can easily make on your own. And what are the remaining two graphs? You can plot a graph between volume and temperature. You can easily plot a grab between and temperature as well. In case of isoothermal process, you just have to show one thing that the

temperature should not change that the temperature should not change. It should remain constant always. It should remain constant always. So this represents your iso. Even this represents your ism. Correct. Yes. Am I clear people? Everyone in the chats Once. Am I clear people? Everyone in the chats once. Everyone done people. Is this particular stuff clear till here? Everyone keep on telling me in the chats quick be active in the chats then only I can feel that you guys are listening to me otherwise the class becomes boring and if the class becomes boring there's only one

option left to end the class which I don't want to Amazing. Let's move on to one more thermodynamic process. What is that? That is simply your adabatic process. So what happens in an adabatic process? What happens in an adabatic process? In an adabatic process when do I say first of all the gas is undergoing adabatic process again pretty much simple whenever you will be changing the state of the gas Whenever you'll be changing the state of the system but you will make sure during the change in the state there is no heat exchange between system

and surroundings whenever Ever during the change in the state of the system, you will make sure you'll make sure you'll make sure that there is no heat exchange between system and surroundings. You'll make sure that heat should not flow from system to surroundings or from surrounding to system during the change in the state. Whenever you will be changing the state of the gas, but you will make sure that there should be no heat exchange between the system and surroundings. At that point of time you will say the gas is undergoing which process adabatic process. So

in short in case of adabatic processes dq or you can say Q is equal to zero. There is no heat exchange between system and surroundings. And what equation the gas follows when it under goes adabatic process? The Equation will be PV raised^ gamma is constant. PV raised^ gamma is constant. But remember this particular equation is called as the equation of state equation of state for reversible adabatic processes for reversible adabatic processes what is meant by that that I'll make you understand in some time what is reversible and what is irreversible first of all you keep

on remembering the things okay now people one more thing One more thing if you want to plot a graph between pressure and volume If you want to plot a graph between pressure and volume here, I can write this equation like this. I can write it as P is equal to K / V ra^ gamma. I can write P is equal K / V ra^ gamma. So pressure and volume are they directly or inversely? Pressure volume they are inversely proportional. And whenever you are supposed to make a graph between two variables which are inversely Proportional, the

nature of the graph is like this rectangular hyper hyperola. And you call this as the ADA back. You call this graph as the adabat. Now my dear students, if you look carefully, if you look carefully, the graph between pressure and volume for adabatic process came out to be hyperbola. And the graph between pressure and volume for isothermal process also came out to be hyperbol. Both the graphs are similar. Pressure Versus volume graph for an ideal gas undergoing isothermal process is hyperbolic. Similarly, pressure versus volume graph for an ideal gas undergoing adabatic process that is also

hyperbolic. So, how can we distinguish which one is isothermal and which one is adabatic? For that, remember one statement. Well, it has got derivation as well. But I'm skipping the derivation because that's of no use to us. Okay? So, I'm writing a Statement slope of the tangent Slope of the tangent at a point. Slope of the tangent at a point on PV curve on PV curve in case of adiabatic process is equal is equal to gamma* is equal to the gamma* the slope of the tangent At a right on PV curve in case of isothermal

process. In case of isothermal process, now what do we have to do? Let's not remember this particular thing. Let me give you one trick to remember all this. Let me give you one small trick by means of which you can remember all this. Say guys for example I'm plotting a graph between pressure and volume I'm plotting a graph between pressure and volume let's say this is one hyperola and this is one more hyperola so I have shown two hyperolas here I'm showing the arrow towards right I'm showing the arrow towards right and towards the right

what is happening to volume towards the right volume is increasing increase in the volume of the gas Increase in the volume of the system is what we call as expansion. I would say the gas is undergoing expansion. Now hyperbolic hyperbolic. Now which will be isothermal and which one would be adabatic. What do we have to do? Understand? Place your thumb at the common point. Curl your fingers in the clockwise direction. The graph which touches the fingers first. That's always your isothermal. So that means the other one is going to be adabatic. Simple. Now for example,

I'm giving you one more scenario like this. Let's say this is pressure. Let's say this is volume. I'm starting the graphs from here and hyperbolas are like this. So two hyperbolas and I'm showing arrow towards left. I'm showing the arrow towards left. And towards left my dear students towards left volume is Decreasing. Decrease in the volume is what you call as compression. Decrease in the volume is what we call as compression. So out of one of these graphs, one is going to be isothermal and one is going to be adabatic. Now what do we have

to do? Place your thumb at the common point. P your fingers in the clockwise direction. That graph which touches the fingers first. That's always your isoothermal and the other one is your adabatic. Is it simple guys? Yes or no? Is it simple now? So don't compare the slopes and all. Just use the right hand thumb rule. Place your thumb at the common point. Curl your fingers in the clockwise direction. The graph that touches the fingers first, that's isothermal. And the other one will automatically be adabatic. Yes. Quick. Perfect guys. Moving on to one more Thing.

Moving on to one more thing. What is that? That is your cyclic process. cyclic process. What is a cyclic process? When do I say that the gas is undergoing a cyclic process? When do I say that the gas is undergoing cyclic process? Pretty much simple. I would say when the system system means our gas which is there in The container. When the system returns to its initial state, when the system returns to its initial state after going after going through a series of steps, When the system returns to its initial state, when the system returns

to its initial state After going through a series of steps, at that point of time we say that the system is undergoing cyclic process. For example, for example, let me show it graphically. Let's say I'm showing a PV graph here. For example, this is the initial state of the gas. The gas is right now at this particular point, point A, where the pressure of gas is something, volume of gas is something. Now the gas is going from A To B. A to B. That means pressure is kept constant but volume has increased. Okay. Now from

B the gas is going to point C. From C the gas again coming to point A. So can I say whatever was the initial state of the gas same is the final state of the gas. So whenever the system returns to its initial state after going through a lot of steps at that point of time we can say that the system is undergoing which process cyclic process. So in case of cyclic Process I would say the initial state. In case of the cyclic process I would say the initial state and the final state of the

system is same. And if the initial state and final state is same, what does that mean? If you talk about your state function, state function depends on initial state and final state. The value of a state function, it depends on initial state as well as the final state. The value of a state function depends on initial state And final state. If the initial state and final state is same that means that means the initial value of state function and the final value of state function would be also same. And if the initial value of state

function final value of state function is same that means that change in the value of a state function which is final minus initial that is going to be zero. So that is the reason in case of the cyclic processes your delta u your delta h your Delta s your delta g all these values in case of a cyclic process will be equal to zero because all these are what all these are state functions. Change in the value of a state function for a cyclic process is going to be zero. Change in the value of a

state function in case of a cyclic process that's always zero. Yeah. Am I clear people? Done. All clear people. I hope all these things whatever things I've taught you till now I believe every single thing is clear till now. Now let me move on to one more important thing what is that is heat and work so I'm marking the heading here as heat which we represent with Q. Heat which we represent with Q. So if I would want to define heat, what is that? How do we define heat? My dear students, heat is nothing but

It is the mode of transfer. It is nothing but it is a mode of transfer of energy between the system and surroundings. The mode of transfer of energy between the system and surroundings. Point number one. Point number two. It flows due to The temperature difference it flows due to the temperature difference from higher temperature to the lower temperature from the higher temperature to the lower temperature. Let's try to understand these two points first. My dear students For example, I'm taking a container here and in this particular container you are keeping an ideal gas. You have

got ideal gas in this container which I'm calling as system. Which I'm calling as system. Outside this container what do we have? We have got surrounding. Outside the container what do we have? We have got surrounding. Whenever exchange of energy Whenever exchange of energy has to take place Between the system and surroundings. Whenever the exchange of energy has to take place between the system and surroundings. Whenever energy has to go from system to surroundings or from surrounding to system. Whenever energy has to exchange between system and surroundings. Whenever energy has to go either from system

to surrounding or from surrounding to system. that exchange of energy can happen via heat. So what is heat? Heat is nothing but the Mode. It is the mode through which energy is exchanged between system and surroundings. So energy can come from system to surroundings or from surrounding system in the form of heat. So heat is nothing but it is the mode of transfer of energy between the system and surroundings. And heat flows due to what? It flows due to temperature difference. It flows due to what? It flows due to temperature difference. Now my dear students

something important. If For example temperature of surrounding is greater than that of temperature of system. If temperature of surrounding is greater than that of system. If temperature of surrounding is greater than that of system that means between these two there is temperature difference. And if between these two there is temperature difference. So heat will automatically flow from higher temperature to lower temperature. Which one is at higher Surrounding? Which one is it lower system? So heat will flow from surrounding to system. What does that mean? That means heat is flowing from surrounding to system. That means

system is absorbing heat. That means system is absorbing heat. And whenever heat is absorbed by the system, whenever heat is absorbed by the system, remember the sign of Q. Remember the value of Q as per the sign conventions in physics as well as in Chemistry is taken to be positive. So whenever you see heat being absorbed by the system, whenever system absorbs heat, Q value is taken to be positive. Okay, Q value is taken to be positive. Now similarly my dear students if for example temperature of system was more than that of temperature of surroundings.

If temperature of system was more than that of surroundings. So again there was temperature difference And whenever there is temperature difference heat is going to flow from high temperature to lower temperature. Now the heat is going to flow from system to surroundings. If system is at high temperature surrounding is at lower temperature. Heat is going to flow from system to surroundings. And if heat flows from system to surroundings that means the system releases heat that means the system releases heat. And whenever heat is released by the system Q value is taken to be negative as

per the sign conventions. Right? Am I clear people? Everybody in the chats now similarly my dear students one more point as I've told you already this heat is it a state function or path function it is a path function this heat is a path function perfect now similarly if I talk about one more Parameter what is that work which I represent with What is work also? What is work also? Work is again the mode of transfer of energy. It is again the mode of transfer of energy between system and surrounding. Work is again the mode

of transfer of energy between the system and surrounding. As I told you whenever the Exchange of energy has to happen between system and surroundings, this exchange of energy can happen either in the form of heat or in the form of work. This exchange of energy can happen either in the form of heat or in the form of work. So heat is sorry work is also what? It is the mode of transfer of energy. Whenever energy has to go from system to surrounding or from surrounding to system, this exchange of energy between system and surroundings, it

can happen Via work as well. So work is again nothing but work is again the mode of transfer of energy between the system and surroundings. Okay. And this work as I've told you already is it a state function or path function? It is a path function. It is a path function. Now let's try to understand this a bit more in detail. Let's try to understand this a bit more in detail. Understand guys? For example, here I'm taking a Container. This is a container which I have taken. This is a container. And here I have kept

a piston. Here I have kept a piston. For example, at point A. The piston can come inwards. The piston can go outwards. It's a movable frictionless piston. It's a mobile frictionless piston. It can come inwards. It can go outwards as well. And imagine that you have kept an ideal gas in this container. Imagine that you have Kept an ideal gas in this container. Now, first of all, if I ask you on this piston on this piston, how many pressures will be exerted on this piston? How many pressures will be experienced by this piston? The answer

is going to be two pressures. What does that mean? See, see the gas which is in the container, the gas which is there in the container, these gas molecules will be colliding with the piston. These gas molecules would be colliding with the Piston. So I would say in this direction you are going to experience pressure of gas. The piston is going to experience pressure of gas. Pressure of the internal gas. Now outside this piston, outside this piston there is atmosphere and atmosphere also has the gases. So those atmospheric gases also will be colliding with this

piston from outside. So in this particular direction you'll find P external external pressure. So on the piston how Many pressures do we have? Two pressures. P gas P external. P gas P external. Now when the piston is at point A let the volume of the container at that point of time be V1. You know ideal gas always occupies the volume of container in which it is kept. Ideal gas always occupies the volume of container in which it is kept. So I would say when the piston is at point A, volume of the container is V_sub1.

That means the volume of the gas is V_sub1. Okay. Perfect. Now my dear students, case number one. If for example P gas is greater than P external for example if P gas is greater than what P external if P gas is greater than P external the piston is going to come outwards and when the piston comes outwards let's assume the piston stops here finally at B if P gas is greater than P external the piston is going to Come outwards outwards outwards and finally the piston stopped at point B and when the piston stopped at

point B let the volume of the container now is V2 that means The volume of the gas right now is V2. So what is happening to the volume of the container? Is the volume of container increasing? Yes. Is the volume of gas increasing? Yes. Remember increase in the volume of the gas, increase in the volume of the gas is What you call as expansion. So I would say the gas is right now undergoing expansion. And whenever the gas under goes expansion, whenever the gas under goes expansion, we always say that the work is done by

the gas. We always say the work is done by the gas. And whenever work is done by the gas, as per the sign conventions in chemistry, W value is taken to be negative. In physics, you take it positive, Right? Am I clear till here? Am I clear till here people? Everybody in the chats. Now similarly point number two. Point number two. What if what if P external was greater than P gas? If P external was greater than P gas, what would have happened? If P external was greater than P gas, I would say the piston

would have come inwards. Decrease in the volume of container, Decrease in the volume of gas. And decrease in the volume of the gas is what we call as compression. And whenever there is compression, we say work is done on the system. We say work is done on the system. And whenever work is done on the system, the value of W is taken to be positive. The value of W is taken to be positive. Okay. The sign of W is taken to be positive. And my dear students, what would be the third case? Third case is

going to be pretty much simple. If for example P gas is equal to P external, if P gas will be equal to P external, will the piston move? The piston will not move. If the piston does not move, there will be no expansion. There'll be no compression. If there is no expansion, no compression. What is the value of W? Then W value work will be Zero. That means at that point of time when the piston does not move, there will be no exchange of energy between the system and surroundings in the form of work. Right.

In the form of work I believe every single thing is clear till here. Yeah. Everyone if P gas and P external will be the same that means the piston does not move and when the piston does not move what will Happen there'll be no exchange of energy in the form of work at that point of time. Okay. Now my dear students how do we calculate this work? Before telling you that how do we calculate the work? Remember one thing for now. Whenever you see piston showing any moment either going outwards or coming inwards. Whenever you

see piston going whenever you see piston going outwards or piston coming inwards. Whenever you see piston Showing any sort of moment understand energy is being exchanged between the system and surroundings in the form of work at that point of time there is one exception as well that is a free expansion which I'll be dealing with separately leaving the free expansion aside whenever you see piston showing any sort of moment either going outwards or inwards exchange of energy between the system and surroundings is happening at that point of time in the form of Work except in

case of free expansion. Okay. Now the point is how do we calculate this work? My dear students, there is a general formula with the help of which we can calculate work. What is that general expression? That is minus integral v_sub_1 to v_sub_2 p external multiplied by dv. This formula has got the derivation as well. But we have got nothing to do with this derivation. W is equal work is equal minus integral v1 to v2 p external mult* dv. Two points I'm Going to tell you right now. two points and for now you'll be remembering those

two points for now you'll be remembering those two points okay and later on in this chapter only I'll be discussing those two points in very much detail right what are those two points have a look my dear students we have got basically two types of processes one is called as a reversible process one is called as irreversible process these are the two types of processes Which we have to discuss in some time in detail But right now we just remember we have got two types of processes reversal irreversible. In reversible processes P gas and P

external will be almost almost the same. And in case of irreversible processes you will find external pressure as constant. Why this happens? You'll understand that when I'll be teaching you reversible irreversible process in detail. But right now you just remember. Right now you just Remember. Right now you just remember we have got two types of process. Reversible irreversible inreversible processes P gas P external almost same in irreversible processes P external is constant. Now how do we calculate work done in reversible process? Work done in reversible process will be equal minus integral v_sub_1 to v_sub_2. Since

in reversible processes p gas p external are almost same. Therefore instead of p external I'm going to write p gas Multiplied by what multiplied by dv. And in case of irreversible processes in case of irreversible processes p external is constant. If P external is constant comes out of the integral integral of DV is V upper limit lower limit. So it's going to be minus P external V2 minus V_sub1 which I represent with delta V. So these are the two expressions which I would want every one of you to remember from now onwards. One is valid

for reversible processes. Another one is valid for irreversible processes. Yes. Right people, everyone Yeah. All right guys, now comes one more thing. If I ask you what are the units of energy, what are the units of energy? As I told you, as I told you already, your heat and work. As I told you already your heat and work they are the modes of transfer of energy. So heat and work will have same units. Heat and work will have same units as of energy. Heat and work will have same units as of energy. Now first of

all if I ask you what is going to be the SI unit of energy. You should know SI unit of energy is nothing but jewel. So SI unit of heat and work will be also jewel. So what is the CGS unit? What is the CGS unit of energy? That's urg. Now I believe you all must be familiar with the fact one jewel is nothing but 10^ 7 URGs. One jewel is nothing but 10^ 7. Okay. And people at the same time remember one more thing. There are other units of energy as well which I'll be

using in this chapter. And what are those other units of energy? What are those other units of energy which I'll be using in this chapter? Those units of energy which I'll be using in this particular chapter. What are those? One of the unit is atm liter atmosphere liter. One more unit of energy I'll be using bar liter. One more unit of energy I'll be using in calories. Okay. I'll be using in calories. These are the other units of energy which I'll be using in this chapter. And my dear students remember one very very very important

thing. 1 atm liter is nothing But 101.3 joule. Okay. 1 bar liter is nothing but that's equal 100 joule. Okay. And 1 calorie is nothing but that's equal to 4.2 jou and 1 atm liter is equal 24.2 calories. Remember these convergence I will have to use these convergence in the questions. Yeah. Am I clear with this everybody? Am I clear with this everyone in the chats people? Yes. Even J aspirants can watch the Session. Nobody's you can watch it. perfect. Okay, let me let me solve like two three questions and after that we can take

a dinner break and come back. Huh? Okay, let me first of all solve certain questions which will make these concepts like very clear to you and after that we can take a small break. your dinner break basically because a lot of students might be hungry and even I am Hungry. Yeah. Okay. So the first question which is going to be on your screen. For example question is like this. An ideal gas under goes an expansion. An ideal gas under goes an expansion from 10^ -3 m cube to 10^ - 1 m cube against against a

constant External pressure against a constant external pressure of let's keep it as 10 raised^ or two newton per meter squared. Calculate work in jewles. This is the question. This is the question. So my dear students, look at the question carefully. Try to note down all the things that are given in the question. An ideal gas under goes an expansion from 10^ - 3 m cube to 10^ - 1 M cube. So I would say initial volume of the gas is 10^ -3 m cube. Final volume of the gas is 10^ -1 m cube. And the

expansion is happening against a constant external pressure of 10^ 2 Newton per meter squared. And few minutes back I told you when the external pressure is constant the process is supposed to be irreversible. When the external pressure is constant the process is supposed to be irreversible. So I would call this Process as the irreversible process. And if the process is irreversible, how do we calculate work done in irreversible process? Simple. Minus P external V_sub_2 minus V_sub_1. What is the value of P external? P external is 10^2. What is the value of V_sub_2? 10^ minus 1

V_sub_1 10^ - 3. Now do a bit of calculation. 10^2 as such. Now out of these two terms, which one is smaller? 10^ - 3 is smaller. I'll take smaller term common. So I'm taking 10^ - 3 Common this becomes 10^ -1id 10^ - 3 minus 10 - 3 divid by 10^ - 3 that's 1. So the value comes out this value becomes 10^ - 1 this becomes 10^ 2 that's 100 - 1 is 99. So the final answer is going to be - 9. The answer is going to be in jewles. Why the answer

is going to be in jewles? Because if you look carefully, you have taken pressure as per SI system in Newton per meter squared. You have taken volume again as per SI system in meter Cube. So the value of W will also come as per SI system that two in jewles. Yeah, perfect. I believe this sort of equation is clear. If this sort of equation is clear, let's try to solve this one. Let's try to solve this one. People following reactions are happening. You have to identify in which reaction in which process work is done by

the gas And work is done on the gas. You need to check whether work is done on the gas or work is done by the gas. So how do we exactly check it? See look at the first reaction. Your first reaction is 4 * NH3 gas plus 7 * O2 gas. It gives 4 * NO2 gas plus 6 * H2O gas. Do you see all the reactants and products in gaseous state? I would say yes. All the reactants and products they are in gous state and when All the reactants and products will be in gaseous

state what kind of process you have to do at that time what kind of procedure you have to do see when all the reactants and products will be in gaseous state at that point of time you'll be just using your ideal gas equation and ideal gas equation says that PV is equal to NRT since this process is carried out at STP at STP that means pressure temperature is kept constant That means pressure temperature is kept constant. Yeah, that means pressure temperature is kept constant. Hence the reaction is carried out at STP. So pressure temperature is

constant. If pressure temperature is constant, R value is already constant. I'll say volume and moles, they're directly proportional. And if volume and moles are directly proportional. What does that mean? Means more the moles, more the volume, lesser The moles, lesser the volume. Right? Now look at the reactant side. On the reactant side, how many gaseous moles do you see? 7 + 4. That means 11. So we have got 11 gaseous moles on the reactant side. On the product side, how many gaseous moles do you see? 6 + 4 that is 10. So when this process

is happening, when this reaction is happening, are the gaseous moles increasing or decreasing? I would say gaseous moles are decreasing. When this process is happening, when this reaction is happening, gaseous moles are decreasing. And if moles are decreasing, if moles are decreasing, that means volume will be also decreasing. And you know decrease in the volume of the gas, decrease in the volume of the system is what you call as compression. And whenever there's compression, work is done by the gas or on the gas. Whenever there is compression, work is done on the gas. And whenever

work is done on the gas, as per sign conventions, the value of W is taken to be positive. Right? So first question is done. Yes. The first question is done people. Look at the second question now. But remember this particular approach you'll be using only when all the reactants and products are in gaseous phase. When all the reactants and products are in gaseous phase what happened? I hope you got to know the answer of the first question is work is done on the gas. W value is positive. W value positive. Okay, look at the second

one. First one is done. Now talk about second one. If you look at the second one, the reaction is carbon monoxide gas plus 2 * H2 gas gives what? It gives CH3 O liquid. Now are all the reactants, products in gaseous phase? No. All the reactants and products are not In gaseous phase. So you are not going to use the ideal gas equation here. Ideal gas equation you'll be using only if the all the reactants and products are in gaseous phase. So right now all the reactants products they are not in gaseous phase. So what

concept I'll be using? See guys pretty much simple. On reactant side you can see gases only on product side we have got liquid. On reactant side we have got gases. On product side we have got liquid. So Gases are getting converted into liquid. Gases are getting converted into liquid. And when gases get converted into liquid, what happens? Do the molecules come close or molecules go far? What do you know about that? When gases get converted into liquid, when gases are converted into liquid, I would say the molecules are going to come close. If molecules are

coming closer, that means the volume of the system is decreasing. And when volume of the System decreases, you call it as compression. And whenever there is compression work is done on the system. And whenever work is done on the system. Whenever work is done on the system W value is taken to be positive. So here also W is positive. In the first case also W was positive. Leave the third case aside. Talk about the fourth one. Talk about the fourth one first. In the fourth case if you see carefully your solid H2O solid is getting

converted Into H2O liquid. Sorry is getting converted into H2O gas. If solid is getting converted into gas, solid is getting converted into gas. That means molecules are going far. Volume is increasing. Increase in the volume is what you call as expansion. And whenever there is expansion, we always say work is done by the system. We always say work is done by the system. And whenever work is done by the system, the value of W will come out to Be negative. Okay? Perfect. Similarly guys, for example, you have got a scenario like this. X solid gives

X liquid. X solid gives X liquid. When solid will be getting converted into liquid. When solid will be getting converted into liquid, that means molecules will be going far. So volume would be increasing increase in the volume. Expansion. Whenever there is expansion, work is Done by the system. And whenever work is done by the system, W value is taken to be negative. But this is a special case. This was a general when solid gets converted into liquid. But when H2O solid gets converted into H2O liquid, when H2O solid gets converted into H2O liquid, what happens

in this particular case? Generally, what should happen? When solid gets converted into liquid, W value should be negative. Expansion Should happen. But there's a special case here. Why is a special case? Because on reactant side, you have got ice. On reactant side, you have got ice. On product side you have got liquid water. On product side you have got liquid water. And remember because of the special structure of ice. Because of the special structure of ice because of the because of the cage-like structure of ice because of the hydrogen bonding Present in ice volume of

ice is more than that of volume of liquid water. Volume of ice is more than that of liquid water. So during this particular process during this particular process what is happening to the volume is volume increasing or decreasing volume is decreasing and decrease in the volume is what you call as compression and whenever there is compression work is done on the system and whenever work is done on the system W value is taken to Be positive. So here work is done on the system W is positive. Here it was expansion so W was negative. So

I believe all these questions are like super clear to every one of you. Yes. If yes, let me know once in the chats. I think let's take a dinner break. It's 8:25 right now. So, I'm giving you a good break. Long break. Dinner break. Be back on time. Okay, be back on time, people. All right, I will be expecting every one of you to be back by 9:5. Get done with your dinner. I'll also get done with my dinner and come back exactly at 95. Okay, see you in a while. Take care. But make sure

everyone is back. Make sure everyone's All right. Is everyone back? Is everyone back? Yes. [laughter] Is everyone back? Tell me quickly in the Chats. We have to start. We need to start. Quick, quick, quick, quick. Quickly people, quickly, quickly, quickly. Are you done with your dinner? Are you done with your dinner? Okay. All right. All right. All right. So, I think everyone is back now. Uh sir, will you take thermmochemistry 2 In this lecture? No. No. Thermochemistry will be separate. This is thermodynamics. There'll be two parts of this chapter basically. Thermodynamics part and thermmochemistry part.

And today I'm discussing thermodynamics with you. Thermmochemistry will be part two. Today we shall be discussing about the dynamics part. All right. So tell me once whatever things we have discussed till now. Is Every single thing clear? Let me know quick. Let me know quick. Let me know quick. Whatever things we have discussed till now is every single thing clear. [clears throat] Live class start one day one chapter. All right, perfect. Perfect. So, let's get going then. Let's get started with one more topic. And The name of this topic is going to be heat capacity.

And this heat capacity is represented by C. This heat capacity is represented by C. Okay. So first of all, how do we define the term heat capacity? My dear students, heat capacity is nothing but it is basically the amount of heat. It is basically the amount of heat required to raise The temperature. It is the amount of heat required to raise the temperature of a substance of a substance by 1°ree. The amount of heat which is required to raise the temperature of the substance by one degree that amount of heat supplied is called as the

heat capacity. Let's try to understand. Let's try to understand. For example, this is an object. Let's say this is an object. You want to increase the temperature of this object by 1°. You want to increase the temperature of this object by 1°ree. Whatever amount of heat is required, whatever amount of heat is to be supplied to this object so that its temperature increases by 1°ree. In order to increase the temperature of this object by 1°ree, the amount of heat supplied is called as the heat capacity. For example, sometimes if I say heat capacity of the

object is 5 gh per deg centigrade. What does that mean? That means in order to increase the temperature by 1°ree, heat supplied is equal 5 ghou. So 5 jewles of heat is to be supplied to increase the temperature of the object by 1°ree. For example, sometimes I'll say the heat capacity of the object is 20 gh per deg centigrade. What does that mean? That means 20 jewles are to be supplied to the object. Then only its temperature will increase by 1°. As simple as that. As simple as that. All right. I believe this is clear.

I believe you got the idea about what this heat capacity exactly is all about. what this heat capacity exactly is all about. Perfect. Perfect people. Is it clear? Say yes or no. Did you understand the definition? Did you understand the definition? The amount of heat which is required to raise the temperature of the substance by one degree. You call that as the heat capacity. And my dear students, let's try to understand few more things. Let's try to understand few more things. For example, this is the object. Let's say you are supplying small amount of heat

dq to this object. Let's say you are supplying small amount Of heat to this object which is dq. Okay. And by supplying small amount of heat dq to this object let's say let's say the change in the temperature of the object was dt by supplying small amount of heat dq to the object okay the temperature of the object changed by how much dt amount so can I say something like this can I say in order to raise the temperature of the object by dt in order to raise the temperature of the object by dt. How

Much heat was required? I would say heat required was how much? Heat required was dq in order to raise the temperature of the of the object by dt. Heat required was dq. So if I use the unitary method, tell me one thing. If I want to increase the temperature of the object by one unit, if I want to increase the temperature of the object by one unit, how much heat is required? I would say heat required will be dq upon dt. So what is this dq upon Dt? Then I would say this is the amount

of heat. This is the amount of heat which is required to raise the temperature of the object by 1°ree. And the amount of heat which is required to raise the temperature of the object by 1°ree. What do we call that as? We call that as heat capacity. So C heat capacity is nothing but dq upon dt. Heat capacity is nothing but it is dq upon dt. Now if I ask you what could be the possible units of this heat Capacity? I believe you can make its units on your own. Now, for example, in the numerator

you have got heat. In the denominator, you have got temperature. Let's say heat I'm taking in jewels. Let's say temperature I'm taking in degrees centigrade. So, this can be one of its probable units. Or for example, heat I'm taking in jewels. Temperature I'm taking in Kelvin. So, this can be one more unit. For example, heat I'm taking in calories. Temperature I'm Taking in degrees centigrade. So, this can be one more probable unit. So, you can make it different units. Yeah. You can make its different units. Okay. Now guys try to understand one more thing. Try

to understand one more thing. Have a look. I told you C is equal DQ upon DT. So can I say DQ is nothing but it is CDT. DQ is nothing but it is C DT. Okay. DQ is nothing but C DT. DQ is the small amount of heat. Small amount of heat That is required to increase the temperature of that object by how much? By DT. Now for example, see you have got this object whose heat capacity is for example C and you want to increase the temperature of the object from T1 to TS2. Now

you have got the object whose heat capacity is for example C and you want to increase the temperature of this object from T1 to TS2. I'm asking you How much heat is required. How much heat is required? What you'll be doing? You'll be integrating on both the sides and here you are integrating with respect to temperature and temperature you are increasing from T1 to TS2. Now integral of dq is q right if c is constant then c comes out integral of dt is t upper limit lower middle so it's t2us t1 t2 - t1 is

nothing but delta t right this is the expression by means of which this is the expression by means of Which we can calculate the amount of heat required to raise the temperature of the object from t_1 to ts2 okay for example you have got an object whose temperature you are changing ing from T1 to T2. Now I want to know how much heat is required to raise the temperature of the object from T1 to TS2. So simple Q is equal to C delta T. That's it. Yeah, I believe it's clear. I believe it's clear. And

at the same time, I've told you already heat Capacity is it extensive property or intensive? It is an example of extensive property. Okay. Now comes point number two. Point number two, what is that? That is specific heat capacity. That is specific heat capacity. And specific heat capacity I'll be representing with S. So what is basically the specific heat capacity? Specific heat capacity is nothing but it is defined as the heat capacity of an Object, heat capacity of the system per unit mass of it. per unit mass of it. Heat capacity per unit mass. Heat capacity

per unit mass is called as the specific heat capacity. That means it is the amount of heat. It is the amount of heat which is required to raise the temperature of a unit mass of substance. It is the amount of heat which is required to raise the temperature of a unit mass of substance by 1°ree. Unit mass means 1 g, 1 kg, Right? That's called as specific heat capacity. So I would say your specific heat capacity is nothing but it is heat capacity per unit mass. And heat capacity C already I've told you that is

dq upon dt. So it becomes dq upon mdt dq upon mdt. Now you can make its units as well. See in the numerator you have got heat denominator mass and temperature. Let's say heat you are taking in jewles. Mass you are taking in kg. Temperature you are taking in Degrees centiggrade. So this can be one of its units. You can make it more units as well. You can make it more units as well. Take this heat in calories. Take this in Kelvin. That can be one more unit. So a lot of units can be made

here. Now I can say since S is equal since S is equal to DQ upon MDT. So can I say your dq becomes equal? DQ becomes equal mdt. dq becomes equal msdt. Now imagine that You have got an object whose mass is m whose specific heat capacity is s and you want to increase the temperature of the object from t1 to t2 you want to increase the temperature of the object from t1 to t2 the how much amount of heat is required how much amount of heat is required to raise the temperature of the system

from t1 to t2 how do I calculate that I'll have to integrate it under the limits t1 to t2 integral of dq m comes out. If s is constant, it comes Out. Integral of dt is t. Use upper limit lower limit. So t2 - t1. T2 - t1 is nothing but delta t. Right? So this is one more way of calculating Q. The amount of heat required to raise the temperature of the object from T1 to TS2 whose specific heat capacity is S. This is the formula I'll be using. Yeah, this is the formula I'll

be using. Perfect. Now people one more thing if I ask you Whether the specific heat capacity is whether it is intensive or extensive since I've told you heat capacity itself is extensive and whenever an extensive property is expressed per unit mass per unit mole or per unit volume the new property becomes intensive right so the specific heat capacity it is the example of which property it is the example of an intensive property Right? Then comes your third terminology my dear students. You call that as the Molar heat capacity. Molar heat capacity which I will be

representing with cm. Now tell me what should be its definition? What should be your molar heat capacity? What should be our molar heat capacity? My dear students, how should we define it? see guys molar heat capacity is nothing but it is defined as it is defined as let me write in short It is defined as the heat capacity of the system per unit mole heat capacity of the system per unit mole or you can define it in simple words you can define it as the amount of heat which is required to raise the temperature of

one mole of the substance by 1°ree amount of heat which is required to raise the temperature of one mole of the substance by 1°ree you can call that as The mole read capacity or in short you can say it is defined as the heat capacity per unit mole of the substance per unit mole of the substance or I can say your specific heat capacity it is nothing but heat capacity per unit mole of the substance now heat capacity you already know that is dq upon dt so it becomes dq upon ndt it becomes dq upon

n dt now you can make its units let's say heat I'm taking in jewles in the denominator you have got Mole with mole let's say temperature you are taking in degrees centiggrade so this can be one of its unit you can make other units of it as Okay. Or I can say since sorry this is not S this is cm. Since I got to know your cm is equal to dq upon n ddt. So I can say dq has to be equal dq has to be equal n cm dt n cm dt. Now again if

I integrate under the limits t1 to t2 I would say integral of dq becomes q. n comes out cm comes out If it is constant. So integral of dt is t. Upper limit is t2, lower limit is t1. So it's going to be q is equal to n cm delta t. Where do I use this result? Where should I be using this result people? Where should I be using this result? Where should I be using this result? See this result can be used like this. For example, you have taken a container. In this container, imagine

that you have got n moles of ideal gas. Let's say the molar heat capacity of the ideal gas is cm. Now I want to increase the temperature of this gas from t1 to ts2. I'm supposed to calculate how much heat is required. How much heat is required to raise the temperature of n moles of the gas from t1 to t2 whose m heat capacity is cm. So q would be nothing but it will be n cm delta t. Clear to here? Tell me once in the chats if it is clear. Tell me once in the

chats if it is clear people. Yeah. Sir, how much you scored in NEAT? I'm not a NEAT student. I have been a PCM student, J student. All right, people. Now if I summarize all of this in one slide, let me quickly summarize this all in one slide. So I'm writing heat exchange between system and surrounding. Heat exchange between system and Surrounding which I'm representing with Q. Heat exchange between system and surrounding which I'm representing with Q. That means heat absorbed or heat released by the system. Heat exchanged between system and surroundings means heat absorbed or

heat released by the system. Okay, there are many types of questions wherein they'll ask you to calculate the amount of heat absorbed or heat released by the system. So Technically they'll ask you to calculate Q value and I give you many results to calculate Q value. Number one, if in the question heat capacity of the system is given, then how do you calculate Q? You calculate Q by the formula C delta D. If in the question specific heat capacity of the system is given then how do we calculate Q? Then Q is calculated by this

result MS delta T. And if in the question they give you the value of molar heat capacity. If they give you the value of molar heat capacity then Q would be equal to N PM delta T. So these are the three results by means of which we can calculate heat absorbed or heat released by the system depending on what is given in the question. If C is given then this result. If S is given then this result. If CM is given then this result. Okay. Now people one more thing one more thing which I want

to tell you. What is that? What is CM? CM is nothing but molar heat Capacity. CM is nothing but molar heat capacity. What is molar heat capacity? the amount of heat required to raise the temperature of one mole of the gas to raise the temperature of one mole of the gas by one degree that's more read capacity. Now the gas can be either kept at constant volume or the gas can be kept at constant pressure. Gas can be either kept at constant volume or the gas can be kept at Constant pressure. And when the gas

is kept at constant volume at that point of time instead of cm you write CV and when the gas is kept at constant pressure instead of cm you write CP okay the amount of heat required to raise the temperature of one mole of the gas by one degree that's mole capacity now if the gas is kept at constant volume then CV if the gas is kept at constant pressure then CP what is CP what is The Meaning of this CP and CV let me write it too. So CP is the molar sorry CV is molar

heat capacity at constant volume. Similarly CP is called as molar heat capacity at constant pressure at constant pressure. Okay. So what I can do here is I can further write two more equations. One equation I'll write when the gas is kept at constant volume instead of Q I'll write QV that will be called NCV delta T. Now if the gas is Kept at constant pressure then instead of Q I can write QP that will be called NCP delta T. Right? So they can give you two formats of the questions. they can give you to calculate

the heat absorbed or released by the system at constant volume. If they ask you to calculate heat absorbed or released by the system at constant volume, that means you have to calculate QV. And if they ask you to calculate heat absorbed Or released by the system at constant pressure at constant pressure, then you have to calculate QP. Okay. over guys. Am I clear? Am I clear? perfectly done. All right. All right, people. Now, now, now there are some important results related to this CP and CV which all of you must remember because I'll be using

those results and the questions Directly. Okay, I'll be using those results and the questions directly. Now, what are those results? I'm not going to derive any result because that's of no use to us. So, I'll be just giving you the results and we'll see the application part of those results. Now, what are the results exactly? The first result my dear students PP minus CV is equal to R for 1 mole of ideal gas. Remember this Particular result PP minus CV is equal to R. If CP minus CV comes out to be R. R is positive

basically its value is 8.314. You know it if it is positive it means that PP value is greater than that of CV. CP value is greater than that of CV. If CP is greater than CV, there is a term called as gamma. Gamma is defined as the division of CP and CV. So if CP is greater than CV, that means the value of gamma would be greater than one. Okay. Now for one mole of gas, CP minus CV is R. For 2 moles of ideal gas, CP minus C will be 2 R. 3 moles of

ideal gas 3 R. So I would say for N moles of ideal gas, CP minus CV is equal NR. or n mo moles of ideal gas. Okay, there is one more result which I would want you guys to remember. What is that? CV is equal r / gamma minus 1. It has got the derivation as well but I'm skipping the derivations. That's of no use to us. Okay. Similarly, CP is equal gamma * r / gamma minus one. This is one More result which you can remember. Similarly, my dear students, imagine you have got a

mixture of non-reacting gases in the container. Imagine that you have got a mixture of mixture of non-reacting gases in the container. At that point of time, how do we calculate CV for the mixture of gas? CV for the mixture would be N1 CV1 plus N_sub_2 CV2 divid by N1 + N_sub_2. Now, similarly, how do we calculate CP for the mixture of gas? PP for the mixture of gas would be N1 CP1 Plus N_sub_2 CP2 divid by what divided by N1 + N_sub_2 and similarly if you want to calculate gamma value for the mixture of gas

it's going to be CP mixture divided by CV mixture okay got it these are some important results my dear students which you have to remember okay now as you can see the value R came here as you saw R is coming here everywhere. What is R? R is nothing but it's a Universal gas constant. It is a universal gas constant and R can have three values. The first value is 8.314 Jew per kelvin per mole or to reduce the calculation instead of 8.314 you can use 25x3. 25x3 also comes out to be 8.314 only. You

can take the value of R as 0.0821 atm liter per kelvin per mole or to reduce the calculation instead of 0.0821 you can use the term 1x2 because 1x 12 again comes out to be 0.081 only. Yeah. Now similarly the third value of R is 1.98 or you can take it as 2 calories per Kelvin per mole. These are the three values of R which you have to remember. Right? And accordingly we'll be using them in the questions. Perfect. If this is perfect, then there is one more table which you have to remember. And what

is that table? Let me make that table for you. First of all, in the first column of the table, I'm Writing nature of gas. Here I'm writing nature of gas. Here I'm writing CV value. Here I'm writing CP value. And here I'm writing gamma value which is CP by CP. The first gas which I'm talking about here the first gas which I'll be taking here that is going to be your monotomic gas. monatomic gas for example helium helium is the example of monotomic gas argon neon they are the monotomic gases for All the monotomic gases

CV value is equal to 3 divid by 2 rp value is equal to 5 / 2 r and gamma is equal cp by cv so it becomes 5 / 3 which comes out to be 1.66 66 number one. Number two, if you will be having a diatomic gas, if you will be having a diatomic gas like like you have got O2, O2 is diatomic, right? Two atoms in one molecule of oxygen. Diatomic. If you'll be having a diatomic gas or if you have triatomic Linear gas. What is the example of triatomic linear? Carbon dioxide. Carbon

dioxide is triatomic at the same time it's linear. At the same time it's linear. So for diatomic gas or triatomic linear gas CV value is 5x2 R. CP value is 7x2R and the value of gamma is CP by CV. So it comes out be 7 by 5. 7x 5 is nothing but 1.4. And similarly if you will be having a triatomic if you'll be having a triatomic Nonlinear gas. What is the meaning of triatomic nonlinear? Like you have got SO2. SO2 is triatomic but it's not linear. It's bent. SO2 is triatomic. It's not linear. It's

bent. Right? So if you'll be having a triatomic nonlinear or in general I say polyatomic nonlinear. At that point of time PV value is 3 R, CP value is 4R and gamma is 4 / 3 that means 1.33. This is a table which you have to remember before solving the Questions. CP and CV value for different types of gases. Now guys, if you look carefully, if you look carefully, I started with monatomic, then I went to diatomic, then I went to triatomic. So on moving from top to the bottom, what is happening to the atomicity

of the gas? Monatomic, atomist 1, diatomic, atomist 2, triatomic, atomist 3. On moving from top to bottom, atomity is increasing. On moving from top to bottom, atomicity is increasing. What is Happening to gamma value? gamma value is decreasing. So as the atomicity of the gas increases the value of gamma decreases. You can generalize this statement as well. You can generalize this particular statement as well. Okay. Let me write the statement here somewhere. As the atomisty of the gas increases, The gamma value for the gas decreases. All clear with this. All clear with this people. Quick.

[snorts] All clear with this. Quickly tell me in the chats everyone. Everyone. Everyone. People >> [snorts] >> Let's move ahead then. Perfect. Let's move ahead. Let's move ahead. The most important topic of the chapter. What is that? That's the internal energy. That is the internal energy. My dear students, how do we exactly define internal energy? This is the most important topic. Okay, one of the very very very important topics of the chapter. Internal energy. Let me first of all tell you internal energy I represent with either U or capital E. Internal energy is represented either

With U or capital E. How do we define it? How do we define it? How do we define it? See guys, internal energy is defined as the sum total the sum total of all the sum total of all the possible kinds of energies present in the system. The sum total of all the possible kinds of energies present in the system. The sum total of all the possible kinds of energies present in the system. What does that mean? What does that mean? Let's try to understand. Let's try to understand people what it means exactly. Let's try

to understand it. For example, I'm taking a container here. Let's say this is the container. And imagine in this container I'm keeping a real gas. Let's say in this particular container we have got what? We have got a real Gas. We have got a real gas in this container. My dear students, since the gas molecules they're always in continuous random motion. You know that already. The gas molecules they're always in continuous random motion. Perfect. Due to which due to which this gas has got some kindic energy. due to which the gas has got some kindic

energy. These gas molecules they are in Continuous random motion molecules would be showing translational motion, rotational motion, vibrational motion etc etc whatever right. So basically I would say since the gas molecules are in continuous random motion so this gas would be having some kindinetic energy which can be translational, rotational work perfect. So let me write this gas has got some total kindinetic energy sum total of all the kindinetic energies in the gas. Now since the gas is real and You should know between the real gas molecules there will be attractions repulsions. Between the real gas

molecules there will be what? There will be attraction repulsions and due to attractions repulsions. Due to attraction repulsions which energy comes into existence? Due to attraction repulsions potential energy comes into existence. So this gas which is there in the container it has got kindinetic energy As well as potential energy. It has got kindinetic energy as well as potential energy. So if I take the sum of if I take the sum of kinetic energy of the gas and the potential energy of the gas if I take the sum of the total kinetic and total potential the

sum of these two terms will give me the total energy present in the gas that total energy present in the gas I'll be calling as the internal energy that total energy present in the gas I'll be calling as The internal energy okay and this is valid for what since the gas which was there in the container that was the real gas. So I was talking about the real gas. Okay. Now imagine the gas is ideal. Now imagine the gas is ideal. If the gas is ideal there'll be no interactions. There'll be no interactions. There'll be

no attraction repulsion between the molecules. And if there is no attraction repulsion between the molecules I would Say potential energy of the gas would be zero because due to attraction repulsion potential energy comes into existence. If there is an ideal gas between the ideal gas molecules there will be no attraction repulsions so there'll be no potential energy. So in case of ideal gas your internal energy is nothing but internal energy is the total kinetic energy present in the gas not the potential energy. Okay I believe this is clear. Now my Dear students if I talk

about this internal energy a bit more in detail what do you think is internal energy going to be extensive or intensive? It is an extensive property. internal energy. It is an extensive property because if you increase the amount of substance in the in the container, if you increase the gas in the container, if you increase the gas in the container, if you increase the amount of substance in the container, The value of total energy present in the container will change. So I would say internal energy is amount of substance dependent. Internal energy is amount of

substance dependent. And any property that is amount of substance dependent, you call that as extensive property. Similarly, this internal energy, it is a state function as well. It is a state function as well. So, its value only depends on initial and final state of the gas. Its value only depends On the initial state and the final state of the system. How we change the state of the system, internal energy has got nothing to do with that. Internal energy only depends on initial state and final state. Okay? How we change the state of the gas? How

we change the state of the system? Internal energy has got nothing to do with that. It just have got something to do with initial state and final state. How state is changed? Nothing to do with that. Okay. Now my dear students, one more thing. If I talk about this internal energy, this internal energy or a fixed amount of real gas or a fixed amount of real gas in the container, this internal energy depends on two parameters. One is volume and one is temperature. It is valid for a fixed amount of real gas. If you will

be having a fixed amount of real gas in the container, its internal energy will Depend on two parameters volume and temperature. But if you will be having an ideal gas, if you take a fixed amount of ideal gas in the container, then internal energy will depend only on temperature. For a fixed amount of ideal gas, internal energy only depends on temperature. Okay? Internal energy only depends on temperature. Now at the same time one more thing For a cyclic process for a cyclic process in case of the cyclic process where initial state and final state is

same in case of the cyclic process I must say delta u that would be equal to zero in case of the cyclic process delta net is equal to what it's equal to z why because in case of cyclic process initial state final state is same that means initial ial value of internal energy and the final value of internal Energy will be the same. So U final minus U initial will come out to be zero because U final and U initial are same. Got it? Now guys one more thing. One more thing. Since in case of

ideal gas internal energy depends on temperature. In case of ideal gas internal energy depends on temperature. Okay. For example, this is the container and in this container, imagine that you have got an ideal gas. You have got an ideal Gas in the container. Let's say let's say the molar heat capacity of this ideal gas at constant volume is CV. Let's say molar heat capacity of this ideal gas at constant volume is CV. Now, for example, I'm increasing the temperature of the gas from T1 to T2. I want to increase the temperature of this gas from

T1 to TS2. That means I have to supply heat. Now how much heat is required or let's say let's talk like this wait for example there is inter There is the ideal gas in the container now you are increasing the temperature of ideal gas from t1 to t2 and if we increase the temperature of the ideal gas internal energy is going to increase and how do I calculate that change in internal energy of an ideal gas that I'm representing delta u so delta is equal it will be equal n cv delta t delta u would

be equal n CV delta T and this particular result is only valid for which gas? It is only valid for ideal Gas. So in case of ideal gas in case of ideal gas if you change the temperature its internal energy will change and that change in internal energy is represented with delta U which is equal N CV delta T. Again it has got a derivation but that's of no use to us right now. Okay. Now guys one more thing. One more thing. When an ideal gas When an ideal gas under goes When an ideal gas

under goes an isoothermal process. When an ideal gas under goes an isoothermal process, what would be its deltaU value? If you have got ideal gas which is undergoing isothermal process, that means temperature of the gas is not changing. If temperature is not changing, if temperature is constant, that means delta t is zero. If delta t is zero, that means delta u value would be equal to zero. For which gas? For an ideal gas? It's clear guys because in case of ideal Gas, internal energy only depends on temperature. If you are not changing the temperature, that

means there will be no change in the internal energy of the ideal gas. If you are not changing the temperature of the ideal gas, there will be no change in the internal energy of the ideal gas. As simple as that. As simple as that. Right. Perfect. Okay. If I write a statement like this, tell me what is going to be the answer of this statement. When a real gas When a real gas under goes When a real gas under goes an isoothermal process. When a real gas under goes an isothermal process will deltaU be zero

or non- zero what do you think if real gas is undergoing isothermal process? Is deltaU going to be zero or non- zero? Real gas undergoing isothermal process that means its temperature is constant. But the internal energy of real gas Depends on two parameters. One is temperature, what is volume. No doubt you are keeping the temperature constant. But volume can change due to which there can be change in internal energy. So delta U is zero or non zero. It is non zero. Yeah. Am I clear with this? Every one of the jazz people. Am I clever

with this? Yeah. All right people. Perfect. At the same time if I ask you internal energy and temperature would they be directly proportional or inversely? Would they be directly proportional or inversely? What do you think? Internal energy and temperature directly or inversely? Simple. If you increase the temperature, the kindinetic energy of the molecules will increase. And if the kindinetic energy of molecules will Increase, that means internal energy will increase. So internal energy and temperature will be directly proportional. Okay, it will be directly proportional. Now let's try to solve a question based on this and then

and then we can move on to our first law of thermodynamics. For example, a simple question I'm giving you 44.8 8 L of helium At STP is heated to 100° centigrade is heated to 100° centigrade. Calculate deltaU in calories and also assume the gas to be ideal. Assume the gas to be ideal. Assume the gas to be ideal. Can we solve this? As far as the question is concerned, you have got 44.8 L of helium At STP. So that means how many moles of gas we have? For example, this is a container. And in this

container, we have got 44.8 L of what? Of helium. And this gas is kept where? This gas is kept at STP. What does that mean? That means the temperature of this gas is right now 0° centigrade. And the pressure of the gas is 18 atm right because it was mentioned that the gas is present at STP. STP means temperature zero pressure 180. Now we Have got 44.8 L of gas at STP. So tell me how many moles of the gas we have given volume divided by 22.4 given volume divided by 22.4 the value comes out

be two. So basically we have got two moles of gas in the container. Basically we have got two moles of gas in the container. And as per the question, you are increasing the temperature of the gas from 0° to 100°. You are increasing the temperature of the gas from 0° to 100°. And since you Know if the temperature of the gas is changed, there will be change in its internal energy. If the temperature of the gas is changed, there will be change in its internal energy. And that change in internal energy is given by the

result delta U is equal to N CV delta T. So deltaU is equal what is the value of N? N value is 2. CV value since we have got a monotomic gas. So CV is 3x2 R. Delta T change in temperature final minus initial the value comes out to be 100. Since I need to get the answer in calories so what I'll be using I'll take the R value in calories that is 2 multiplied by what? Multiplied by 100. So this term this term gets canceled. DeltaU value comes out be 600 calories. So change in

the internal energy of the gas on increasing the temperature from 0° to 100° is nothing but 600 calories. I believe this is clear to everyone. I believe this is clear to everyone. Yes. Am I clear guys? Am I clear to everyone? Am I clear to everyone? Wake. All clear. If you take temperature in Kelvin also, change in temperature is also going to be same. Only it's anyways going to be same. Whether you take change in temperature in degrees centigrade or change in temperature in Kelvin, that's going to be same always. For example, I'm writing let's

say the Temperature is changing from 0° to 100°. So what is the change in temperature here? Final minus initial that will be 100. Now 0° means 273 Kelvin. 100° means 373 Kelvin. Here also change in temperature will be 100 only. Do you see that? So if you take change in temperature either in degrees centiggrade or in Kelvin, it's going to give you the same value. Perfect. All right, let's move on now. Now, now First law of thermodynamics. First law of thermodynamics, it is basically an energy conservation principle. It is basically an energy conservation principle. Okay.

Now what this first law of thermodynamics states what this first law of thermodynamics states try to understand. First law of thermodyamics says that when the state of the system when the state of the system change When the state of the system changes there will be there will be change in its internal energy. which is something we already know. When the state of the system changes, internal energy is going to change which can be brought about which can be which can be brought about in two ways. Which can be brought about in two ways. Number one,

When heat is absorbed or released by the system, when heat is absorbed or released by the system. Number two, when work is done by the system or on the system. What does this mean? Let's try to decode this. Okay, let's try to understand this because the questions are frequently asked from this particular topic which Is first law of thermodynamics. I do understand guys. For example, this is a container. This is a container, for example. And this is its piston. This is its piston. The piston can go up. It can come down as well. Okay. Now,

in this container, for example, you have kept an ideal gas. In this container, you have kept an ideal gas. Let's say this ideal gas in The container is right now present at its initial state. Let's say this ideal gas kept in the container is right now present at its initial state. And let's assume the internal energy of this ideal gas is equal to U1. The internal energy of this ideal gas right now is how much? It is U1. The total energy of this gas right now is how much? It's U1. Now for example, for example,

you are supplying some heat to this gas. You are supplying For example some heat to this gas. How much? For example, you are supplying Q amount of heat to this gas. When you supply Q amount of heat to this gas, tell me one thing. On supplying Q amount of heat to the gas, what would be the total energy of the gas? Now, earlier it was U1. Now, it will be U1 + Q. This would be the total energy of the gas. U1 + Q. Okay? For example, let's do one more thing. For example, you are

bringing the piston down. If you are bringing the piston down that means the volume of the container is decreasing and if volume of the container decreases that means volume of the gas decreases and decrease in the volume of the gas is what you call as compression and whenever there is compression work is done on the system. So if I'll be bringing the piston down some amount of work W will be done on the system will be done on the gas. Now tell me one thing. If work is done on The gas, if work is done

on the gas, what is work? Work is the mode of transfer of energy between system and surroundings. So if energy has to be exchanged, if energy has to be exchanged between system and surroundings, that exchange of energy can happen via work. So right now work is done on the gas that means energy is flowing from surrounding to system. That means energy is going from surrounding to system. How much energy? W amount of energy is going From surrounding to system. So if I ask you now after bringing the piston down tell me what will be the

total energy of the gas? You would say total energy of the gas now is going to be earlier it was U1 + Q. Earlier it was U1 + Q. Now some amount of energy has traveled from surrounding to system in the form of W. So the total energy possessed by the gas right now is U1 + Q plus W. Okay. Now assume that the gas is now present at its final state. If the gas is present At its final state, so I would say this is the final value of internal energy of the gas U2.

This is the final value of internal energy of the gas U2 which is U1 + Q + W. Okay. So if U2 is equal to U1 + Q plus W, take U1 on that side. So I would say U2 minus U1 is equal to Q + W. U2 - U1 is delta U. So I got to know delta U is nothing but Q + W. This is the result which you have to remember and we have to use this result my dear students in The questions. Am I clear with this? Am I clear with this?

Let me know quickly in the chats everybody. Yeah. Am I clear with this? All clear guys. Whatever I've taught in mole concept that's all important. Okay. So study that. All right. Let's try to do some questions on first law so that you can understand the significance and you can understand where do we use this first Law of thermodynamics. My dear students, the first question which I'm giving you a theoretical question deltaU is equal to deltaU is equal to I'm giving you first option isocoric work number two isobaric work number three adabatic work number four Isothermal

work. Tell me the answer of this question. Will you write its answer in the chats? Uh till then I'm going to write one more question here. Till then I'm going to write one more question here. The answer of these two questions. The first question which I gave you. The first question which I gave you. Quick. Delta U is equal to isocoric work, isobaric work, adabatic work, isothermal Work. It's pretty much simple. Use the first law. First law says that delta U is equal to Q plus W. First law says that deltaU is equal to Q

plus W. Now you tell me when can be delta U equal to W. Delta U will be only equal to W if Q is zero. Right? So delta U is equal to work. Delta U is equal work. In which process? Where Q is 0. Q is zero in which process? Adabatic process. I would say delta U is equal adabatic work. Yeah. Delta U is equal adabatic work. Simple question. But a lot of students do mistaken these. Yeah. What about the second question? Quick. What about the second question? What about the second question guys? Tell me

tell me tell me tell me as far as the second question is concerned during the compression of a spring. So we are compressing a spring. We are compressing a spring. So spring is becoming my system. Spring is becoming my system. We are compressing the spring. So spring is under investigation right now. Spring is under investigation right now. We are compressing the spring. If we are compressing the spring, is work done by the system or on the system? I would say work is done on the system. I would say work is done on the system. And

whenever work is done on the system, W value is taken to be positive. So how much work is in on the system? 10 kg. So W is going to be + 10 kiloj. Okay. As far as the question is concerned, 2 kgj escaped to the surroundings as heat. 2 kgj escaped to the surroundings as heat. What does that mean? That means system is releasing 2 kgj of heat. System is releasing heat and whenever heat is released by the system, Q value is taken to be negative. So Q value here I'll be taking as -2 Kgj.

Now if delta U is 10 kJ, Q is equal -2 KJ. You know delta U is equal to Q + W. Since Q is -2, W is 10. The value comes out to be 8 kg. So delta U is nothing but 8 KJ which I was supposed to calculate. Am I clear with this one as well? Yes, I believe this is clear. I believe this is clear. Look at this one. Look at this one guys. Read the question Carefully. A system expands from 5 L to 10 L. A system expands from 5 L to 10

L. That means the initial volume of the gas is given to me as 5 L. Final volume of the gas is given to me as 10 L. A system expands from 5 L to 10 L against a constant external pressure of 2 ATM. So external pressure is given to me as constant. How much? 2 ATM. And as I have already told you when external pressure is constant the process is taken to be irreversible. When external pressure is taken constant the process is called as irreversible. So as far as the question is concerned the gas is

undergoing expansion against a constant external pressure of 28. If it absorbs if it absorbs 800 jew of heat. So during the expansion during the expansion the gas is absorbing 800 jw of heat and when heat is absorbed by the gas Q value is taken to be positive. So Q is plus 800. What do we have to calculate? We have to calculate deltaU In jewles. We have to calculate delta U in jewles. That is the question. That is the question. Now you know deltaU is equal to Q plus W. Since the process is irreversible. So let

me write it as W irreversible. So in order to calculate deltaU in jewles, I would say Q has to be in jewels as well as W has to be in jewels. Q has to be in jewels as well as W has to be in jewels. Now people think properly. How do we Calculate W? W irreversible is nothing but minus P external delta V change in volume. What is P external? P external is 2 atm delta V change in volume final minus initial 10 - 5 is 5. So the value comes out to be -10 since

pressure I have taken in atm volume in liter. So W is equal - 10 atm liter. W is equal to - 10 atm liter. Now my dear students if if w is equal to - 10 atm liter I have to calculate delta u. I have to calculate delta u in Jewles. But here w is in atm liter. So you know 1 atm liter is nothing but 101.3 jw. So w is equal to minus 1013 jou. This is the value of w. This is the value of W. W is equal to -1013 Jew. If W is

equal to -1013 Jew, I'll be simply using the first law of thermodynamics. And first law of thermodynamics says that delta U is equal to Q + W. Delta U is equal to Q + W. What is Q? Q is already 800 Jew W. W is -1013 Jew. The value comes out be -213 Jew. This is The value of deltaU which I was supposed to calculate. So let me know once in the chats if all the things are clear. Let me know once in the chats if all the things are clear to you. Quick. Is it

clear tell me. Tell me quickly in the chats. all right, sir. I'm here. Good, good, good. Study, study thermodynamics. Yeah, study study thermodynamics. All right, let's move on to one more question then. Let's move on to one more question. Look at this particular question carefully. Let's see what this question exactly is all about. See guys, for a system undergoing a process, For a system undergoing a process, deltaU is equal to 300 Jew. In the question they have mentioned deltaU as 300 Jew. Okay. And it absorbs 400 Jew of heat. So system is absorbing heat. And

when System absorbs heat Q value is taken to be positive. So it's plus 400 Jew. During the expansion against a constant external pressure. So external pressure is given to be as constant as how much? 0.5 bar. All right. And since you know when the external pressure is constant the process is called as irreversible. So this is an irreversible process. Calculate the change in volume. So we have to calculate delta V and delta V has to be calculated in liters. That's The question. Delta V has to be calculated in L. That is the question. So how

do we calculate it? How do we solve this question? I'll be simply using the first law of thermodynamics. And first law says that deltaU is equal to Q plus W. Since the process is irreversible, so I'll write W irreversible. Now my dear students, deltaU delta U is given to me as 300 Jou. Q is given to me as 400 Jew. Right? W Will be minus P external. P external is 0.5 bar minus P external delta V minus P external delta V. Now here I have taken P external in bar delta V I have to calculate

in liters. So the units of this term units of this particular term will be right now in bar liter. Units of this particular term will be right now in bar liter. Now this is in jewels this is in bar liter. So you cannot subtract them. Right? I would say I would say I would say all the units are supposed to Be same. Now this is bar liter. So what I'm supposed to do? I'm supposed to multiply it with 100 then only this will also come out to be in jewles because as I've already discussed one

bar liter is equal to 100 jewles. One bar liter is equal 100 Jew. Now all the parameters are in jewles. Now solve it. So get the value of delta V and delta V value will come out to be 2 L. Right? Change in volume. Delta V is equal to L. Clear to everyone? Clear to everyone Okay. So now guys we are going to enter into one more one more topic. What is that going to be? That is going to be enthalpy function. Enthalpy function. Let's take a break for 10 to 15 minutes and get back.

Okay. So that you'll also feel refreshed. I'll also feel it refreshed. It is 10:20 right now. 10:30. 10:35. So, break till 10:35. Okay. Break till 10:35. But be back on time, guys. Be back on time. Okay? Be back on time. 15 minutes break it is. Be back on time. I'll be joining back in some time. But everyone has to be here. Okay? Everyone has to be here again. Is everyone back? Yes. Everyone back? Everyone back. All right. So guys, tell me quickly if all the things discussed till now are like super clear to you or

not. Tell me once in the chats. Tell me once in the chats. Once in the chats. >> [laughter] >> simar late. Yeah, sir. Nice, nice, nice. Good one. Try next time. Yeah. Okay, let's get going. Let's get started with the term enthalpy. Okay, enthalpy. So my dear students first of all What is this enthalpy all about? Let me tell you firstly enthalpy which we represent with H is nothing but it is U + P V. U + P V is what you call as H. U is your internal energy. P stands for pressure. V stands

for volume. U + P V is what you call as H. Simple. Now the point is what was the need to introduce this term H? What was the need to introduce this term H? Let's have a look. My dear students, as per your first law of thermodynamics, you know deltaU is equal to Q plus W. As per first law of thermodynamics, delta U is equal to Q plus W. Okay, if I write the same equation at constant pressure, imagine you have got an ideal gas which is kept at constant pressure. Imagine you have got an

ideal gas which is kept at constant pressure. Now for that ideal gas which is kept at constant pressure. If I write the same equation, I'll be writing deltaU is equal instead of Q, I'll write QP plus W. What is QP? QP represents heat absorbed or heat released by the gas at constant pressure. Heat absorbed or heat released by the gas at constant pressure. Okay. Now people in deltaU I can write U2 minus U1 is equal to QP. In stop W I can write minus P delta V. Delta V2 minus V_sub_1. Okay. So I can say

U2 minus U1 is equal to QP. It becomes minus P V2 Plus P V1. Perfect. Now take P V2 and P V1 on this side. So it becomes U2 + P V2 minus U1 minus P [clears throat] V1 is equal to QP. And if I do one thing, if I keep this Q2 + P V2 in the bracket minus if I initiate and bracket here, it becomes U1 + P V1 is equal to QP, correct? Is equal to QP. Now my dear students, do you see U + PV here? Absolutely. Do you see U plus

PV here? Absolutely. This U plus PV, this U Plus PV, this U plus PV was called as H. So basically, so basically this U plus PV was called as H. Just to reduce the calculation, U + PV was given one more name that was H. Right? So basically just to reduce the calculation this U plus PV was given one more name that is H. Now this is U2 plus Pv2. U2 plus PV2 means H2. U2 U1 + P1 means H1 is nothing but QP. And you know h2 minus h1 is nothing but delta h. So

I got to know delta h is equal qp. And what is delta H? Delta h stands for enthalpy change. Delta h stands for enthalpy change. So if I ask you how do we define the enthalpy change. If I ask you how do we define the enthalpy change? I would say enthalpy change is nothing but it is the heat absorbed. It is the heat absorbed or the heat released. It is the heat absorbed or the heat released by the system by the system at constant pressure By the system at constant pressure. So whatever heat whatever heat

would be absorbed or released by the system at constant pressure whatever heat would be absorbed or released whatever heat would be absorbed or released by the system at constant pressure that's called as delta H right Perfect. Is it clear? Is it clear people? So from now onwards if I have to define delta H if I have to define enthalpy change enthalpy change is nothing but it is the amount of heat absorbed or the amount of heat released by the system at constant pressure. Okay. It is the amount of heat absorbed or amount of heat released

by the system at constant pressure. Perfectly done. >> [clears throat] >> Now guys, let's talk about this enthalpy A bit more in detail. Let's talk about the enthalpy a bit more in detail. First of all, enthalpy which I represent with H. Let me tell you enthalpy. Enthalpy gives an information. Enthalpy it gives an information. It gives an idea about It gives an idea about It gives an idea about the heat content present in the system. It gives the idea it gives the information about the heat content present in the system. Point number one. Point number

two, if I talk about enthalpy a bit more, this enthalpy is an extensive property. It is an extensive property. Number three, this enthalpy is a state function. It is a state function. Number four, for A cyclic process, for a cyclic process, delta h is equal to zero. For a cyclic process, delta h would be equal to what? It would be equal to z. Yeah. Okay. So for a cyclic process delta H value would be zero. Now let me tell you in case of the real gas this enthalpy in case in case of the real gas

in case of the fixed amount of real gas this enthalpy it is the function of pressure and temperature. It depends on two Parameters pressure and temperature. But in case of ideal gas enthalpy only depends on temperature. In case of ideal gas enthalpy only depends on what? It depends only on temperature. And how enthalpy is related to temperature? Enthalpy is directly proportional to temperature. If you increase the temperature, enthalpy of the system is going to increase. If you decrease the temperature, enthalpy the heat content present in the system That's going to decrease. Okay? Now guys, one

important thing. What is that? For example, you have got n moles of ideal gas in the container. You have got n moles of ideal gas in the container. Let's say molar heat capacity of the ideal gas is CP. Let's say molar heat capacity of the ideal gas at constant pressure is how much? It is CP. Okay. Now, for example, I want to increase the temperature of This gas from T1 to TS2. I want to increase the temperature of the gas from T1 to TS2. And if I'm increasing the temperature of the gas from T1 to

T2, I want to calculate the enthalpy change of the gas. How do I calculate this enthalpy change of the gas when its temperature is changed from T1 to TS2? Result is delta H is equal NC cp delta T. NCP delta T. Okay. result is ncp delta t right Now my dear students one more thing for an ideal gas for an ideal gas undergoing undergoing an isoothermal process. If you'll be having an ideal gas which will be undergoing an isoothermal process. Isothermal process means temperature constant. Temperature constant means delta t0. If delta t that means delta

h has to be zero as well. So whenever you have got the ideal gas which will be undergoing an isothermal Process its delta h would be equal to zero. Okay, it delta H would be equal to zero, right? And at the same time guys if I want to write the same statement for a real gas. So I would say for a real gas for a real gas undergoing For a real gas undergoing an isothermal process for a real gas undergoing an isothermal process will delta H value be zero or Non zero what do you think

for a real gas undergoing an isothermal process will delta H be 0 or non- zero Since for a fixed amount of real gas enthalpy depends on pressure as well as temperature. No doubt you are talking about isothermal temperature constant but real gases enthalpy depends on pressure as well. So no doubt you're keeping temperature constant but pressure can change due to which enthalpy can change. So delta H for the Real gas undergoing an isothermal process will be zero or non-0ero it will be non zero. Done. Okay. I believe all the things still here are clear to

you. I believe all the things clear all the things still here are very much clear to you. Okay. Now comes few more things. Let's try to Understand those things in detail. Okay. As I told you what is enthalpy change. The amount of heat absorbed or released by the system at constant pressure is what we call as enthalpy change. And enthalpy change of an ideal gas is nothing but NC cp delta t is nothing but ncp delta t. So let me write one simple question for you. Let me see if you can solve this or not.

The question is 11.2 L of helium at STP Is heated to under a deg centigrade. Calculate enthalpy change in calories. Calculate enthalpy change in calories and do one thing. Assume the gas to be ideal. Assume the gas to be ideal. Assume the gas to be ideal. So first of all if I ask you how many moles of helium do we have since we are given with 11.2 L of helium. So imagine that This is your container and in this container we have got 11.2 L of helium. Right? And this gas is present at STP. If

the gas is present at STP, its temperature will be 0° centigrade. Its pressure will be 18 atm. Now, as per the question is concerned, you are increasing the temperature of the gas. You are increasing the temperature of the gas from 0° to 100°. If you are increasing the temperature of the gas from 0° to 100°, that means there will Be change in the enthalpy of the gas. And that change in enthalpy is represented by delta H, which is nothing but np delta T, right? The first thing how many moles of the gas do we have?

N is equal to given volume divided by 22.4 0.5. So we have we have got basically 0.5 moles of ideal gas and we are increasing its temperature from 0° to 100°. We have to calculate the enthalpy change in the ideal gas. N value is 0.5. Since helium is monotomic so it is CP is 5 / 2 R. Take the R value in calories. Delta T change in temperature final minus initial will come out to be 100. So this two this two gets cancelled 55 this is 2.5 2.5 into 100 comes out be 250 calories. So

this is the value of delta h for the gas. This is the value of delta h for the gas. Right? live evil. So whenever you are supposed to calculate delta h, delta h is nothing But n cp delta t for an ideal gas. That's all. Okay. Now there is one more thing which I want to discuss with you. What is that? See guys, that is going to be a relation between delta H and delta U. A relation between delta H and deltaU. A relation between delta H and deltaU. Now first of all you know h

is equal to u + pv h is equal to u + pv. If I multiply this equation by d it becomes d is equal d u + d of pv. If I integrate also integral of dh becomes delta h. Integral of du becomes delta u and integral of d is delta. So this is delta of pb. So delta h is equal delta u + delta of p v. delta h is equal delta u plus delta of pv. This is the general result that relates delta h with delta u. This is the general result that relates

delta h with deltau. Okay. Now my dear students If for example I write the same result when pressure of the gas is kept constant. If pressure of the gas is kept constant then I would say delta H is equal delta U plus pressure constant take it out. So P delta V. This is the result I'll be using when in the question they'll write the pressure of the gas is constant. So when the pressure of the gas is constant then delta H will be equal delta U plus P delta V. If in the question they'll Give

you volume of the gas is constant. If volume of the gas is constant then I would say delta h would be equal delta u plus delta u plus take volume out it becomes v delta p. This result I'll be using when volume of the gas is given to me as constant. Okay. Now for example if pressure and volume both will be changing. If pressure and volume both will be changing. If pressure and volume both will be changing at that point of time What result is used between delta h and delta u. Delta h is equal

delta u plus this will be written as p2 v2 minus p1 v1. This result I'll be using if pressure and volume both are given both are changing in the question. And similarly for a gaseous phase reaction for a gaseous phase reaction carried out at constant temperature carried out at constant temperature. What relation between delta H and delta is used? Delta H is equal deltaU plus Delta N GRT. This is the result used when you'll be given with a reaction which is carried out at constant temperature. So for a reaction which is given at constant temperature

or a gaseous phase reaction which is carried out at constant temperature delta H would be equal delta U plus delta NG RT. Now what is this delta NG? Delta NG is nothing. It represents number of moles of gaseous products minus number of moles of Gaseous reactants. Number of moles of gaseous products minus number of moles of gaseous reactants. Okay. So you you just have to remember this particular result. Delta H is equal delta U plus delta of PV. Now if in the question they give the pressure constant then you use this result. If in the

question they keep the volume constant use this result. If pressure and volume Both would be changing you use this particular result. If you'll be given with a gaseous phase reaction which will be carried out at constant temperature then you'll have to use this particular result. Right? Now how do we use all these results? How do we use all these results? For example, this is the first question on your screen. Check it out. This is the first question on your screen. One mole of ideal gas under goes a change in State. 1 mole of ideal gas

under goes a change in state from 2m 3 L to 2MA 7 L. So when the pressure of the gas was 28, volume was 3 L, I would say the gas was kept at initial state. The gas was present at its initial state. Now the pressure of the gas is 280m, volume is 7 L, I would say the gas is present at its Final state. The gas is present at its final state. So look at the scenario carefully. Is the state of the system changing? Yes, the state of the system is changing. And when the

state of the system change there will be change in internal energy as well as enthalpy because both are state functions. Now as per the question when the state of the system is changing change in internal energy delta u is given to me as 30 atm liter is given to me as 30 atm liter. What I need to calculate? I want to calculate change in enthalpy delta h. I want to calculate change in delta h. So delta h delta u is given delta h is to be calculated. Delta U is given Delta H is to be

calculated. If you look carefully the initial pressure of the gas is 2 atm. The final pressure of the gas is also 2 atm. That means no doubt we are changing the state of the system. No doubt we are changing the state of the gas but we are keeping the pressure Of the gas constant. We are keeping the pressure of the gas constant. So the first thing in the question is that the pressure of gas is kept constant. And when pressure is kept constant at that point of time which result do we use? Delta H is

equal delta U plus P delta V. Correct? Now I have to calculate delta H. So delta H would be equal delta U value is T atm liter plus P value is 2 ATM. Change in volume final minus initial 7 - 3 is 4. Pressure Is atm volume in liter. So this is also an atm liter. So the value comes out be 38 atm liter. This is the value of delta h which I was supposed to calculate. Got it? Did you understand how these sort of questions are to be solved? Yes. [clears throat] Did you understand

how these sort of questions are to be solved? [clears throat] Since you have to calculate delta H in atm liter that's why I took this term in atm liter as well as this term in ATM liter yeah Okay, let's try to solve one more question. Look at this particular question, guys. Which of the following is correct for the following exothermic reaction carried out at constant pressure? Carried out at constant pressure. All Right, the reaction is given. Okay, see let me first of all tell you whatever reactions would be given to you. Whatever reactions will be

given to you my dear students generally the reactions are carried out at constant temperature which is generally room temperature majority of the reactions I would say majority of the reactions they are carried out at constant temperature this is the reaction this is the reaction that's given to me okay so this Particular reaction is also carried out at constant temperature now apart from temperature being constant it is mentioned that pressure is also constant apart from temperature being constant It is also mentioned that pressure is constant. So first of all the reaction is carried out. All the

reactions are carried out at constant temperature itself. Now apart from that pressure is also kept constant. Pressure is constant. Temperature is already by Default constant. Okay. Now first of all in the question they have mentioned that in the following exothermic reaction. So this particular reaction is exothermic. If the reaction is exothermic, tell me whether heat will be absorbed or released. In case of exothermic reactions, heat is released. And whenever heat is released by the system, Q value is taken to be negative. Whenever heat is released by the system, Q value is taken to be negative.

But Here Q value is given to me as positive. So this cannot be the answer since the reaction is exothermic. In exothermic reactions, heat is released. And when heat is released by the system, Q value is taken to be negative. But here Q is positive. So this cannot be the answer. Number one. Number two. Since at constant pressure and temperature, if pressure and temperature is constant, if I use the ideal gas equation, PV is equal to N RT. Pressure Temperature constant. R constant. Volume and moles directly proportional. So if moles are increasing, volume is increasing.

Moles are decreasing, volume will be decreasing. Now tell me on reactant side, how many gaseous moles do we have? 3 + 1/4. On product side, how many gaseous moles do we have? 1 plus 1 2. So when this reaction is happening, are the moles increasing or decreasing? When this reaction is happening, moles are decreasing. If moles are decreasing, Volume would be decreasing. If moles are decreasing, volume would be decreasing. Decrease in the volume of the system is what you call as compression. Decrease in the volume of the system is what we call as compression. Whenever

there is compression, work is done on the system. And whenever work is done on the system, W value is taken to be positive. But here W value is given to me as negative. So this cannot be the answer. This cannot be the answer. Okay. Now one more Thing. If I ask you to calculate delta NG for the reaction, delta NG for the reaction will be number of moles of gaseous products which is 1 + 1 2 minus number of moles of gaseous reactants which is 3 + 1 4. So 2 - 4 is -2. If

delta ng is minus2 then delta h would be equal delta u plus delta ng rt right so I can say delta h is equal deltau u delta ng is minus2 this is r this is t so I would say delta u is equal delta h + 2 rt so this equation Tells you that delta u is greater than delta h this equation tells you that delta u is greater than delta h Okay, delta is greater than delta. So this is the correct answer. That automatically means this one is the incorrect one. So it is option

B that is going to be the correct answer of this particular question. Yeah. Am I clear? Am I clear with this everyone? Am I clear with this? Everyone quick. All right, look at this particular question, guys. Then look at this particular question. Calculate deltaU when 2 moles of liquid water vaporizes at 100° centigrade. As far as this particular equation is concerned, you are vaporizing liquid water. That means H2O liquid you are converting into H2O gas. You're vaporizing liquid water. H2O liquid is getting converted into H2O Gas. As far as equation, delta H for the vaporization

is given to me as 40.66 kiloj. What we have to calculate? We have to calculate delta U. We have to calculate delta U. So in this question, delta H is given. Delta U is to be calculated. As I already told you, whenever you are given with a gaseous phase reaction which is carried out at constant temperature, which is carried out at constant temperature, what is the result between Delta H and delta U that we are going to use? The result is going to be delta H is equal delta U plus delta NGRT. Delta H is

equal delta U plus delta NG RT. So I would say I have to calculate delta u. So I would say delta u is equal to delta h minus delta ng rt. Correct? Now as far as the question is concerned we have to calculate delta u in kilogj. If delta u has to be calculated in kilogj that means your delta h should also be in kiloj. And your delta ngrt This also has to be in kiloj. This also has to be in kilogjles. Correct? So how do we solve this now? How do we solve this now?

First thing I will be calculating delta ng for the reaction. Delta ng for the reaction is number of moles of gaseous product which is one minus number of moles of gaseous reactant. There is no gous reactant. So delta ng is one. Okay. Delta ng is one. Now put the values put the values in this expression. I would say delta U Which is to be calculated in kiloj is equal delta h delta h is given to me in the question 40.66 66 kg right delta ng value comes out be 1 R value is 8.314 Jew per

kelvin per mole R value is 8.314 Jew per kelvin per mole temperature is 373 since I have taken the value of R in jewles that means this particular term it will also be in jewles right now this particular term will also be in jewels right but I want this term not in jewels I want this term in kilogjles so what I'll be doing I'll be dividing with th00and Then only this term also will come in kiloj. And when you solve this the value will come out be 37.5 kJ. Do you see any option like this

37.5 kgj? Yes the answer is 37.5 kgj. Okay but but this is not the answer of the question. This is not the answer of your question. This is not the answer of the question. Why? Because this is the value of deltaU When you are vaporizing one mole of liquid water. This is the value of deltaU when you are vaporizing one mole of liquid water. But as per the question is concerned are we supposed to vaporize one mole of liquid water? No. We are supposed to vaporize 2 moles of liquid water. So done and dusted. When

one mole of liquid water is vaporized deltaU value is equal to 37.5 kg. Therefore, when two moles of liquid water are vaporized, deltaU would be 37.5 * 2. The Value would be 75.1 kiloj. This is the final answer of the question. This is the final answer of the question. Yeah. Correct guys? Is it clear to everyone? Tell me that quickly. Tell me that quickly. Look at this particular equation guys in the combustion of benzoic acid at 300 kel and 180 atl. So we are doing the Combustion of benzoic acid at 300 kel and 180. First

of all I told you already the reactions are carried out by default at constant temperature. Now here what we have to do I mean here we have we have kept the pressure constant as well. By default temperature is already constant but here in this question we have kept pressure constant as well. So the reaction is technically carried out at constant pressure. Temperature is already constant that is by default. Okay. But where in this particular question we are carrying out this particular reaction at constant pressure. Now as per the question heat release is 300 kgj. So

when this reaction is happening, heat released is 300 kgj. Heat released is 300 kgj. If you look carefully guys, since the reaction is carried out at constant pressure, since the reaction is carried out at constant pressure, since the reaction is carried out at constant Pressure and heat absorbed or heat released at constant pressure is what you call as QP and I believe you know it already. QP is nothing but delta H. We have discussed it. Heat absorbed or released at constant pressure is nothing but delta H. So technically delta H is given to you in

the question. How much? 300 kg. Since heat is released. So Q value has to be negative. So it is - 300 kg. Okay. What are we supposed to calculate? We are supposed to calculate the heat released at constant volume. We are supposed to calculate heat released at constant volume. QB. QB. If you remember your heat capacities, I've told you QP was equal QP was equal NCP delta T. QV is equal NCV delta T and NCV delta is nothing but that deltaU. Basically NCV delta T is nothing but delta U for an ideal gas. Okay. So

technically if you look carefully QP is given, QV is to be calculated. QP is given QV is to be calculated. Okay. So basically delta h is given delta u is to be calculated. Now the question is simple. If delta h is given delta u is to be calculated. So I'll use the result simple. Delta h is equal delta u plus delta ng rt. What do we have to calculate delta u? So delta u would be equal delta h minus delta ng rt. So calculate delta ng first of all delta ng would be equal to number

Of moles of gaseous product which is 7 minus number of moles of gaseous reactant which is 15 by2. The value comes out be minus one. Delta U we have to calculate in kilogjles is equal delta H is given to me as minus 300 kiloj right delta ng is equal to minus 1 by2 R value 8.314 Jew per kelvin per mole temperature 373 kel sorry 300 kel so since I have taken the R value in jewles so this term is in jewles right Now but I want it in kilogjles so divide with th00and now the term

is going to be in kilogjles. Solve this and get the value of deltaU. That's all. Perfect. Clear guys. Is it clear to everyone? Tell me quickly in the chats. Yes. Is it clear? Okay guys, so this was the relation between delta H and deltaU and the kind of the questions That can be asked from it. Hope this particular question is clear. I hope this particular equation is clear. I hope this particular question is clear. Yes. I hope this one is clear to everyone. In this particular question, delta H was indirectly given. Delta U was indirectly

to be calculated. Okay, one more thing if you want to have A better understanding of this particular part, you can have the better understanding of this particular thing through first law of thermodynamics. Well, first law says that delta is equal to Q plus W. If I write this equation at constant volume, If I write this equation at constant volume, when volume is constant, that means W value is zero. If W is zero, so deltaU becomes equal to QV, heat absorbed or released at constant volume. And for an ideal gas, delta U is nothing but n

CV delta T. So whenever in the question they ask you to calculate heat absorbed or released at constant volume. Whenever in the question they ask you to calculate the heat absorbed or released at constant volume that is nothing but delta U and deltaU for the gases NC cv delta T. Right? Similarly if I again use the first law delta U is Equal to Q plus W. Q plus W. W is minus P delta V. If I write the same equation at constant pressure. If I write the same equation at constant pressure, I can write delta

U is equal QP minus P delta V. So minus P delta V. If I take on the other side, so QP would be equal delta U plus P delta V. A few minutes back only we have discussed deltaU plus P delta V is nothing but delta H. So heat absorbed or released at constant pressure is nothing but delta H. And delta H for the ideal gas is nothing but NCP delta T. Okay. So remember these two results as well. So sometimes in the question they won't directly ask you to calculate delta H. They won't directly

ask you to calculate delta U. They'll ask you to calculate heat absorbed or released at constant volume. They'll ask you to calculate heat absorbed or released at constant pressure. Understand? QP delta H QV deltaU. That's all. Okay. I hope all this is clear guys. I think this chapter I'll be taking in two parts. Okay. because my throat started hurting. I don't want my throat to be dead. Let's take the second part of this chapter on Saturday. Okay. All right. So tell me once in the chats if all the Things that we discussed till now are

they clear. Are they clear to you? Ah I'll take it live. I'll take it live. I won't I won't post it record it. Don't worry. Okay. I'll take that live. With thermmochemistry. Yes. With thermmochemistry. Don't worry. With thermmochemistry. Perfect. Then so I'll schedule one more Session on Saturday or Monday. I'll keep you updated though from the in the telegram group don't worry chemical bonding also will be done everything will be done don't worry ch guys then I'll take a leave okay I'll see you in the next session and you know next session is going to

be your thermodynamics part two including Thermmochemistry Perfect.