 Hello guys. Good evening on. Can you hear me? Yeah. One sec. Yeah. So I think we started last class thermodynamics. Correct. So we were discussing terms involved in thermodynamics and I think we stopped that heat. Correct. We're talking about heat capacity. Let's check your notes and confirm once. Yes. So we'll continue with this heat just a second. I'll just pause this. Okay. So we had discussed about heat capacity last class and we have seen the, you know, the assumption also that we had a UPSC, you know, assumption and it says just a quick recap of it. Sorry. Heat we say that heat is represented by Q. Okay. Heat is given to the system, given to the system is positive, right? It is positive and heat released by the system, by the system. When system is releasing heat, it is assumed to be negative. This is the IUPSE convention we have. Okay. So what is heat? Heat we have discussed. It is the energy transfer that takes place because of difference in temperature. We also have seen the total heat capacity. Total heat capacity. It is the heat required to change the temperature by one degree. Okay. This is the total heat capacity. Okay. Heat is Joule per Kelvin. Joule per Kelvin is the unit for this. In this only, we have two types we have, which is molar heat capacity and specific heat capacity. Molar heat capacity and specific heat capacity. Correct? To raise the temperature by one degree. Change the temperature. Both are the same thing. Since one degree difference in temperature you must have. Either it is raised, obviously you are providing heat, so temperature will raise. So change also the same thing. Correct? Molar heat capacity like Joule per Kelvin is total heat capacity. Molar heat capacity as we discussed last class, it is for one mole. Right? One more term we have here. That is specific heat capacity. Molar is defined for one mole. Specific is defined for one gram. So unit for this is Joule per mole Kelvin and this is Joule per gram Kelvin. This is the unit we have. Okay. This is the thing that we had discussed. Now, if I write down, till here I think we had discussed last class. I think some of you haven't joined yet. Do you have any exam in the school going on or about to start? Guys? Because I can see only few of you have joined like 9 or 10 of you. Where are others? Any information, any exam in the school? About to start or going on? Okay, fine. Not a problem. I see if DQ is the amount of heat required, raise the temperature by DT. Okay. This is the assumption we are making. So to change in changing temperature by DT, the amount of heat required is DQ. If one degree change is required, then amount of heat required is DQ by DT. So this is the specific heat we have. Or we can say the total heat capacity. Total heat capacity. Okay, DQ by DT by definition. One degree rising temperature, the amount of heat required. If we define this for molar heat capacity, so per mole we have, so one mole substance you need to take. Okay, you see Q here or DQ if I write down, DQ is equals to the total heat capacity C into DT, total heat capacity C into DT we can write. Further, we can write this as for n number of moles, n into n into C DT we can write or we can write MS, MS DT. Okay, where this C is the terms you must understand here, C is the molar heat capacity because we have n number of moles here, molar heat capacity. S is the specific heat capacity. Sometimes a specific heat capacity also represented by C, capital C it is. S or capital C. Molar capacity is small C. I'll just write down for your reference. This is capital and this is small. With this also we represent it. One question on this they have asked in need exam. I'll tell you what was the question actually but this is the formula we have. This becomes molar heat capacity and specific heat capacity. Okay, now did you copy this? So further we can write Q is equals to, Q is equals to C, C, M, suppose the molar heat capacity small C it is. It is delta T and N, C or S simply I'll write down. MS delta T we have here. Yeah, I'll just go back one second. So this is the relation of Q we have with total heat capacity or molar heat capacity or specific heat capacity. This is the relation. Now you see one very important thing I'm going to tell you over here. From this relation we get this. How do we get this? By integrating them and put the limit. So what we are doing? We are integrating all these expressions. We are integrating all these expressions and we are taking this heat capacity outside the integral sign. And this we can do only when it is independent of temperature. So when you write after this, when you write this expression here, this expression it means it means what heat capacity whether it is molar or specific heat capacity is independent of temperature. Yes, independent of temperature. Most of the time it is independent only. If it is not mentioned, then also we consider it as independent. Okay, so note down this point. Change in heat capacity, change in heat capacity with temperature is negligible with temperature is negligible. So if it is not mentioned, so if it is not mentioned, we assume it as constant. If it is not mentioned, we assume it as constant. Okay, this is one thing. So what I said if it is not given the relation of specific heat capacity and temperature, you have to take it as constant. Now, you see under different, different condition, what is the value of heat capacity? Okay, based on this relation, this relation you must keep in mind. Okay, suppose I am assuming. We just started a few minutes back, I was doing some revision. And I think this is the only thing that we have done new, which is this till now we had discussed last class. I have done the revision of it. And then we did this thermodynamics we are continuing understood. Okay, now we'll do the calculation of heat capacity based on the process. So heading you write down all of you calculation of heat capacity, heat capacity. Okay, if the process is isothermal, then what happens? Could you tell me in isothermal process, what is the value of delta t delta t for isothermal processes? Zero, no changing temperature is zero. And when delta t is zero, what is the value of C or CM or S? It is infinity. When the process is adiabatic, what is adiabatic process? Delta t is zero. So what is C? Q by zero, infinity. That's what I said, although this is zero delta t. So what is C from the seed just focus on this relation, we are going to calculate C with respect to this relation only. So if delta t is zero, Q is C is what Q by zero, infinity adiabatic process. We know in adiabatic process, delta Q or Q is equals to zero. When Q is zero, what is the value of C? Could you tell me C value? Zero again. adiabatic isothermal, they are not going to ask you these two things. Okay, just for understanding, we are doing this copy You know what is first law of thermodynamics? Yes, keep that in mind. We haven't done this. No, first law we haven't done. I'll take the reference of it. Yeah, I'll take the reference of it just to make you understand one thing. After some time we'll discuss first law of thermodynamics also, because we are discussing Q will discuss you then we'll discuss work done in Talfi. And then we'll see the first law of thermodynamics. Okay. So this is the two things we have. Now the third one is, third one is isochoric process. What is isochoric process? Volume constant? Yes. So we have volume constant. So molar heat capacity at constant volume, molar heat capacity constant volume is represented as Cv. Okay, molar heat capacity at constant volume is represented as Cv. Okay, and this C is the small one, small c. Okay, capital if you write down capital C, then that would be capital C I'll write down like this. Okay, capital C that would be the specific heat capacity. Okay, so dq is equals to Cv dt and Cv is equals to dq by dt. And remember this relation we have at constant volume. So if you look at the relation of first law of thermodynamics, du is equals to dq plus dw, pressure we know pdv, right? If volume is constant, dv is zero, means work done is zero over here, work done is zero. And hence dq is equals to du. So if I substitute here du, we'll get Cv is equals to du by dt. Cv is equals to du by dt, isochoric process. Similarly, if you think of first you copy down this, I'll go to the next slide. Okay, now in the next one you see isochoric we are done. Now the next one is isobaric process, isobaric process. And molar heat capacity we know isobaric is constant pressure, constant pressure. And molar heat capacity, constant pressure is Cp, small one. Okay, Cp. And the relation is again Cp is equals to dq by dt at constant pressure, right? And at constant pressure, q is nothing but enthalpy that we'll discuss later. For now you just write down this formula dq is equals to dh by dt. So at constant pressure, the energy is nothing but enthalpy, dh by dt. We'll discuss this enthalpy, we'll discuss separately and then you will understand this. For this just you copy this down now. Yes. Now, so we had discussed work, we have discussed heat that is q, w and q we have discussed. Now we'll see internal energy and then we'll see the first law of thermodynamics on this. The third term you write down internal energy. It is represented by either U, capital U, or capital E, internal energy. Internal energy is what? We just discussed this internal energy on micro level. Okay, it is the energy of the molecules, atoms, which is present there in the system. Okay, usually what we do, we write internal energy U is equals to, it is the sum of all kind of energy actually. It is the sum of kinetic energy, potential energy and chemical energy. What are different kinds of energy you can think of? All these comes under internal energy. Okay, of the system. If you see kinetic energy, kinetic energy is the function of temperature. Right, we know kinetic energy depends only upon temperature. So kinetic energy is a function of temperature. Potential energy is the function of volume. Right, because if the volume is more, then the molecules are far apart, their potential energy will be less. In the molecules are close, their potential energy will be high. So potential energy is because of the position of the molecules, but atoms, particles present. Hence it is a function of volume. Yes, did you understand my volume? If volume changes, the relative distance between the molecules changes. Relative distance changes, then what happens? Then potential energy also changes. Okay, so basically from this point, what we can write, that internal energy for any system, U, is the function of temperature and volume. Temperature and volume. It is a function of temperature and volume. And if you differentiate this, du is equals to Islet's differentiation will do. We'll have du by dou u by dou t at constant v into dt plus dou u by dou v at constant t into dv. This you don't, you know, you don't require this. It is not there in your syllabus. Okay, but just for, you know, to get the understanding of internal energy, we are doing this. Okay, because one relation we'll get, which is very important here. du by dt at constant volume is what? du is equals to cv dt plus it is dou u by dou v at constant temperature into dv. Is it clear? Yes. This is for one mole. If you have n number of moles here, du by dt, we can write ncv dt. Or if I write down here, it is ncv dt for n number of moles we can write down. In general expression is this. For n number of moles, du by dt is ncv. So ncv into dt. Right? So this is the formula for the change in internal energy. Have you seen this formula before? I'll explain. Yes, I'm just one second. Have you seen the formula of change in internal energy? What is change in internal energy formula? Change in internal energy formula you must have seen. That is ncv dt. Have you seen that? ncv dt? Yes, yes, correct, correct. Only ncv dt part was there. Where did you see this formula? Okay. Yes, now you'll listen to me. See, the actual formula for du is this. Change in internal energy is this. Okay. How do we get this formula? Again, I'll explain this. From this relation, we understood that internal energy is a function of temperature and volume. Correct. One more question you just answer me. Where this formula is applicable? du is equals to ncv dt. Where this formula is applicable? Always applicable. Correct. This formula is always applicable. Constant volume is not the any criteria over here. First of all, this thing is understandable. So you must know when we're using ncv dt, what's that? What's doh? Doh? Doh is, yeah, I'll explain this a second. So you need to understand, you're writing down du is equals to ncv dt is for constant volume. Cv is defined at constant volume. Then why that formula is not applicable only for constant volume? It is applicable for all processes. Whether the volume is constant or not. That's what you will understand it from here. Doh is partial derivative. See, if you have one function, I'll come back to this expression. This is, if you have a function, suppose randomly I'm taking m is a function and it depends upon two variables. It's a function of, suppose, p and q. Okay. So m depends upon p and q. So we can find out dm is equals to, by partial derivative, dm by doh m by doh p. And when we are differentiating m with respect to p, we are keeping q as constant. So doh m by doh p, partial derivative of m with respect to p, keeping q constant into dp plus, again, another variable. Doh m by doh q. We are, you are differentiating with respect to q, keeping p as constant into dq. This is Euler's formula. Euler's formula. Okay. Don't need to know. Yes. Yes. Hello. Can you hear me? Yeah, it was Parkett. Parkett or Parkett was there. Yeah. It's fine. Okay. Yeah. Yeah. So this is the Euler's formula we have. One, you know, it is a dependent variable. These two are independent variable. This variable depends upon two variables here, p and q. So we can find out the change in m, dm, by this formula. Okay. You are not going to use this formula anymore. In engineering mathematics, you will have the use of it, right? But we'll have this expression. I'll go back to the change in internal energy expression we have here. And we get this. So this is the expression for change in internal energy right down here. It is applicable for, applicable for all conditions, processes. This is the in general formula and it is applicable for everything. Now, so this is the formula we have. All of you have copied down this. Have you written this? Okay. Now, like you said, du is equals to Ncv dt. Now, what is this formula? Right. And how do we get this? This is what we need to understand here. Okay. Now you'll see what is this term, du by dv at constant temperature. We'll try to understand this. So let us see this term first. We have du by du v at constant temperature. This is the expression we have. Okay. So u is what? u is energy and v is volume. So we can also write this as it is energy per unit volume, energy per unit volume. And we know energy is work, work is p into v. So further, we can understand this as p into v by v, which is pressure. So du by du v, this term is the pressure or for gases, it is the internal pressure actually, internal pressure we have here. Okay. Just for your understanding. Okay. So du by du v into dv is the internal pressure. Now, while taking one condition here, what is that condition? Condition is for ideal gas. Could you tell me what is the internal pressure for ideal gas, P internal? What is the internal pressure for ideal gas? Internal pressure is zero. No, we have discussed, no, there's no interaction in the ideal gas. We do not have any interaction, correct? So internal pressure is zero, which means what? Which further means what? Which further means this du by du v is equals to zero, as it is the expression of internal pressure. So in this expression here, you see, in this expression here, yes, in this expression here, for ideal gas, this term would be zero. And hence the formula is du is equals to ncv du. So the formula that you get here, hence for ideal gas, the change in internal energy du is ncv dt, the formula we have. And this formula you see, this formula is applicable for all processes, for ideal gas. Wherever we have ideal gas, and mostly we are dealing with ideal gas only. So for ideal gas, this formula is correct, whether the volume is constant or not. Did you get it? Yes. So this formula that we use, it is applicable for ideal gas at all processes. It's not like ncv dt we have, it's only for constant volume, we'll apply this. No. If ideal gas is there, we can apply this formula. Now one more thing you see, second condition in this. If you have a closed rigid container, for closed rigid container, what is dv for closed rigid container? Close and rigid container, dv, dv is again zero. So when dv is zero, look at this expression here. Look at this expression. If this dv is zero, again du is equals to what? ncv dt, right? So if you have a closed rigid container, means non-ideal gas if it is present in a closed rigid container, then also the formula for du is equals to ncv dt. This formula is applicable for all ideal gas or for a closed rigid container. So if a real gas is present in a closed rigid container, then also we can apply this formula. Tell me, any doubt in this? Now you understood where we got from where we get this formula of du and why du is equals to ncv dt is applicable for all processes where whether the volume is constant or not. Next, write down first law of thermodynamics. Okay. First law of thermodynamics is basically the conservation of energy, conservation of energy. It is based on this actually, based on the conservation of energy. Suppose we have state one and from this, we are going to state two by some process. We don't know what process it is, but there's a change in the state by some processes. So here's a pose. We have initial internal energy ui and this is the final internal energy uf we have here. Okay. So that the change in internal energy delta u is equals to uf minus ui. Now this change in internal energy we can do by heat flow, by allowing heat flow into the system, heat flow into the system or we can also do it by work done, by work done on the system, on the system or by the system both way we can do it on or by. Okay. Both way we can change the internal energy. Okay. So now suppose we have internal energy ui initially and you provided q amount of heat into it. So plus q since you are giving heat plus q and work is done on the system. If I write down this theory over here just a second, if I write down this, if heat absorbed, heat absorbed and work done, work done on the system. So we are doing work on the system. So this will increase the internal energy. So the final internal energy uf is equals to we have ui initial internal energy plus q amount of heat absorbed and work done on the systems of plus w we have. So uf minus ui is equals to what? Uf minus ui delta u is equals to q plus w. So this expression is we have delta u is equals to q plus w where w is the work done on the system. Always keep that in mind. It is the work done on the system, on the system always. Now we can also have this possibility that if heat absorbed and work done by the system, system is doing work. Okay. Copy this down first. Okay. Next you see if heat is absorbed, work done and work done by the system. So what we can write? System is doing work. So work done is what? This is negative. Heat absorbed, right? Heat is absorbed. So it is positive. So ui is the initial internal energy, q amount of heat it absorbs and out of this much amount of energy, w amount of work is done. So it is done by the system. So minus w, this would be the total internal energy. So uf minus ui that is delta u is equals to q minus w and further we can write q is equals to delta u plus w. So when you write this expression, it means work done by the system we have. This is the wd work done by the system. Then okay. So we have the expression here and the expression is if this is the expression work done by the system, if the expression is this delta u is equals to q plus w, this means work done on the system. Clear? So this is the difference in the two expression we have on the system by the system convention difference we have. Now few conclusions of first law of thermodynamics you see. First one if delta u is equals to zero. This means what? The change in internal energy is zero and the change in internal energy will be zero for we can say cyclic process. Any other process in which change in internal energy is zero? Any other process in which change in internal energy is zero? Apart from cyclic process, internal energy depends upon what? Internal energy depends upon temperature. So yes, isothermal is the another process we have in which the change in internal energy is zero. So when change in internal energy is zero, then what we can write dq is equals to minus dw, dq is equals to minus dw. One way is this we can write and if you write this expression, it means what? Minus dw is what? Work done on the system or by the system. It is wd work done by the system. Understand it carefully. It is work done by the system. This is plus dq. Plus dq means what? Heat absorbed by the system. Isn't it? Heat absorbed by this. Let me know if you're not getting it. Is it clear? Yes. Now, if you try to understand this mathematical relation we have, if you try to understand it this way, suppose you have a system, system has some internal energy, which we don't want to increase or decrease. Means we want this delta u is to be constant. Delta u is zero and we want this to be constant and system is taking 10 joule of heat from surroundings. 10 joule of heat system is taking from surroundings. We want you to be constant. What you need to do so that this u is constant while we are taking this 10 joule of energy, what we need to do? The amount of energy that you are taking in equal amount of work has to be done. Right. That's very good. 10 joule of energy you are taking and if equal amount of work if you do, 10 joule of work if you do, it means there is no net gain or loss of energy for the system and hence delta u is constant. And that is what the mathematical relation we have here. The amount of energy absorbed equal amount of work done by the system. Any doubt? Any doubt? We can also write this as minus dq is equals to plus dw. This is also possible. Means heat released by the system in this type. This is not a process, Anushka. Anushka. Okay. This is not a process. We are just trying to understand in different, different condition what is the possibility we have. Okay. Yeah. Minus dq is equals to dw. It means heat released by the system is equals to work done on the system. If 10 joule of heat, if you release, it means 10 joule of work you need to do on the system, so that u remains constant. Yes. Now the second condition you see, suppose we have a process, we have a, you know, a process in which the work done is zero. Second one you see. If work done is zero. This means what? Work done is zero means du is equals to dq we can write. du is equals to dq. This is plus heat absorbed by the system is equals to increase in internal energy because this is also positive. So we can say what plus dq is heat absorbed by the system. And this is increase in internal energy, increase in internal energy. If internal energy decreases, then heat is lost by the system to maintain zero work done, then understood. Yes, understood. Now the second case I'll write down the third one I'll write down quickly. This is what you can observe like this. If the process is adiabatic, adiabatic process. So we know in adiabatic process, delta q is equals to zero. So we have delta u is equals to delta w means adiabatic process means what? There is no exchange of energy. When suppose you have a system, system can't take energy from surroundings because the process is adiabatic and it is doing work. Means if it does work, it means it does work on the cost of its own energy, means internal energy will decrease if work is done by the system. Correct? Work done by the system, internal energy decreases, work done on the system, internal energy increases. Is it fine? So all these different different condition, you can think of whether internal energy decreases or increases accordingly. Understood, clear? Okay, we'll see some questions on this first law of thermodynamics. One second guys. Okay, I'm not getting it here. See this question. The question is, a gas occupy 2 liter at STP. It is provided 300 joule heat so that its volume becomes 2.5 liter at 1 atm, 1 atmospheric. Calculate delta u, change in internal energy. Try this one? No, no, no, you cannot take the exact value. No, that's not right. 350 is wrong. What is the answer Anusha? No, that's also wrong. 150 is wrong. No, 300 is also wrong. 450 is wrong. 470 is wrong. 250 is right. Yes, 250 approximately it's right. Okay, so here work done is equals to, we'll write minus P external delta v minus P external delta v. P external is one atmospheric. So minus one into 2.5 minus two, that is minus 0.5 unit is liter atm. Okay, liter atm. So this is liter atm. We need to convert this into joule, right? So we know one liter atm. This is a relation. We have one atm liter is equals to 101.328 joule approximately we have. Or if you don't remember this, I have given you last class the relation that is 0.0821 atm liter is equals to 8.314 joule. These are the relation we have. So one atm liter you can easily find out 8.314 divided by 0.0821 atm liter, joule, sorry. Okay, with this also we can change 0.5 atm liter into 101.328 joule approximately when you solve this you'll get minus 50.631 joule of work done. Now q is equals to, you see a gas occupy this provided this joule of heat volume becomes this. So there's expansion, right? So work done by the gas we have here, right? So work done by the gas what we'll write delta u is equals to q plus w. We need to find out what change in internal energy. So q is 300 joule of heat provided and then work done is minus 50.631 joule. So when you do this you'll get 249.317 joule. So change in unit you must remember. One more question you see on this. This question you see 2.8 gram of N2 gas at 300 Kelvin and 20 atmospheric pressure was allowed to expand isothermally against a constant external pressure against a P external one atmospheric calculate delta u, q and w for the gas. Try this one.