 Yeah. Shall we start? Yeah. So we were discussing enthalpy. Mathematical definition is u plus pv. It is a heat content of the system at constant pressure. Correct? So du is equals to dq plus dw, the first law of thermodynamics. And then if you find out this du, so du is what? dw is what? Minus pdv. So I forgot to write down one minus sign here. We'll have minus pv here. So we'll have here pdv minus of pdv minus vdp. This is what we get. So here also we'll get the negative sign minus pdv. So dq is what? dq is the heat content at constant pressure because we have assumed this at constant pressure and then only we have taken dp is equals to 0. So at constant pressure, the heat content of the system, no, was still 620 only. Break was still 620 only. Yeah. So at constant pressure, the heat content of the system is dq is nothing but dh. That's what the definition we have over here. Correct? So from this, you know, relation we can write dh is equals to du plus pdv. And that is what the mathematical definition of enthalpy change we have here. Now, you see this enthalpy is a function of pressure and temperature mainly. To some extent, it depends upon number of moles also and volume also, but main factor is this pressure and temperature. Those are negligible. So again, we'll write down the Euler's formula here. So dh is equals to doh by dohp at constant temperature into dp plus doh by dohd at constant pressure into dt. So if you see doh by dh by dt at constant pressure, we have done this formula. This is nothing but cp ncp for n number of moles. Right? So if you substitute this here, so dh is equals to we have ncp dt, ncp dt plus doh by dohp at constant temperature into dp. Now this is the formula of enthalpy change for all processes. This is the formula of entropy change for all processes. But we have to see for ideal gas because mainly we will be discussing with where we have negative. Where is negative sign? Anurag, no, work done is minus pdv, minus pv. pdv is positive in physics and estimates negative. Yes, Anurag, tell me. See, work done is always minus p into p, negative of p into volume change. That is work done always. On the system, by the system is different. Like it will come automatically. Because volume change is there, no. On the system is there, contraction will be there, volume will decrease. By the system expansion, volume will increase. So accordingly, we will get the sign. But formula of work done is always minus p external into delta v. Understood, right? So this is the relation we have. Now we will apply the condition of ideal gas you see what we get. That is the formula we have. So for ideal gas you see, we have this relation h is equals to u plus pv. We know this internal energy depends only upon temperature. So it is a function of temperature and pv is equals to nrt, which is also a function of temperature. So what we can say in Thalpy depends only upon temperature for ideal gas. Correct? So when we write down this expression, doh h by doh p at constant temperature. So it depends upon temperature only. And we have constant temperature. So there is no change in this. So it will be zero. And when you substitute this in the previous expression, the formula that we have we will get dh is equals to what? Ncp dt. So this is the formula we have applicable for ideal gas. Real gas because in Thalpy is a function of temperature. Correct? This means what? This expression? I am just checking out this expression. What happens in this? This means what? Differentiation of h with respect to p. But it is independent of p. You see it is a function of temperature only. In Thalpy is independent of pressure. So this differentiation would be zero. Temperature is constant. So this is for ideal gas. Even for real gas also we get the same expression. But the condition for real gas is right down for real gas at constant pressure. At constant pressure. So real gas if you have constant pressure, so dp is equals to zero. Which means doh h by doh p at constant temperature into dp is equals to zero because dp is zero here. So when this is zero, dh is equals to Ncp dt. This is the for real gas. This is the condition. Here. So when you write Ncp dt, you have taken the formula for ideal gas only. Most of the time we will be dealing with ideal gas only. That is why the formula we have. Clear? No doubt. Now this is done. One other thing you see in this particular in Thalpy concept. At constant pressure, at constant pressure the relation we have dh is equals to du plus pdv. Right? Both sides will divide by dt. What is this expression? dh by dt? You see? dh by dt for one mole. For one mole. dh by dt is nothing but cp and is one. du by dt is nothing but cv plus. This is one. This is r. Yes. Hence the formula is cp minus cv is equals to rv. You must have done degree of freedom in physics. Right? Okay? So cp minus cv is equals to r. One formula, very important. These formulas you have to memorize. And we also have one formula. We call it as cp by cv. The ratio of cp by cv is gamma. Right? All these are molar heat capacity, not specific heat. Okay? Gamma. Gamma is called Poisson's ratio. Gamma is the Poisson's ratio. Okay? With this two formula, you can find out cv and cp. cv is equals to gamma by cp. Sorry, cv is equals to cp by gamma. You substitute here. So cp minus cp by gamma is equals to r. So we'll have cp is equals to gamma r by gamma minus one. And cv, if you calculate here, it will be gamma minus one by r. cp by cv. cv is equals to r by gamma minus one. This is the relation we have. Okay? cp cv. cv, we have one more formula in terms of degree of freedom. And cv is equals to f by two. F is the degree of freedom. Okay? In general, what happens on the basis of the atomicity of the gas, you can have the idea of degree of freedom of the gases molecule. F value. If the atomicity is one degree of freedom, atomicity is one degree of freedom is three. Atomicity is two. Degree of freedom is five. Atomicity is greater or equal to three. Degree of freedom is six. Okay? We can discuss this also, but this is not a part of chemistry. You have done this in physics, so I'm not discussing. Okay? Just you need to show the all possible, you know, motions. You haven't done this in physics, degree of freedom. Have you finished this chapter? Yeah. So it will, you will do this in physics. Okay? In physics, you will do this degree of freedom. So I will do that. But just quickly, if I tell you, degree of freedom is a number of independent ways by which a system can exchange energy. Okay? Number of independent ways by which a system can exchange energy. Okay, just a second. So one, two, three, quickly, I'll tell you how to do this. But mainly you have to memorize this value. Even if you don't memorize this degree of freedom, just you need to know the value of Cp and Cp. We'll discuss that value also. But degree of freedom is actually quickly, I'll just, few things we'll discuss over here and then we'll move on because we'll do this in the syntax, right? Degree of freedom F, like I said, write down, it is a number of independent ways. It is a number of independent ways by which, by which a system can exchange energy. Exchange energy. Okay? So there are three types of degree of freedom we have. We have translational, translational concept. Okay? We have translational DOF, degree of freedom. Then we have rotational DOF. I'm not going into that detail, okay? Because you will do this in physics. Little bit of idea I'm just giving you. And then we have vibrational, vibrational degree of freedom. Translational is FT. It is FR. It is FB, vibrational. This is the three things we have. Now, depending upon the atomicity of the gas, okay? Depending upon the atomicity of the gas, we can find out a degree of freedom. The CV formula that we have, the formula of CV, you have to memorize this. We have FT plus FR. Or you let it be, this formula is not required. You will get confused on this. Just one formula you keep in mind. CV is equals to F by 2R, which I have given you already. F is the total degree of freedom, means this plus this plus this gives you F for any gas. Now you see how do we calculate this? One or two will do. Suppose the gas is monoatomic, monoatomic gas. Right? So for monoatomic gas, the translational motion is what? We have only one atom. Suppose helium. It can move in x-axis, y-axis and z-axis. Right? Three possibilities we have. So FT, if you see, translational degree of freedom is 3 here. Are you getting it? Which table? Which table, Madan? Previous one. Once again. Which table tell me? I did not draw any table. I just said for, this one you are talking about. Once again. This one. Right? Understood? So mostly what we do, we vibrational degree of freedom we don't consider. Vibrational motion we don't consider here. In our syllabus, we have mainly translation and rotational. But if you talk about the degree of freedom, means basically for a given molecule, we are considering all types of motion by which it can exchange energy. So one motion is translational, other one is rotational and other one is vibrational. Vibrational we are not considering. Okay? We are ignoring this. It is not in our syllabus. Vibrational. So monotomic gas can exchange energy by moving along x-axis, y-axis and z-axis. So translational degree of freedom is 3. Rotational degree of freedom if you see, rotational degree of freedom is 0. Vibrational we are not assuming. Why rotational is 0? Because it cannot rotate. Right? So total degree of freedom is what? F value is 3. That's why the value of Cv for monotomic gas is 3 by 2R. This F will substitute here. Cv is equals to 3 by R. 2R, you must have done this, you know, value, this formula also. That's Cv value for monotomic gas 3 by 2R. Clear? Guys? So rotational, rotational, vibrational we are not considering. It is not in our syllabus. Okay? So vibrational motion you let it be. It's difficult to visualize this now. So just ignore this. Okay? Why rotational is 0? This molecule can't rotate where translation is not 0. Translational is 3. Rotational is 0. See, very simple one. A single atom cannot rotate. Okay? It can spin, but it cannot rotate. Right? So spinning is different. Rotation is the rotation is the thing when the molecule rotates about an external axis. Right? Single molecule can only spin. We don't say off-rotator. Off-spinner both. The leg is spinnable because the ball can spin around its own axis. It cannot rotate. Good example. Right? So yeah, keep that in mind. We have off-spinner. We don't have off-rotator, but leg rotator. Correct? So single molecule we do not have rotation. Spinning is possible. Right? Molecule can spin. That's why FR is 0. FT is 3. Total degree of freedom is 3. This 3 will substitute here. CV is 3. Once you have CV, could you find out CP from this? We can find out because we know CP minus CV is equals to R. Correct? You know CV, you can find out CP. So CP is what? 5 by 2R. What is the Poisson's ratio for this? It is CP by CV. So 5 by 3, which is 1.66 approximately. So in chemistry, we just know this value. This value, this value, this value for monotomic gas, for diatomic gas, for polyatomic gas. All this derivation is not required. Just you need to know CP, CV and gamma value. You must keep this in mind. Now, since we have done this much, I'll tell you how to find out for diatomic molecule also. If you have diatomic molecules, then degree of freedom, the translational one is always 3. Diatomic molecule. FT will be 3, because it can rotate, you know, it can exchange energy along x, y and z axis. Diatomic molecule, the rotational degree of freedom is 2. Why? Because suppose we have molecules. Once again, I'm going back and forth. So suppose you have x, y and z axis, x, y and z axis. So diatomic molecule we have. So if the molecule is along x axis, so it can rotate along y axis. We can have an axis along y and another axis along z. x axis, we do not have any rotation. Spinning will be there along x axis. So if molecule is along y axis, then it can rotate along like, you know, rotate with respect to x axis and z axis. That's why we have 2 degree of freedom here. Total degree of freedom is 5 and CV value is what? Could you tell me? It's 5 by 2R. CP value, you add R into this, 7 by 2R and gamma value is CP by CV, which is CP by CV, which is 1 point or 204. This value, you must remember for monotomic, diatomic and polyatomic gas. Because if the molecule is along x axis, no. So xk along, we have a spin, not rotation. With respect to y axis, we have rotation. With respect to z axis, we have rotation. So 2. Copy it down. Because the molecule is linear, diatomic molecule, no. The molecule is linear. You have h2 molecules, for example. Okay. Along this axis, we have rotation because the molecule lies along this axis. And these 2 axis, we have rotation. It can rotate around this. Sk along, we have a spin. Sk along, we have rotation. So in triatomic molecules also, in triatomic molecules or polyatomic, 3 or more than 3 triatomic or polyatomic molecule. In this, we have 2 types. Either the molecule is linear or the molecule is non-linear. Linear or non-linear. If it is linear, then in this case, the degree of freedom here you see is same as we have in diatomic molecule. FT is 3. Logic just now I explained. FR is 2. And then degree of freedom is 5. Everything you can find out. CPCV, everything you can find out. Similar to that of the diatomic molecule. If it is non-linear, for example, SO3, we have the molecule. Then FT is again 3. Translational degree of freedom. FR is also 3. Because it is non-linear, XYZ, every axis will have rotation possible. Hence the degree of freedom will be 6. CV is 6 by 2R. That is 3R. So CP is 3R plus R. That is 4R. And gamma value, Poisson's ratio is 1.33. This you need to memorize for all the 3 gases. This is for non-linear. If you have CO2 molecule, it is also polyatomic. But it is linear. So for this, the degree of freedom is 5. Understood? Yes, you just need to remember the value. Degree of freedom is not the portion of chemistry. You will study this in physics. Since you asked, so I discussed this now. You just need to remember CPCV and gamma value. Then copy. Gamma value, we have one more relation. Gamma and degree of freedom you can find out. Gamma is equals to 1 plus 2 by F. This value also, relation also you must keep in mind. What is the degree of freedom for monoatomic gas? Tell me. The degree of freedom for monoatomic gas? 3. If you substitute F is equals to 3, what you are getting gamma values? 1.6. And that is what we get in the calculation. So with this also, if you keep in mind, you can find out gamma value if you forget in the exam. Understood? Yeah. Now, you see one more concept, a small concept of enthalpy formula of enthalpy we'll see. And then we'll move on to the question. Okay. Probably the last concept for today. You see, if you talk about the enthalpy change in chemical, all these things we are doing under enthalpy only in a chemical reaction. Enthalpy change in a chemical reaction. Generally, what happens chemical reaction takes place at constant pressure and constant temperature, any reaction means at a given temperature and pressure the reaction takes place. So look at this work done. We know delta H is equals to what? Delta H is equals to delta U plus delta W. Delta W is P delta V. Correct. Because constant pressure we have. So P delta V. P delta V is what? Sorry, minus of P delta V. Work done is minus of P delta V. Okay. Delta V is what? Delta U P minus I'll take outside P V F final volume minus P V I initial volume. Can we write this further? What we can write delta U is equals to negative sign. P V F is N F final number of moles RT minus N I RT because PV is equals to NRT. And this is for the gaseous species only for the gaseous species. Why? Because we are applying PV is equals to NRT here. And this is applicable only for gases. So this whatever we are discussing it is true for chemical reaction involves gaseous species either on the reactant side or on the product side. So further it becomes delta U plus minus RT N F minus N I okay. N F minus N I we call it as delta N G. G stands for gas. This would be minus of delta N G RT. No, no, it's not like that. If gas is not involved, then we can have other kind of work done. Pressure volume work done is there. Right. If gas is not involved in this formula, we cannot use. That is what the meaning is. It's not like suppose if gas is not there, then we cannot do any work. No, work done is possible, but formula we can use when the gases are involved. So the formula is delta H is equals to delta U plus delta N G RT. Minus sign will take it inside. Delta N G how do you find out? Delta N G will find out number of gaseous product gas on the product side minus number of gaseous reactant. So product minus reactant will do delta N G minus sign will take it inside here. Hence it is it becomes gaseous product and gaseous reactant. If you write minus sign here, it is reactant minus product. So with this formula delta N G you will calculate like this gaseous product minus reactant. So you see one more thing here. All of you are done. Yes, one more thing you can understand from this. What is that? Work done what we get delta W is equals to we get here minus delta N G by RT. This is what the work done we get. Instead of P delta V, we simply have delta N G RT. That is it. So if work done, if suppose in a given reaction what all things we can conclude you see. See that this is what the difficult part in this chapter. There are so many conditions. One thing we have then we can apply some conditions on that and we can conclude things like that. Like suppose and one example I will take. The reaction is given 302 gas converts into 203 gas. Could you tell me delta N G value here? What is the value of delta N G? It is gaseous product 2 minus gaseous reactant 3. So that would be minus 1. It means delta N G is less than 0. When delta N G is less than 0, it means delta W is negative or positive. Delta W if delta N G is negative, so delta W is positive. Delta W is positive means what? Work done on the system. Delta W positive means see how we are concluding things. One thing was given and how we are concluding things from this. Delta W greater than 0. Work done on the system. Work done on the system means compression or expansion. It is a case of compression. So all these data all these informations you have once you know this delta N G value. So if you have this reaction CaCO3 solid or gas state of this solid. When it dissociate what it forms CaO. Solid or gas solid yes and it forms CO2 gas. So what is delta N G for this? Delta N G is 1 because all are solid. So 0 0 0 1 greater than 0. Work done negative. Work done negative means buy the system. Buy the system means means expansion. Our DO will be 0 when T is constant. Internal energy is a function of temperature. We can predict based on the value of delta W. If delta N G is 0 then what happens? Work done is 0 in that case. So this formula delta H is equals to delta U plus delta N G. We can use when chemical reaction is given which involves gaseous species. We can use this and find out the data. Let's see some questions on this. So we can't hear you if you're talking. I wasn't mute. What I was talking about I just used I said I just we just we just have to use this formula del U and this. So delta H is given 176 kilojoule. Unit is also very important here. Okay. So unit you must take care of. Kilojoule is given here for options also given in kilojoule. So we'll keep this in kilojoule. R value we'll take in Si that is joule. Okay. If calorie is given over here then we'll take 0.082 and now we take 1.92 calorie here the value of R. Okay. So this we convert into kilojoule. This and this will get cancelled. It's subtracted. So roughly around 168. Closest option is 167.7 kilojoule per mole. 37 what is the answer? 37 is C. Is it C? In which reaction delta H is less than delta U? So for this data for the relation to be this delta H should be less than delta U. What is the value of delta Nd? Greater than 0, less than 0, what? Delta U is more, right? So it should be less than 0, right? Less than 0, right? It is less than 0. So in which reaction it is less than 0? For this one it is 2 minus 1 minus 1, 0. It is again 2 minus 1 minus 1, 0. In this one it is 1, in this it is minus 1 and 2 plus 3 plus 1. So only option is this. 38th one. What is the answer? 38th one. Did you get C? Sharad and Sashav got C. One mole of a non-ideal gas undergoes a change in state. Pressure volume temperature is given from this to this. With a change in internal energy is this. Change in enthalpy of the process. What can we do in this? Delta H is equals to? Yes, yes. So we can use delta H is equals to delta U plus P delta V. P delta V we can use P1, P1, P1 minus P2, V2 we can use, right? Delta Ng RT we won't use here because reaction is not given. Reaction is not given. So you cannot find out the value of delta Ng, right? You cannot find out the value of delta Ng. That's why delta Ng we won't substitute here. But what we can write? We can write it is delta PV in fact, one second because P is also changing. No. So I'll write down here delta PV. So this would be delta U plus P2 V2 minus P1 V1, okay? So delta U value is given. What is the value of delta U given here? Let me see the question. Delta U value is 30 liter ATM and options also in liter ATM. P2 V2 is 5 into 4. That is 20 minus 6. They're getting 14, 20 minus 6, 14. So 30 plus 20 minus 6, 14. So we are getting approximately 44 liter per unit. Option C is correct. Yes. In that case, if it is reversible, then delta W you have to calculate separately. You have to do the integration and find out delta W and then substitute it. This one. Quickly, tell me the value. Tell me the answer. Now, in the last question, my other pressure was not constant. It is given that the pressure is changing now. So once it is given that pressure is changing, so there's no point of assuming it to be constant. No, the question was, you see the question was not like that. The question was we have a process which is going from state one to state two. So in that, we have a certain work done. Chemical reaction was not given. No. So from that PV value, you will get the work done. And what is the enthalpy exchange in that process or internal energy involved that you can relate then from delta H and delta U plus P delta V. Okay, done. 26th one is A you are getting. 27th one. Combustion of this obviously 27th one, what is we are getting? We are getting delta NG as negative, right? So obviously delta U should be more than delta H. So C is correct, 27th one. I hope all of you have got this. 27th one is this. 26th one, you need to find out the difference between delta H and delta U. This difference you need to find out. Delta NG RT you need to find out. That is what the question is. Which one? Question number? Why delta U would be zero? I'll just go back. Wait, let me finish this. Delta NG. What is the value of delta NG? Which one? That's what I'm doing. Delta NG is 12 minus 15. That is minus three. R value, you see the option. We need to find out in kilojoules. So R value is 8.314 into 10 to the power minus three into temperature is given 25 degree Celsius to 298 Kelvin. This you need to solve. No, it's not like the heat of the reaction. It's a difference between the heats of reaction at constant pressure and constant volume. So what is the heat at constant pressure? Clearly it says heat at constant pressure is delta H enthalpy. Heat at constant volume is delta U. That is what the question is, right? Indirectly it says. So what is the answer? I think A you are getting. No, A is correct. I guess minus 7.43. Yes, minus 7.43 you are getting. Question number 30. What was the doubt and go back with it? 30 ways, 30. The previous one. 36, 36. Enthalpy change for this at 1000 Kelvin is this, the value of delta U for the reaction. Why dt will be zero here? See, the reaction is taking place at this temperature. See internal energy depends upon temperature, correct? In the chemical reaction, which usually happens at a constant temperature and pressure, if you go by this method, then always delta U will be zero. Delta U here won't be zero because it is a chemical reaction. One molecule is converting into other molecule, right? So chemical reaction, you see the third part of internal energy, kinetic energy, potential energy and chemical energy. The third part is coming here into the picture. That's why usually we cannot take delta U is equals to zero simply for any chemical reaction. There will be exchange of energy between the molecules, which is present there in the reaction mixture. And internal energy involves all these energies, the energy associated with molecules. On a micro level, the energy associated with the molecules present in the system, that also we count. Obviously CO2 will have some energy. So in general, we say for processes, for any process that we have, but for chemical reaction, we don't apply the same case over here. At constant temperature, delta U will be zero. Otherwise for all chemical reactions, delta U would be zero. Here we don't apply because we are getting different molecules from a given molecule. Let's see if you have a few more questions. One or two questions you can do from this. Question number 22, you let it be. We haven't done this. We'll do this next class. 19 also you let it be. Question number 21, you try. 21. That's what the definition matter. You see, you go back and see the definition of delta U. When at constant volume, while a condition you see, when the volume is constant, work done is zero. Work done is zero. So du is equals to d cube. So heat is nothing but the internal energy at constant volume. Use first law of thermodynamics. Condition you take delta, you know, delta V is equals to zero since constant volume. So d cube is equals to du. Ideal diatomic gas is heated at constant pressure. The fraction of the heat supplied with increases the internal energy of the gas is ideal diatomic gas heated at constant pressure. Okay. Okay. So heat, it uses to increase the internal energy du 21. I'm doing du is equals to cv dt. One more is it. So one more. I'm assuming cv dt. And at constant pressure, the heat is DH. It is cp dt. Ideal diatomic gas we have is heated at constant pressure. The fraction of heat supplied the fraction of heat supplied to increase in internal energy. Right. So that would be what du by dh once again. So what is du by dh? The fraction of heat supplied with increased internal energy of the gas. So du by dh you need to calculate. So it is cp by cv. What is the value of cp by cv for diatomic gas? We have cv by cp. Sorry. The cv is 5 by 2R and cp is 7 by 2R. So when you solve this, you'll get 5 by 7. So option D is correct. Correct. Got it. So this question also we'll see 19, 20, 22. Next class we'll start with work done calculation. Okay. And then we'll see some more numericals on this. Okay. So I will share one DPP that is based on first law of thermodynamics. Finish that. Okay. Okay guys. Thank you so much. Take care. Bye. See you next class.