 Welcome back to the next lecture on statistical thermodynamics. Today we are going to discuss another very important thermodynamic quantity which is heat capacity. Heat capacity is a connector between temperature dependent thermodynamic quantities. That means, if I have a knowledge of heat capacity and I have experimentally measured some thermodynamic quantity, then I can calculate the same thermodynamic quantity at a different temperature without doing experiments provided I have information on heat capacity. That is why I said heat capacity is a very important thermodynamic quantity. Usually we talk about heat capacity under constant volume conditions or under constant pressure conditions. And we have also discussed from time to time that maintaining constant pressure conditions is far easier than maintaining constant volume conditions. For example, if I have a sample let us say consisting of ice and ice melts, when ice melts the volume decreases. So, I cannot keep the volume of the system constant, but if I have ice at a certain temperature I can keep the pressure constant and I can change the temperature. We must keep the phase changes in mind when we discuss the heat capacities. So, let us get started with discussion on heat capacity. Heat capacity under constant volume conditions and under constant pressure conditions both are important. And both are interrelated also. C p minus C v is equal to n r for a perfect gas. C p minus C v can be expressed in different forms for non ideal systems, non ideal gases, non ideal systems. But we will not get into that kind of discussion at present. And in our canonical ensembles we have usually been discussing constant volume conditions. So, therefore, let us stick to the discussion on constant volume heat capacity. And then we will discuss how to get the constant pressure heat capacity from the knowledge of constant volume heat capacity. In layman's language heat capacity is the amount of heat required to change the temperature of a system by 1 unit, 1 degree centigrade or 1 Kelvin. And if the conditions are constant volume we will put the constant of constant volume. And in that case the derivative will be represented as a partial derivative. And sometimes we represent temperature in terms of that undetermined multiplier that is beta for which we earlier have discussed that beta is equal to 1 by k d. So, in that case if we are required to differentiate internal energy with respect to beta we should have an expression for that. Why it is important? It is important because certain derivations for thermodynamic quantities have been expressed in terms of partial derivative with respect to beta. So, let us discuss over here C v is equal to del u by del t at constant volume. Now, let me represent d d t I can write this as d beta d t d d beta mathematically this is permitted. And we know beta is equal to 1 over k t therefore, d beta over d t is equal to minus 1 over k t square right differentiation of 1 over k t 1 over k is constant. So, t raise to power minus 1 derivative of t raise to the power minus 1 with respect to temperature is minus 1 over t square k t square this is the d beta by d t. And I can further write this as minus k over k square t square and beta is 1 over k t therefore, minus 1 minus k beta square. So, therefore, when I substitute this over there what I get is I can write d d t as equal to minus k beta square into d d beta. And I can make use of this now in writing an expression for heat capacity at constant volume and I come up with this expression that is C v is equal to minus k beta square del u by del beta at constant volume. You may ask why did I do that why did not I leave it to a simple expression which is a derivative with respect to temperature. I already gave you the reasoning that some of the formulae have been derived in which you are required to take the derivative of certain quantity with respect to beta. So, therefore, in order to provide a simpler solution to those kind of issues it is advisable then to use this expression for constant volume heat capacity alright. Now, let us discuss little more we know that C v can be expressed in terms of partial derivative of u with respect to temperature or partial derivative of u with respect to beta. Let me write another the second definition also here for our simplicity that is C v is del u del t at constant volume. Now, total energy of a system can be written as u minus u 0 we have discussed many time what is the reason for u minus u 0 and that means, I can write u minus u 0 is equal to or if I divided by n if I divide by n then this is what this is mean energy ok. So, in other words I am saying u minus u 0 I can write this as n times mean energy why this is mean energy and is the total number of molecules or total number of atoms or particles in the system. Now, the temperature derivative of this or the derivative with respect to beta which is also essentially temperature derivative that means, C v then C v will be equal to partial derivative of let us write del u by del t at constant volume and derivative of u 0 it is a constant number it is 0. So, we do not care this will be equal to n times del e del t at constant volume very simple derivation. So, now, pay attention to what is written over here C v for a any mode where m here represents mode what is this m m can be translational mode m can be rotational mode m can be vibrational mode m can be electronic mode or whatever mode you can consider. So, therefore, C v for a mode is equal to the total number of molecules into partial derivative of mean energy with respect to temperature at constant volume. And if you come across the expressions which involve beta then it will be easier to use this alternate form of the derivative essentially both are same. Once we have understood this then we are all set to now evaluate different contributions which are translational rotational vibrational or electronic ok. Let us move ahead and see. So, what we have done is we have discussed this form where C v m is equal to total number of molecules into partial derivative of mean energy. The total number of molecules n is equal to number of moles into Avogadro constant this is how we calculate total number of moles. Therefore, this n can be written as number of moles times Avogadro constant and same thing I can apply to the other form. Now pay attention to this this is the heat capacity at constant volume for a particular mode of motion. This is the heat capacity of actual size of a system. Actual size means for a given number of moles right if I for example, if I take 5 gram of something which has a molecular weight of 200 that means, number of mole is 5 divided by 200. So, this C v represents heat capacity for actual size of the substance. I can change it to a property which is not dependent upon size of the system that is molar molar properties. So, if I divide C v by n I get molar constant volume molar heat capacity for a particular mode and now here that small n is gone because this is consumed on the left hand side. So, therefore, constant volume molar heat capacity is n a times partial derivative of mean energy with respect to temperature at constant volume. Now the problem is simple once we know this formula I only need to know the expressions for mean energies. We have already discussed mean energies for translational contribution rotational contribution vibrational contribution we have already discussed this of mean energies. So, let us keep this expression in mind this one and proceed ahead. We will now start talking about individual contributions. When we consider a gaseous molecule or gaseous atom the temperature is always high enough for the translational contribution to be completely active. Translational mean translational energy according to equipartition theorem here according to the equipartition value it is 3 by 2 k T. We are interested in finding C v m translational and this will be equal to Avogadro constant into derivative of mean translational energy with respect to temperature at constant volume. Now you see here 3 by 2 k T this is the equipartition value. Let us take a shortcut and directly substitute this equipartition value and see what do we get C v m translational is equal to n a into partial derivative of 3 by 2 k T with respect to temperature at constant volume 3 by 2 is constant k is constant and we have n a and delta T delta T that is 1 partial derivative of temperature with respect to temperature is 1 and k times n a is R 3 by 2 R. So, therefore, mean translational energy which we used in this expression of heat capacitate constant volume 3 by 2 k T is the value that we remember from equipartition theorem and we have already discussed that the mean translational energy which is u minus u 0 by n mean translational energy u minus u 0 is always equal to minus n by q del q del beta at constant volume and if we are considering the translational contribution or translational participation of a molecule or of a particle of an atom in 3 dimensions that means we have to use q T q T here this is all t is equal to v upon lambda q. Now I am not going to go into details because we have already done this once you substitute this q T is equal to v upon lambda cube here and here you will get that u minus u 0 upon n will be equal to 3 by 2 k T I am not doing it because we have already done in earlier lectures and this is equal to the mean translational energy. What I have done here is shown you by both methods by using the equipartition value directly use the equipartition value and if you are asked not to use equipartition value, but use the concepts that you developed in statistical thermodynamics in that the case that you will be going through this route and by this route u minus u 0 by n is equal to 1 over q T into del q T by delta beta at constant volume by this means by using q is equal to v upon lambda cube you can further show your steps and eventually come up to 3 by 2 k T. Now the derivative of this which is done over here is same as 3 by 2 k times n a and this is equal to 3 by 2 r. Therefore the mean translational energy for a molecule which is free to move in 3 dimension is equal to 3 by 2 r. Keep in mind that the temperature is always high enough provided look at this comment provided the gas is above its condensation temperature then only the full translational contribution will be there that value is equal to 3 by 2 r. So, I hope that this translational contribution to constant volume heat capacity is clear. Now from the knowledge of C v we can get C p let us have some discussion on this. We know that C p minus C v is equal to n r we have been studying from class 11 or 12 onwards and if I divide throughout by n I can write C p molar minus C v molar is equal to r, but remember that these 2 equations are only applicable to ideal gases. If the gases are non ideal then these 2 equations need further modifications then C p minus C v is not simply equal to n r there will be additional terms. When you switch over from ideality to non ideality what is the difference between an ideal gas and a non ideal gas in ideal gas you assume that there are no intermolecular interactions. But in a non ideal gas you assume that there are intermolecular interactions these intermolecular interactions can be of attractive interactions in nature or it can be repulsive interactions in nature. Therefore, depending upon the extent of deviation from ideality the new terms can come into C p minus C v, but so far we have been focusing our discussion on ideal system ideal gases. So, therefore, let us further discuss for ideal gases at present. We have shown that C v m is equal to 3 by 2 r. I am again and again emphasizing on the fact that I am writing the left hand side as a molar quantity therefore, there is no need to include n on the right hand side and I can use now C p m is equal to C v m plus r from this and we already know it is 3 by 2 r plus r is equal to 5 by 2 r. Therefore, K ma which is the ratio of C p m by C v m is 5 by 2 r divided by 3 by 2 r therefore, it comes out to be 5 by 3. Remember the value of C v will depend upon the state of the system. What is that? You have monatomic gas, you have diatomic gas, you can have multiatomic gas as long as you keep the system under ideal conditions for monatomic 3 by 2 r. For diatomic now you remember our previous discussion that you not only have translational contribution, you also have rotational contribution because once you have 2 atoms rotation is possible and that contribution can come in. Vibrational contribution can come in, electronic contribution can come in. So, therefore, this C v m is subject to change depending upon is it a monatomic system or a diatomic system, triatomic system or whatever is the state of the system and accordingly your gamma value is going to change. So, what we have discussed in this lecture is on a very important thermodynamic quantity heat capacity. We have seen by using statistical means how to get heat capacity at constant volume and once we have the information on heat capacity at constant volume, we can easily get heat capacity at constant pressure at least for the systems which form ideal system or ideal gases. If the system is non ideal then other terms will come in. For example, pi t I am taking you back to your chemical thermodynamics discussion pi t which is equal to del u del v at constant temperature. This pi t gives information on how the internal energy will change when the system is subjected to either compression or expansion at constant temperature. And if a system does not have intermolecular interaction and the temperature is constant then whether you compress the system or you expand the system there is no change in internal energy. That means in that case pi t is equal to 0 and this is the definition of an ideal gas. Why I discussed this? Because we are restricting our discussion to ideal systems and if you were to make some transformations in the in the expressions or if you have to bring some non-ideality into the system then this factor will also come in. However, if the gas is ideal then you follow this discussion and gamma for a monatomic perfect gas turns out to be 5 by 3. And the statistical means of obtaining this number we have discussed in this lecture. We have covered translational contributions, the remaining contributions we will cover in the next lecture. Thank you very much.