 Welcome back to the next lecture on Fundamentals of Statistical Thermodynamics. So far we have developed equations which connect internal energy with molecular partition function and entropy with molecular partition function. Both these thermodynamic quantities internal energy and entropy are very important in thermodynamics. Internal energy is the total energy of the system added up in all the forms and entropy is a measure of degree of disorder or ordered state of a system. Why both these thermodynamic quantities are important? Because work or heat which both are equivalent form of energy when these are done on the system or these are done by the system these affect the internal energy. Work is done at the cost of internal energy. Similarly entropy if there is a change in entropy that can be connected with the creation of more order or more disorder. Going back to the second law of thermodynamics where we discussed that the change in entropy of an isolated system is positive for a spontaneous process. When I said I meant in change in entropy of the system plus change in entropy of the surroundings has to be positive if the process is to be spontaneous. Therefore connecting internal energy with molecular partition function and entropy with molecular partition function is very important and it tells us how to determine these values by using spectroscopy. Today we will use the derived equations to further discuss their applications. Let us take up the first example in one of the previous lectures we talked about heat capacity. So again revisiting heat capacity is a very important thermodynamic quantity because it connects one thermodynamic signature or one thermodynamic quantity at one temperature to the same thermodynamic quantity at another temperature. So therefore it is a connector for the thermodynamic quantities at different temperature. Not only that a literal definition of heat capacity is the amount of heat required to change the temperature of a system by one degree one Kelvin. That means it also directly talks about the strength of the system. If we impose constant volume constraints then the definition that we need to use is C v is equal to del u by del t at constant volume. And in the previous one of the previous lectures we have shown that u is equal to u 0 plus 3 n by 2 beta. We also then derived this expression in terms of Boltzmann constant temperature etc. And then we equated it equal to 3 by 2 nRT. This is for the translational motion of a particle of a molecule or of a molecule. I am talking about a monatomic type of gas in three dimensions. Substituting the numbers we got C v equal to 12.5 joules per Kelvin per mole. Let us highlight some of the comments. This value agrees almost exactly with the experimental data on monatomic gases at normal pressures. This is theoretical prediction. Second is experimental data. And we have earlier talked about how to determine heat capacity experimentally. We need calorimeters. Under constant volume conditions bomb calorimeter. Under constant pressure conditions the other type of calorimeters which can work under constant pressure conditions. Second point that we need to be careful in this case about is that the value that we got 12.5 joules per Kelvin per mole. It applies only to monatomic gas, monatomic perfect gas. And for more complex molecules means for diatomic molecule for triatomic molecule. There are other modes of motion which we need to consider. Monatomic only translational degree of freedom is there. Electronic can also be there but that can be discussed later. But when you go to diatomic and we go to triatomic and more complex molecules then not only translational degree of freedom is there. There is rotational degree of freedom. There is vibrational degree of freedom in addition to electronic contributions. So, this is a very simple example. Now let us go to some other type of questions. The question is to show that heat capacity of n two level systems that is C v is given by the following expression. C v is equal to n k into beta e square exponential beta e divided by 1 plus exponential beta e square. And we know beta is equal to 1 by k t. So, that means the expression that we are going to derive can allow us to calculate the value of C v as a function of temperature and from the knowledge of energy separation. The given system is a two level system. And let us say separation between the ground state and the upper or the excited state is e. So, that is why C v basically by definition C v is equal to del u del t at constant volume. Now, either we can differentiate with respect to temperature and result can also be expressed in terms of beta, beta and temperature anyway are connected to each other. And sometimes instead of differentiating with respect to temperature it becomes easier to differentiate with respect to beta. So, you see d by d t I can write d by d t as d beta by d t into d by d beta mathematically that is allowed. And since beta is equal to 1 over k t therefore d beta over d t is going to be minus 1 over k t square. Therefore, in place of d beta by d t I can write minus 1 over k t square and then I can write differentiation with respect to beta. Why this kind of transformations? Why this kind of derivatives? Because sometimes it becomes easier if we work in terms of d by d beta. You can work in terms of d by d t and then convert into beta choice is yours. But as I said that sometimes if we choose an appropriate derivative it is easier to get results. Now, let us come back to the problem that we need to solve. We have this two level ground state and upper state. The first step will be to write partition function which will be 1 plus exponential minus beta. For ground state 1 for upper state the contribution is exponential minus beta. C v is a derivative of internal energy. So, therefore first of all let us talk about internal energy. u minus u 0 is equal to minus n by q del q del beta at constant volume q is 1 plus exponential minus beta i. So, therefore let us now work on this minus n by q minus n by q q is 1 plus exponential minus beta i and into derivative of q with respect to beta which is exponential minus beta e and into it is going to be minus e. So, what I have now is u minus u 0 is equal to minus minus is plus. So, it is n e exponential minus beta e divided by 1 plus exponential minus beta e. I can further simplify it I can multiply numerator and denominator and numerator with exponential plus beta e exponential plus beta e why because if I multiply by exponential plus beta e this upper term becomes 1. So, I have n e into 1 that I am not writing divided by exponential beta e plus 1 just to make the things little more simpler. We have now the expression for internal energy and if we take it is derivative with respect to temperature or with respect to beta then we will get the heat capacity. What we have is now u minus u 0 is equal to n e minus u 0. So, this is n e over exponential beta e plus 1 that is what we have c v is equal to del u del t at constant volume and we just showed that it is equal to minus 1 over k t square into I can write del u del beta at constant volume. We just showed that now let us act upon this. So, what we have c v is equal to minus 1 over k t square into let us take the derivative of internal energy with respect to beta n e is anyway there constant into minus 1 over exponential beta e plus 1 square into exponential beta e into e to the power of e. Let us further simplify it. So, c v is equal to what we have minus minus becomes plus. So, I have n then I have e square I have exponential beta e over k t square into exponential beta e over k t square into exponential beta e plus 1 whole square. I have combined the terms and I have n into e square into exponential beta e divided by k t square exponential beta e plus 1. We can now further simplify it by noting that beta is equal to 1 over k t. Now, I can write this as beta over t is equal to 1 over k t square. So, I want to simplify my result I have c v is equal to n instead of 1 by k t square let me write beta by t into e square exponential beta e divided by exponential beta e plus 1 square. Now, remember this is the derivative that I can write another term from here 1 over k t 1 over k t is equal to beta which means 1 over t is equal to k times beta. I can use it over here 1 over t is equal to k times beta that means what I have now is once I substitute over here c v is equal to 1 over t is k times beta. So, I have n k and another beta term will comes beta e square let me combine all this beta e square into exponential beta e by exponential beta e by exponential beta e plus 1 square. So, I have this result c v is equal to n k into beta e whole square exponential beta e divided by exponential beta e plus 1 this is what we were asked to show remember that beta is equal to 1 over k t. Now, we can briefly talk about the effect of temperature on this result we will use this expression that we derived and let us see what happens when temperature approach is 0 when temperature approach is 0 then beta is equal to 1 over k t that means beta will approach a value of infinity and you see you have infinity in the denominator also. So, c v will approach a value of 0 you have infinity in the denominator and when t approaches infinity then beta is 1 over k t that means beta approaches a value of 0 again you see here you have 0 then also that means c v will approach a very small value and if you work out actually take a maxima also it will show a maxima. So, in general what I want to say here is that depending upon the system given to you for example, here we were talking about n 2 level systems and we discussed that each level is a non degenerate and therefore, we draw an expression for the molecular partition function for that and then we developed further equations alright. This is showing that the heat capacity of n 2 level systems is given by some kind of expression which permits us to calculates c v as a function of temperature from the knowledge of energy levels or energy separation I hope this derivation is clear. Now, let us go to another type of question we have a similar system now n 2 level systems evaluate the molar entropy of n 2 level systems and then plot the resulting expression what is the entropy when 2 states are equally thermally accessible in the previous question we talked about internal energy and then we talked about connection of internal energy with partition function and eventually an expression for the heat capacity. Now we will be talking about entropy and entropy is connected to molecular partition function through this equation s is equal to u minus u naught by t plus n k log q and remember that u minus u naught is also connected to molecular partition function u minus u naught for the same system which is given to us n 2 level systems we have just discussed that u minus u naught is equal to we derived the expression which is n times e over n minus exponential beta e plus 1 this is the expression that we derived in the previous lecture let I can take you back to the previous lecture and this is where look at the this expression that u minus u naught we have derived this expression is equal to n times e divided by exponential beta e plus 1 let us make use of this now. So, that means u minus u 0 by t will be equal to what which will be n e over t times exponential beta e plus 1 and since we decided that instead of t let us talk in terms of beta that means beta is equal to 1 over k t or we use 1 over t is equal to k times beta we will use that that means u minus u 0 by t that means u minus u 0 by t is equal to 1 over t is k beta. So, n k k beta I am writing instead of 1 over t e over exponential beta e plus 1 this is u minus u 0 by t now we can write the expression for entropy s is equal to u minus u 0 by t is n k beta e divided by exponential beta e plus 1 plus n k log q plus n k log q q is what it is a tool I am going to write the expression for entropy s is equal to n k into beta e divided by exponential beta e plus 1 that is u minus u 0 by t plus n k log q n k log q n k n k is common. So, therefore, I can take it on the left hand side and then express in terms of s by n k once we do that what we have is the following expression s by n k is equal to beta e over 1 plus exponential beta e plus log 1 plus exponential minus beta e that is what I was saying that this n k and n k is common and you can take it on the other side and eventually the expression that you get is s upon n k is equal to beta e over 1 plus exponential beta e plus log into 1 plus exponential minus beta e we have now entropy as a function of temperature it is not a linear or very simple function you see beta 1 over t is appearing here 1 over t is appearing here 1 over t is appearing here. So, therefore, we can easily talk about in the extreme limits that is when temperature approach is 0 or when temperature approach is infinity, but remember that the expression that we have got here is for n two level systems a ground state and an excited state which is separated by energy e. We have now connected the entropy of such a system with temperature energy separation by this expression we need to further discuss the variation of entropy with temperature in this case in the limits of t approaching 0 and t approaching infinity, but that we will do in the next lecture. Thank you very much.