 Let us continue our discussion on solving numerical problems based on internal energy and entropy. Internal energy and entropy are directly connected to each other. Go back to concepts of chemical thermodynamics where you remember that d s which is equal to d q by t allows you to connect under constant volume conditions d u is equal to t d s. So, therefore, internal energy and entropy can be directly connected. In the previous tutorial session we talked about evaluating the molar entropy of n two level systems and we were also asked to plot the resulting expression. First let us discuss that we have already shown the dependence of entropy upon temperature or beta through this expression s by n k is equal to beta e divided by 1 plus exponential beta e then plus log 1 plus exponential minus beta e. And when you plot this look into the upper figures it is plotted as s by n k versus k t by epsilon. And you see when the temperature is increasing the value is starting from 0 and eventually rising towards a final value of log 2. So, this is how the entropy dependence upon temperature can be expressed. So, definitely we can exactly find out the value of entropy when temperature is approaching 0 and the second is when temperature is approaching infinity. Now I want you to take a look at the statement of the question that is given to us. Evaluate the molar entropy of n two level systems and plot the resulting expression we have plotted what is the entropy when two states are equally thermally accessible. So, if we are considering only two states two levels then at what temperature the two states are equally thermally accessible we have already discussed that when temperature approaches infinity. Remember that when we talked about the fractional population of a system which has only two states as the temperature approaches infinity the both the states are equally thermally accessible. We also derived a general conclusion at that time that whatever is the state under consideration when temperature approaches infinity all the states are equally thermally accessible. But let us talk about what happens when temperature approaches 0 when temperature approaches 0 then beta over k t beta is equal to 1 over k t then beta approaches infinity. But here there is a denominator infinity also and here minus infinity that means this 1 over exponential infinity this will become 0. So, log 1 log 1 is also 0 this is 0 because there is infinity in the denominator that means s approaches a value of 0 that is what you observe over here when temperature approaches 0 s approaches a value of 0. Now the second condition when the two states are equally thermally accessible that means we are talking about a temperature which is very very high approaching infinity. Let us work on that when temperature approaches infinity then beta which is 1 over k t approaches a value of 0. So, this factor the first one this becomes 0 and this exponential 0 this becomes 1. So, log 1 plus 1 then you have 2. So, when beta approaches 0 this first term is 0 and then log 2 s by n k is equal to log 2 that means s approaches a value of n k log 2. So, therefore, such an expression it allows us to express entropy as a function of temperature. We can talk very easily the situations when temperature approaches 0 and the temperature approaches a very high value at the intermediate temperatures one can use this expression and calculate the value of entropy. Now let us talk about another case consider a system which has doubly degenerate ground state non-degenerate first excited state at E 1 and a doubly degenerate second excited state at E 2 derive expressions for its internal energy and entropy. Let us try to understand this problem consider a system which has doubly degenerate ground state let us write that there is doubly degenerate ground state a non-degenerate first excited state let us say this is a non-degenerate first excited state and the energy here is E 1 and a doubly degenerate second excited state. Let us say there is a second excited state which is doubly degenerate and this is at E 2 this is E 0 E 0 is equal to 0 derive expressions for its internal energy and entropy whether you want to derive expression for internal energy or you want to derive expression for entropy we first need an expression for its partition function. Let us write down an expression for its partition function partition function is summation j g j exponential minus beta E j apply to the system q is equal to g 0 g 0 degeneracy of the ground state is 2 plus g 1 degeneracy of the first excited state is 1 into exponential minus beta E 1 and then we have degeneracy of the second excited state which is 2 exponential minus beta E 2 this is the expression that I have for the molecular partition function. Once I have this expression for the partition function now I can write an expression for internal energy remember u minus u 0 this is the internal energy is equal to minus n by q del q del beta at constant volume we already have an expression for q. So, therefore it becomes easier for us now to write an expression for internal energy u minus u 0 is equal to minus n by q q is 2 plus exponential minus beta E 1 plus 2 exponential minus beta E 2 into derivative of this with respect to beta this is going to be exponential minus beta E 1 into minus E 1 plus 2 times exponential minus beta E 2 into minus E 2. So, what I have now is u minus u 0 is equal to I can consume all the negatives with this negative and turn it into positive. So, what I have is n and then I have E 1 exponential minus beta E 1 plus 2 E 2 exponential minus beta E 2 in the denominator I have 2 plus exponential minus beta E 1 plus 2 exponential minus beta E 2 this is the expression for internal energy. You can calculate internal energy for this system as a function of temperature carefully note that in the previous example when we considered only two level systems q was different here we are taking n systems where the ground state is doubly degenerate first excited state is non-degenerate second excited state is doubly degenerate the q is getting modified and as a result of this modification your expressions for the internal energy are also getting modified therefore the temperature dependence of internal energy on temperature is going to be this complex function you can always talk about temperature approaching 0 and temperature approaching infinity as extreme cases. Now that we have the expression for internal energy you can also now derive an expression for heat capacity at constant volume from this expression because remember that c v is del u by del t at constant volume or is equal to minus 1 over k t square into del u by del beta at constant and you can have an expression of heat capacity as a function of beta or as a function of temperature which is also going to be relatively a complex function but the question that we are supposed to solve is to derive expressions for internal energy and entropy we have expression for internal energy ok what we derived just now is u minus u 0 is equal to what I had there was let us take a look back what we had we had this n times e 1 exponential this terms and over q let us take it to the next one this is equal to n then we have e 1 exponential minus beta e 1 plus e 2 exponential minus beta e 2 and then in the denominator we had the molecular partition function 2 plus exponential minus beta e 1 plus 2 exponential minus beta e 2 this is u minus u 0 u minus u 0 by t will be equal to n e 1 exponential minus beta e 1 plus e 2 exponential minus beta e 2 divided by t into 2 plus exponential minus beta e 1 plus 2 exponential minus beta e 2 and we can use the usual transformations that beta is equal to 1 over k t that means 1 over t is equal to k times beta I can use that. So, what I have is now u minus u 0 by t is equal to 1 over t I am writing k beta. So, I have n I have k into beta x inside I have e 1 exponential minus beta e 1 plus e 2 exponential minus beta e 2 and here I have is 2 plus exponential minus beta e 1 plus 2 into exponential minus beta e 2 keep a factor of 2 over here which we missed. So, I have this ok. So, we have u minus u 0 by t. So, s is equal to u minus u 0 by t which is n k beta inside I have e 1 exponential minus beta e 1 plus 2 e 2 exponential minus beta e 2 divided by 2 plus exponential minus beta e 1 plus 2 exponential minus beta e 2 plus n k log q n k log q q was 2 plus exponential minus beta e 1 plus 2 exponential minus beta e 2 n k n k is common. Therefore, the resulting expression is s by n k is equal to beta into e 1 exponential minus beta e 1 plus 2 e 2 exponential minus beta e 2 divided by the partition function that is 2 plus exponential minus beta e 1 plus 2 into exponential minus beta e 2 plus log 2 plus exponential minus beta e 1 plus twice exponential minus beta e 2. This was the expression to be derived. Look at the nature of the expression, dependence of entropy on temperature. It is a complex function, but if you know the values of temperature, if you know the values of the first state energy state and the second excited energy state, these numbers, if you know, then you can definitely calculate entropy at a given temperature. But remember that you can always calculate these values at extremes of temperature for the sake of simplicity. What I mean is, let us discuss what happens when temperature approaches a value of 0. When temperature approaches a value of 0, then beta which is equal to 1 over kT, when temperature approaches 0, this will approaches a value of infinity. So therefore, what you have is, here in the denominator you have infinity. So, first term is 0 and the second term, because it is infinity, exponential minus infinity, that means 1 over exponential infinity, this becomes 0, this becomes 0. So therefore, what you have is s over nk is actually approaching a value of log 2. Now, the second one is when temperature approaches a value of infinity, then beta is equal to 1 over kT, it will approach a value of 0. That means, what we have now, we have this beta is equal to 0, so that will set everything equal to 0, no contribution from first term. But second term, you will have 1 from here and 2 from here, because exponential minus 0 or exponential 0 that is 1. So that means, what we have now, s upon nk will approach a value of log 2 plus 1, 3, 3 plus 2, 5. Points to be noted here. The third law of thermodynamics says that every substance has a positive entropy which may become 0 at absolute 0 and it does become 0 for perfectly crystalline substances. It did not say that all the substances have to have a value of entropy equal to 0. Here is one such case, which you come across that when the temperature is approaching 0, your entropy is approaching a value of nk log 2. And when temperature is approaching infinity, then s is approaching a limiting value of nk log 5. So, when there is some amount of entropy even when the temperature is approaching 0, that means even at 0, there is some sort of disorder there. So, at absolute 0, that some sort of disorder can come from configurational disorder and that configurational disorder can give rise to some value of entropy even at absolute 0. The concept of entropy at absolute 0, which is called residual entropy we will discuss at a later stage. In today's lecture what we have done is we have combined the connection of internal energy with molecular partition function and the connection of entropy with internal energy and molecular partition function. So, this discussion allowed us to derive expressions for the variation of internal energy with temperature and variation of entropy with temperature. I hope that the numerical problems discussed in these two sessions have enabled you to appreciate the relationship of internal energy with molecular partition function and the relationship of entropy with molecular partition function in a more simpler and clearer manner. Thank you very much.