 Welcome back, we have been discussing mean energies and follow up of mean energies, we started discussing heat capacities. Basically, we discuss the applications of mean energies in further discussing heat capacities, because we connected heat capacity at constant volume with the temperature derivative of mean energy. Mean energy which we discussed a couple of lectures before is given by minus 1 by q m del q by del beta at constant volume, where this m mode it represents translational rotational vibrational electronic. And then we discussed in details how to connect heat capacity at constant volume with temperature derivative of mean energies. And we concluded that for a gas at room temperature, the translational contribution to constant volume heat capacity is 3 by 2 r. And then we discussed rotational contribution for a linear rotor is equal to r. And for a non-linear rotor it is 3 by 2 r. For each normal mode of vibration, the contribution is r provided all the modes are fully active. And based upon that discussion, we came up with this expression that is the constant volume molar heat capacity is represented by 1 by 2 into 3 plus nu r star plus 2 nu star r. And just to revisit that nu r star is equal to 2 when the rotor is linear and it is equal to 3 when the rotor is non-linear. And nu v star is equal to 1 for each normal mode of vibration. This equation gives a very good estimate of constant volume heat capacity as long as the temperature is much above its relevant or related characteristic rotational temperature or vibrational temperature. And if the temperature difference is very huge, for example, if the vibrational mode is not active that means, if the experimental temperature is much lower than the characteristic vibrational temperature, then you can even put nu v star equal to 0. After that we discussed this plot that initially 3 by 2 r contribution is there. And for a linear rotor as the temperature increases, the full r contribution is added. So, 3 by 2 plus 1 is 5 by 2 r and then you further increase the temperature. And when the temperature is much higher than the characteristic vibrational temperature, another r will add up. So, you have 7 by 2 r, further increase in temperature leads to dissociation of the molecule. And after dissociation, a diatomic molecule will form 2 atoms and each atom will have translational degree of freedom which is 3 by 2 r. So, 3 by 2 into 2 is equal to 3 r. This I have shown for a very simple diatomic molecule. Now, suppose if you are given an exercise to make a similar plot for a linear diatomic molecule or a non-linear triatomic molecule, you should be able to make similar plots. You have to take into account whether the molecule is linear or the molecule is nonlinear because the rotational contribution will be decided by that. And then after the dissociation again you have to explain your answer with justification. Now, let us discuss an application. Estimate the molar constant volume heat capacity of water vapor at 100 degree centigrade. Vibrational wave numbers are given which is some example, but this vibrational wave numbers you can see here. I have included that in table. Now, once the vibrational wave numbers are there, as I will discuss with you, you can calculate the vibrational temperature. Rotational constants are also given 27.9, 14.5 and 9.3 centimeter inverse. Now, if you carefully look at the problem statement, it says estimate it does not say calculate, it just an estimation that means an approximate number for constant volume heat capacity. And if I go back, you can use this expression to estimate. What you need to know is nu r star what should be the value of nu r star and what should be the value of nu v star ok. So, what we have is C V m is equal to 1 by 2 3 plus nu r star plus 2 nu v star multiplied by r. So, this is the constant volume heat capacity per mole. Now, first of all we need to now judge what will be the values for nu r star and what will be the values for nu v star. Let us talk about first vibration because generally we say that the vibrational energy levels are far separated. So, their contribution may be very less. First of all first step is to calculate the characteristic vibrational temperature. How do you calculate characteristic vibrational temperature? Use h c nu bar is equal to k theta v. We are interested in finding out the value of theta v. From the knowledge of Planck's constant speed of light, wave number and Boltzmann constant. Now, you can look at this table. The wave numbers first one 3656.7, second one 1594.8 and third one 3755.8. This is water. Water is a non-linear molecule. So, therefore, normal modes of vibration is 3n minus 6 which is 3 into 3 and is 3 9 minus 6 is equal to 3. And these 3 normal modes of vibration take place at 3656.7, 1594.8 and 3755.8. Now, by using this expression we can calculate 3 vibrational temperature. The characteristic vibrational temperatures by using this method turn out to be 5300, 2300 and 5400 Kelvin. These are roundabout figures. So, you see 5300 correspond to 3656.7. 5400 correspond to 3755.8. The higher the value of wave number, the higher is the vibrational temperature. But if you look at the lowest wave number 1594.8 is the lowest wave number. Corresponding to that the vibrational characteristic vibrational temperature is 2300 Kelvin and the experimental temperature is 373 Kelvin 100 degree centigrade. 2300 is much much higher than 373 Kelvin. So, I can write that the temperature which is 373 Kelvin is much much less than the vibrational temperature even for the normal mode which is the lowest in wave number. So, automatically then the wave numbers which are higher that those are ruled out. So, that means, vibrations are not excited at 373 Kelvin simply because the temperature which is 373 Kelvin is much much lower than that characteristic vibrational temperature. So, we do not need to worry about the vibrational contribution to constant molar volume heat capacity. Now that means, I can write nu v star I will put to be 0 because the vibrational are not active. Now we also have rotational constants given water non-linear molecule it can it is a triatomic molecule it can undergo rotational motion. And use h c b is equal to k theta r that means, this rotational temperature theta r is h c b upon k b is h cross by 4 pi c i these all we have discussed earlier. And now if you look at the rotational constants given 27.9 14.5 9.3 what I will do is I will first calculate the rotational temperature for highest rotational constant. Because if the characteristic rotational temperature corresponding to highest rotational constants is much lower than the experimental temperature then the other two rotational constants are automatically covered. So, the strategy here should be first you calculate corresponding to 27.9 when you calculate corresponding to 27.9 the rotational temperature comes to 40 k. So, that means, if I put 40 k experimental temperature is 373 k of course, this is much higher that means, T is much higher than theta r this is corresponding to the highest rotational constant. That means, rotations are fully active all three rotational constants will give a characteristic rotational temperature which is much lower than the experimental temperature that means, all these rotational modes are fully excited. So, that means, what value should I put mu r star it is a non-linear molecule I will put a value of 3. We are using this expression C V m is equal to 1 by 2 3 plus nu r star plus 2 nu V star into r. As we just discussed vibrations are not active at that temperature right the vibrational contribution will be 0 nu V star is equal to 0 nu r star we just discussed all three rotations or rotational motions around three axis is fully active. Therefore, this value nu r star will be equal to 3. So, therefore, what we have this will be C V m will be equal to 1 by 2 3 plus 3 r which is equal to 3 r r is gas constant. So, r is 8.3145 joules per Kelvin per mole once you put that number you are going to get a value close to 25 joules per Kelvin per mole. Translational contribution 3 by 2 r rotational contribution 3 by 2 r. So, 12.5 plus 12.5 is 25 joules per Kelvin per mole this is predicted. The experimental value is 26.1 joule per Kelvin per mole. So, there is a difference if you compare there is a difference of about 1.1 joule per Kelvin per mole which is not very large, but we should be able to assign some reason to this discrepancy. The discrepancy is probably due to deviations from perfect gas behavior. What we are considering here water molecule like a perfect gas water vapors, but water vapors are not actually like an ideal gas. So, therefore, this deviation from ideality could be one reason for this discrepancy in the theoretical as well as the experimental value. So, what is important here? It is important to know what value we should put for nu r star and what value we should put for nu v star and for that what we need to do is to know characteristic rotational temperature and characteristic vibrational temperature. So, that we can decide whether to put nu r star equal to 3 or 0 or 2 or and nu v star equal to 1 or 0 that is by the comparison of experimental temperature and characteristic temperature of that mode of motion. But remember that this expression can only be used to estimate the molar constant volume heat capacity. It is not an actual calculation it is simply an approximation it is simply an estimation of the value. Now, let us discuss a relevant numerical problem. So, that we can understand these derivations more importantly and more clearly consider a system with energy levels E j is equal to j times E and there are n molecules question part one question part a is show that if mean energy per molecule is a e then the temperature is given by this beta is equal to 1 over e log 1 plus 1 by a. Let us go step by step first we will solve this part what is given to us is that the energy levels are given by E j is equal to j times E and if the mean energy per molecule is a e then the temperature should be given by this expression of beta fine. Let us try to solve this first and other parts we will come to them later. So, what is given to us is E what is given to us is E j is equal to j times E that means q will be equal to summation j exponential minus beta E j which is equal to summation j exponential minus beta is equal to summation j exponential minus beta j times E because E j is equal to j E this is equal to when start putting now j from 0 onwards 0 1 2 3 4. So, 1 plus exponential minus beta E plus exponential minus 2 beta E. Plus exponential minus 3 beta E and so on which I can simplify to q is equal to 1 plus exponential minus beta E plus exponential minus beta E square plus. So, 1 this reminds you of uniform ladder of energy levels you remember harmonic oscillator when we discussed this is like a uniform ladder of energy level. And this is a series geometric progression series g p series and sum of the g p series is equal to 1 over 1 minus exponential minus beta. This is the expression for the molecular partition function, but what is our problem statement we have to show that this beta is given by 1 by E into log 1 plus 1 by A provided the mean energy per molecule is A times E. So, let us write down an expression for mean energy mean energy per molecule will be equal to U minus U 0 by E. This is mean energy and U minus U 0 is equal to minus n by q del q del beta at constant volume this is mean energy this is a U minus U 0 and there is 1 by n factor here already there. So, what we have is minus 1 by q del q del beta at constant volume. So, mean energy per molecule is equal to minus 1 by q minus 1 minus exponential minus beta E this is minus 1 by q I am using this into derivative of this q with respect to beta. This will be minus 1 over 1 minus exponential minus beta E square into minus exponential minus beta E into minus E ok. This and this can cancel and there are 4 negatives which become positive. So, you have E exponential minus beta E over 1 minus exponential minus beta E. Let us carry it forward mean energy per molecule what we are getting is equal to E times exponential minus beta E over 1 minus exponential minus beta E. This is what we just got this one. Now, let us further try to solve this mean energy will be equal to E times exponential I can write this as exponent E divided by exponential beta E minus 1 I multiply and divide the numerator and denominator by exponential beta E. So, I have come up now with this expression for mean energy. Let us go back to the problem statement now. Now, we are told that treat mean energy to be equal to A times E mean energy per molecule treat it equal to A times E. Let us do that I will write this to be equal to A times E. So, what I have now E and E cancel and I have exponential beta E minus 1 is equal to 1 over A. So, exponential beta E is equal to 1 plus 1 by A that means, beta E is equal to log 1 plus 1 by A. In other words now I have beta is equal to 1 by E log 1 plus 1 by A. We were asked to show this beta is equal to 1 by E log 1 plus 1 by A and that is what we have shown beta is equal to 1 by E into log 1 plus 1 by A. And this beta is equal to 1 by k T that means, the temperature whether you express in terms of T or you express in terms of beta we should be able to get this from the knowledge of the value of E which is the energy separation and the number A. So, therefore, as you might have noted that we started with the definition of partition function. Since we were given a scheme that E j is equal to j times E we use that and we found that the overall expression for the partition function was coming to be equal to sum of a GP. So, by using that we got an expression for partition function and by using that partition function we put that expression into expression for mean energy that is by using internal energy and then by mathematical manipulations we were able to get this expression for temperature. There are still some more parts of the question that is evaluate the temperature for a system in which the mean energy is E by taking E equal to 50 centimeter inverse. The second calculate the molecular partition function q for the system when the mean energy is A E and third issue that the entropy of the system is S over n k is equal to 1 plus A log 1 plus A minus A log A and evaluate this expression for mean energy E. So, there are 3 4 parts still remaining for this question, but if you look at what is requested what is asked is expression for molecular partition function we already got that. The second one is entropy that means, then you have to now think how to connect entropy with molecular partition function that is where our previous knowledge will be used when you connect entropy with the canonical partition function or molecular partition function, but we will be discussing all these in details in the next lecture. Thank you very much.