 Welcome back to the next lecture of statistical thermodynamics and today we are going to discuss mean energies. By now we have connected molecular partition function or canonical partition function with almost all thermodynamic quantities. And now we will start discussing further meaning of derivable properties. For example, today's discussion will be on mean energies. So, when we talk about energies total energy added up in all the forms we are talking about internal energy. So, is this the mean energy which can be connected to internal energy that will be the topic of discussion let us find it. Before we switch over to mean energies let us have a recap of the overall partition function. You remember that from time to time I have discussed that partition function due to different contributions is multiplicative. That is the overall partition function will be the product of partition functions of each contribution. For example, let us take a look at what is presented over here. The overall molecular partition function will be equal to if I just write the overall partition function will be product of translational, rotational, vibrational, electronic and if you can think of any other contribution. Out of these let us take 1 by 1 the translational contribution to the partition function is given by V by lambda cube, where V is the volume of the container and lambda is the thermal wavelength. Rotational contribution will depend upon whether the rotor is a linear molecule or it is a non-linear molecule. For simplicity let us now consider a linear molecule. For that this is the expression k t by sigma h c b, k is Boltzmann constant, t is absolute temperature, sigma is a symmetry number, Planck's constant, speed of light and beta represents temperature that is 1 by k t. A small correction read this as b k t over sigma h c b. The vibrational contribution vibrational contribution will be given by 1 divided by 1 minus exponential minus beta h c nu bar. Here again each term has its own meaning, beta is 1 over k t h is Planck's constant and this nu bar is the wave number where that particular vibration takes place. Finally, this electronic contribution usually it is equal to degeneracy of the ground state. So, therefore, the overall expression for the molecular partition function is G e that is the ground state term into V upon lambda cube into 1 upon sigma h c beta b which in other words is k t by sigma h c b into 1 over 1 minus exponential minus beta h c nu bar. Now, you will remember that whenever we derived these expressions specifically rotational and vibrational, we brought in the concept of characteristic rotational temperature or characteristic vibrational temperature. Because in deriving these relations we made some assumptions. So, therefore, the integrated form of the molecular partition function is applicable only if the experimental temperature is higher than the characteristic rotational temperature in rotational constant case and characteristic vibrational temperature in vibrational partition function case. So, therefore, we can also write the molecular partition function in terms of rotational temperature or vibrational temperature. Bringing in that the overall partition function now becomes G e that is the ground term into V by lambda cube V is the volume lambda is thermal wavelength and the rotational contribution now is represented in terms of rotational temperature and the vibrational contribution is represented in terms of vibrational temperature. So, it is up to you whether you want to use this form or you want to use this form they are equal. Now, read the comment carefully what is written over here overall partition functions obtained in this way are approximate because they assume that the rotational levels are very close and that the vibrational levels are harmonic further these approximations are avoided by using the energy levels identified spectroscopically and evaluating sums explicitly which is not that easy by evaluating the sum explicitly that means, then you have to know how many terms you have to add and the overall partition function which we just described written in this way is actually approximate and the reason for this approximation are given in these comments which we just discussed. Now, let us start discussing about what do we understand by mean energies internal energy relative to its value at absolute 0 is given by minus n by q del q by del beta at constant 1 this is the internal energy. What is internal energy? Internal energy is the energy possessed by a molecule or on a system added up in all the forms. So, mean energy energy per molecule how do we get this u minus u 0 is basically total energy E is equal to minus n by q del q del beta at constant volume and if I divide E by n this E is the total energy possessed by n molecules. So, in order to obtain mean energy which I will define as mean energy will be equal to minus 1 by q del q del beta at constant volume. We can further tag it mean energy for a particular mode is equal to minus 1 by q for that particular mode into del q beta at constant volume where m can be translational it can be rotational it can be vibrational it can be electronic or any other that you are considering. So, therefore, mean energy can be obtained from the knowledge of molecular partition function by using this expression. Once we have established this let us now switch over to different modes as we just said m is translational or rotational or vibrational or electronic. So, 1 by 1 we will substitute the expression and see what the new expression that we get mean translational energy ok. So, mean translational energy is equal to minus 1 over q translational into del q translational del beta at constant volume. Now, let us consider q translational in one dimensional which is equal to x upon lambda it is in one dimensional movement x is the length of the container lambda is thermal wavelength right lambda is equal to beta h square over 2 pi m square root. Remember that we are talking here about one dimensional movement translational movement of the molecule or of the particle. Now, let us substitute. So, what do we have E t is minus 1 by q minus 1 by q means it is lambda divided by x this is minus 1 by q from here into del del beta of q is x upon lambda at constant volume. So, minus lambda by x into x is out into minus 1 over lambda square into del lambda del beta constant volume right lambda raise to power minus 1 which is minus 1 over lambda square into del lambda del beta at constant volume. So, therefore, let us write now translational mean energy is it is 1 by lambda into delta lambda by delta beta at constant volume let us apply to this will be 1 by 2 into beta h square 2 pi m minus 1 by 2 into h square over 2 pi m this is what is the derivative of lambda with respect to beta. Now, I will play a trick over here I will multiply by beta and divide by beta no harm done and by doing. So, what I am getting now is 1 upon 2 beta into 1 upon lambda into now if you combine this 2 this is beta h square by 2 pi m square root when you combine these 2 which is lambda. So, basically this is lambda this into this is beta h square by 2 pi m square root which is lambda. So, your answer is 1 by 2 k t therefore, for 1 dimensional system of length x the mean translational energy is given by half k t and if you use it for 3 dimension then q is equal to V upon lambda cube and in a similar manner you can derive an expression for 3 dimensional system of volume V in that case the mean translational energy will be 3 by 2 k t details we have discussed here for 1 dimensional system you can yourself then work out for a 3 dimensional system. So, we are discussing that for a molecule free to move in 3 dimensions that is we are extending our discussion to 3 dimensions in that case mean translational energy is equal to 3 by 2 k t and you remember that these results were the same or the same as we that obtained from classical equipartition theorem. Furthermore look at the comment the fact that the mean energy is independent of the size of the container 3 by 2 k t size is not there is consistent with the thermodynamic result that the internal energy of a perfect gas is independent of its volume at a given temperature. I hope you got it that since the mean energy does not contain volume term therefore the internal energy of a perfect gas is independent of its volume these results are consistent. After having discussed the mean translational energy now let us switch over to mean rotational energy rotation of a molecule and as we discussed earlier the rotor can be linear and the rotor can be non-linear. And we already know what is the expression for the mean energy, but in order to use any expression the first thing here what you require is an expression for the partition function. So, let us use the expression for partition function rotational partition function is given by summation j equal to 0 to infinity 2 j plus 1 is the degeneracy term exponential minus beta h C b j into j plus 1 where j is the rotational quantum number. Once again the rotational partition function is given by summation j starts from 0 to infinity 2 j plus 1 which is the ground state degeneracy which is the degeneracy of the system exponential minus beta h C b j into j plus 1 and j can vary from 0 towards infinity ok. Let us expand it q r substitute j is equal to 0 j is equal to 0 means this is 1 and j is equal to 0 exponential 0 is 1. So, first term is 1 second when j is equal to 1 2 in plus 1 is 3 exponential 1 plus 1 2 that means minus 2 beta h C b this is for j equal to 1. Now put j equal to 2 2 into 2 4 plus 1 5 exponential j is equal to 2. So, 2 into 3 6 6 beta h C b plus so on. So, we are trying to express here the rotational partition function in explicit summation mode, but then there has to be some point where this summation ends whether you use high temperature result integration result that depends upon rotational temperature. Now suppose the rotational temperature is not very high and the actual experimental temperature is less than the rotational temperature or close to the rotational temperature then you cannot use the integration form. You cannot use the approximation result we have to go by explicit summation. So, that means now q r which is represented here has to be used if the temperature if the experimental temperature is not very high. So, let us use what we have mean rotational energy will be equal to minus 1 by q r del q r del beta at constant volume and we need the expression for q r. How we get q r? As we discussed in the previous slide q r is equal to summation j equal to 0 to infinity 2 j plus 1 exponential minus beta h C b j into j plus 1 ok. So, the q r as we just discussed it is when j is equal to 0 you have 1 plus when j is equal to 1 it is 3 exponential 1 plus 1 2 minus 2 beta h C b plus when j is equal to 2 this is 5 exponential 2 into 3 is 6 beta h C b plus you keep on going and the mean energy is minus 1 by q r I will put minus and I will put q r 1 plus 3 exponential minus 2 beta h C b plus 5 exponential minus 6 beta h C b plus keep on doing and we have to take derivative of q r. So, what is it going to be first is 0 then 3 exponential minus 2 beta h C b into what I am going to get minus 2 h C b plus 5 into exponential minus 6 beta h C b into minus 6 h C b plus so on. Now, you see that the negative term in the numerator is in each so this we can all take it out. So, mean energy will be equal to I can take h C b out h C b out and inside what we have is 6 into exponential minus 2 beta h C b plus 30 into exponential minus 6 beta h C b plus so on and divided by why have 1 plus 3 exponential minus 2 beta h C b plus 5 exponential minus 6 beta h C b this is the same expression in this expression beta represents temperature h is constant C is constant b is constant the only variable here is beta beta is 1 by temperature. So, therefore, it will be interesting to note interesting to see how this mean rotational energy is varies when the temperature is increased. So, that means, in that case you have to plot a figure in which e r is on the y axis and temperature is on the horizontal axis. Let us first consider a general temperature beta is 1 by k t. So, you plot e divided by h C b no problem because h C b is a constant here t divided by theta r theta r is a constant. So, you are essentially showing the temperature dependence and you see the variation when the temperature is very low the contribution is almost 0. It is only when the temperature star t by theta r is higher than 1 then you see how this mean rotational energy contribution increases. So, when the temperature is not very high when the temperature is moderate then you have to use this expression b is rotational constant beta is temperature 1 by k t expressed in that. So, therefore, this expression allows you to evaluate mean rotational energy as a function of temperature even when the experimental temperature is not high. That means, if t by theta r is less than 1 you see how the temperature is how this contribution is lower at low value of t by theta r and then it sharply starts increasing eventually it will take up a constant value. So, what we have discussed so far is how to obtain mean energies from the molecular partition functions. We started with the definition of internal energy and then eventually we reached an expression for mean energy and we found out that in order to obtain mean energies first of all we should be able to write an expression for the molecular partition function and then take its derivative with respect to beta at constant 1. For translational it was easy 1 by 2 k t or 3 by 2 k t, but for rotational you see it can be a complex equation, but this complex equation is valid only when the temperature is low. When the temperature is high much larger than the characteristic rotational temperature then we can use the high temperature result, but that we will discuss in the next lecture. Thank you very much.