 Welcome back. Today also we are going to solve some numerical problems. After having established relationship of almost all thermodynamic quantities with the molecular partition function, now we should be able to apply those developed concepts in solving the problems and addressing the problems. Let us continue with different types of problems and see that how the derived equations, the derived concepts can find their applications. The first question that we would like to solve today is on heat capacity. The question is the heat capacity of a gas determines the speed of sound in it through the following formula. The formula is C s is equal to gamma R t by m raised to the power 1 by 2. C s is the speed of sound and the heat capacity determines that means in the given expression there should be something which is connected to heat capacity and that is given to us that gamma is the ratio of C p by C v, m is the molar mass of the gas, deduce an expression for the speed of sound in a perfect gas of a diatomic, linear triatomic and non-linear triatomic molecules at high temperatures. And it is also given to us that consider only translation and rotation activity. The question is estimate the speed of sound in air at 25 degrees centigrade. Let us first of all try to understand the question. According to the given question to us the speed of sound depends upon the heat capacity of a gas. That means you are talking about what is the speed of sound when it travels through a gas. The heat capacity in the given question is represented as ratio of C p and C v and we need to deal with diatomic, linear triatomic and non-linear triatomic molecules at high temperatures. Then it says estimate the speed of sound in air at 25 degrees centigrade. Let us see how to proceed. This information that C s equals gamma R t by m square root is given to us. We also know that C p minus C v or C p m minus C v m is equal to R and this is for an ideal gas. Therefore, the disclaimer here is that we are dealing with the ideal gas. The ratio of C p to C v is called gamma which is also given to you and from here we have to now proceed. Now let us go back to our discussion when we connected heat capacity with mean energy and then from there we came up with an approximation. Why I am saying an approximation is that the equation which is written over here C v m is equal to 1 by 2 into 3 by nu R star plus 2 nu v star. By using this expression you can estimate the value of C v. Carefully see what I am saying I said you can estimate the value of C v. I am not saying that you can calculate exactly calculate because there are approximations involved here. For a gas translational contribution is always there and in three dimensions that is why this 3 by 2 R there is an R over here also. This 3 by 2 R contribution is always going to be there for a gas at room temperature. Now then as the temperature increases the rotational contribution also comes in. So, therefore, we have nu R star and nu R star is equal to 2 for linear molecule and 3 for non-linear molecules we have discussed it earlier. Then we have 2 nu v star and this nu v star is equal to either 1 or 0. If the vibrational modes are not active we are taking the value as 0. If the vibrational modes are fully active then we are taking the value of 1. We are not taking any in between value over here that is why I said you can estimate heat capacity by using such a formula. So, for vibrational modes remember that this is for one normal mode of vibration. If there is only one normal mode of vibration nu v star is 1 if there are 2, 3, 4 you have to substitute that number. Now let us proceed. So, we have discussed that the constant molar volume heat capacity can be given by 1 by 2 into 3 plus nu R star plus nu 2 nu v star into gas constant. And we just discussed that for diatomic molecule nu R star is equal to 2 because diatomic molecule is linear. For example, A, B and given to us let us look at the statement. What it is given to us? Only translation and rotation active that means the temperature is not that significantly high enough for the vibrations to become active. Fine with this knowledge translation is active we will keep 3 here. Rotation is active so nu R since we are dealing with diatomic molecule linear molecule then nu R star is equal to 2. So, then 3, 3 plus 2 is 5 5 by 2 R nu v star is 0 because vibrational modes are not active. So, 3 plus 2 5 5 by 2 R so CVM is 5 by 2 R and as we just discussed that CPM minus CVM is equal to R right. I am not writing N over here because we are talking about differences in the molar heat capacities. So, using this equation CPM is equal to CVM plus R. So, if CVM is 5 by 2 R then CPM is going to be 7 by 2 R. Gamma which is the ratio of CP and CV then becomes 7 by 5 7 by 5 right 7 by 2 divided by 5 by 2 which is 7 by 5 which is 1.40. Therefore, substituting in this equation the value of gamma you get C S is equal to 1.40 RT by M square root. Now, you can evaluate this at any temperature and you can estimate the speed of sound alright. I hope it is clear. Now, the second part of the question was for a linear triatomic molecule, linear triatomic molecule something like this A B C carbon dioxide. As long as the system is linear the molecule is linear nu R star is to be used 2. That means, CVM remains same 5 by 2 R. How 5 by 2 R? 3 plus 2. CPM also remains 7 by 2. Gamma remains 1.4 and the expression for the speed of sound also remains the same. Only what is differing here is compared to the previous one your molar mass is going to be different. When you talk about diatomic molecule versus triatomic molecule the molar mass is going to be different and by these expressions you can now estimate the value of speed of sound ok. Now, let us talk about linear triatomic molecule. Linear triatomic we have already we have already done I will modify it for non-linear triatomic molecule non-linear. Non-linear one of the example is HOH type. When non-linear you are talking about nu R star is equal to 3. That means, this is going to be 1 by 2 into 3 plus 3 nu V star is 0. So, therefore, you can put R this is equal to 3 R. CVM is 3 R obviously, then CPM means add another R to 3 R it becomes 4 R and gamma which is the ratio of CP and CV that becomes 4 R divided by 3 R which is equal to 4 by 3. Substitute gamma is equal to 4 by 3 over here. So, you have 4 RT by 3 M square root. Now, you can substitute the numbers and estimate the value of speed of sound in this medium the medium here is gas. So, to solve these kind of problems what is important is to remember this expression which can be used to estimate the value of CV. This value will depend upon the temperature. If temperature is high enough for all the rotational modes to be active you substitute the value of nu R star. And if the temperature is not high enough for the vibrational modes to be active then nu V star is 0, but if they are active then you put the number depending upon how many normal modes of vibrations are fully active. So, I hope that with this example it is clear that how to obtain or obtain an estimate of heat capacity at constant volume. Now, let us take another example a different type of example the question here is what is the ratio of the number of molecules with V equal to 1 and J is equal to 2 to those with V equal to 2 and J is equal to 6 for nitrogen at 1000 Kelvin. For nitrogen the frequency is given 7.06 into 10 to the power 13 per second and the moment of inertia is given. The question is to calculate the ratio of the number of molecules with a different set of vibrational or rotational quantum numbers. If it were only belonging to one set of quantum numbers things would have been very very easy, but here you have to deal with the molecules in which the vibrational quantum number V is equal to 1 and rotational quantum number J is equal to 2. And for the other state the vibrational quantum number V is equal to 2 and the rotational quantum number J is equal to 6. How to now address this kind of problems? When we talk about how to evaluate the number of molecules we have an expression n i upon n is equal to exponential minus beta e i over q. Now remember that if there are more than one energy states if there are more than one energy states corresponding to a particular level then that particular level is 2 fold, 3 fold or in general g fold degenerate. In that case g i will come here, but to address this kind of problem first of all we need to know energy levels and we need to know degeneracies. These two things are required to solve this problem. One is energy levels and second is the degeneracy. The question given to us talks about vibrational and rotational levels. Therefore, we will be dealing with the vibrational energy and rotational energy. Vibrational energy is given by e V is equal to V plus half h nu or you can also write V plus half h c nu bar where V can take value from 0, 1, 2 etc. And the rotational energy is given by h cross by 8 pi square i into j into j plus 1. When we discussed the rotational partition function we talked in terms of rotational constant b a b c where b is h cross by 4 pi c i and the total energy then we calculated in terms of b, but in terms of moment of inertia this is the expression. You can get from that by substituting b is equal to h cross by 4 c i. Now the first step is for this set of quantum numbers V is equal to 1 and j is equal to 2. We will add these and this. We will add for e V and we add for e j. For first set V is equal to 1 so therefore 1 1 plus 1 by 2 into h nu we are given information in terms of the frequency plus h square by 8 pi square i. So h square by 8 pi square i j into j plus 1 j is 2 so 2 into 3. When you solve everything you get 7.04 into 10 to the power minus 20 joules this is the energy. Similarly you repeat this exercise for V equal to 2 and j equal to 6. So you put V equal to 2 and j equal to 6. Do the calculations and you will get then now the calculate the energy is little higher. This is 1.186 into 10 raise to the power minus 19 joule. We have the energies. Now for the set of quantum numbers given to us then after having solved this and this I just said that n i is equal to g i exponential minus beta e i over q and similarly I can write n j is equal to g j exponential minus beta e j over q. When you take the ratio n i upon n j then you get n i upon n j is equal to g i upon g j into exponential minus e i minus e j by k t. We have to calculate n i upon n j that means we need to have information on g i we need to have information on g j. This difference e i minus e j that is the difference between these numbers 7.04 into 10 raise to the power minus 20 and 1.186 into 10 raise to the power minus 19. Now the degeneracy for rotational levels for rotors g j is equal to 2 j plus 1. So, in the first case v is equal to 1 j is equal to 2. So, 2 j plus 1 is 2 into 2 plus 1 for the second one v is equal to 2 and j is equal to 6 ok. So, 2 j plus 1 what we have is 5 this is 2 plus 1 is 3 into 2 6 6 divided by 6 plus 1 is 7 into 2 is 4 14. So, 6 by 14 into exponential minus delta e delta e is difference between these two numbers and it turns out to be 12.6. What was the question given to us? What is the ratio of the number of molecules with v equal to 1 and j is equal to 2 to those with v equal to 2 and j is equal to 6 for nitrogen at 1000 Kelvin. Since these quantum numbers belong to different states of motion or modes of motion. Therefore, the problem becomes little complex you need to involve degeneracy and you need to include the differences in their energy states. First you have to calculate the energies which they are occupying for v equal to 1 and j is equal to 2 it was easy to compute it was easy to calculate that it is 7.04 into 10 raise to the power minus 20. Similarly, for the second it came to 1.186 into 10 raise to the power minus 19. The degeneracies were also possible to evaluate because we know that for rotational levels the degeneracy is 2 j plus 1. Therefore, by writing such a number what we have is we have this ratio of number of molecules with v equal to 1 and j is equal to 2 to those with v equal to 2 and j equal to 6 the answer is 12.6. So, you might have noted in today's discussion that when you are dealing with such type of questions you need to know the rotational constants in simple terms for example, b is equal to h cross by 4 pi c i and if I use i is equal to mu r square then even mu I can express in terms of masses. So, therefore, do not get confused if these expressions are written in terms of the moment of inertia or in terms of the masses molar masses. So, I hope that solution of these kind of problems has brought little more clarity on how to use the derived equations on molecular partition function in solving the numerical problems of this type. We are going to solve some more numerical problems, but those we will do in the next lecture. Thank you very much.