 Welcome back. Let us continue our discussion on heat capacity. In the previous lecture we have discussed heat capacity at constant volume and heat capacity at constant pressure for a monatomic perfect gas. Monatomic perfect gas will have translational contribution, rotational and vibrational contributions are not there. Now in case the system which is beyond the monatomic system there are rotational contributions and if there are vibrational contribution then how to account for that that we are going to discuss in this lecture. Now you consider rotation of a molecule. There are two type of rotors we have discussed one is a linear rotor like carbon monoxide, carbon dioxide or a non-linear rotor example is water molecule. We have to again use the expressions which are for the mean energy CVM is equal to NA tell E I will put R here because we are talking rotational contribution at constant volume. So, therefore, I need the expression for the mean rotational energy. Same way we can use directly equipartition value that is when the temperature is very very high. High temperature means you remember that we have discussed that H C B is equal to K theta R and this theta R is the rotational temperature. When temperature is much higher than the rotational temperature then we have already discussed that the mean rotational energy is equal to K T which matches with the equipartition value. So, let us use CVM rotational is NA del the value is K T del T at constant volume which is equal to K times NA which is equal to R. This is an easy way of directly putting the equipartition value otherwise E R which is equal to U minus U 0 by N can be calculated from U minus U 0 is equal to minus N by Q del Q R del beta at constant volume. And if the temperature is high much higher than the rotational temperature then for a linear rotor your sigma which is the symmetry number will factor into Q R is equal to 1 upon sigma H C beta and once you substitute this Q R into this expression you are going to get this equal to K T for a linear molecule. So, therefore, for a linear rotor linear rotor some examples as I just said CO HCl carbon dioxide nitrogen oxide these kind of systems the CVM R will be equal to R provided the temperature is much higher than the rotational temperature. However, for a non-linear rotor you can use the similar exercise you can do similar exercise or if the temperature is very very high then you can use the equipartition value and show that the constant volume heat capacity for a non-linear rotor is equal to 3 by 2 R. The mean energy rotational mean rotational energy is 3 by 2 K T and the constant volume heat capacity is 3 by 2 R. So, whenever you are talking about rotational contribution you have to worry about linear rotor versus non-linear rotor. For a linear rotor the value will be equal to R and for a non-linear rotor it will be 3 by 2 R provided the temperature is much higher than the rotational temperature alright. Now, if the temperature is not high then how do we deal with it? If the temperature is not high if the temperature is lesser than the characteristic rotational temperature then the expression for mean energy is this cumbersome expression which we have discussed earlier in the previous lectures. You have to take temperature derivative of this and then the expression for heat capacity will be further cumbersome. Remember that this expression has been obtained based upon the rotational partition function by direct summation method. Once you substitute this expression here you will get a complicated kind of equation for CVM and that equation when you plot then CVM against R versus temperature T by theta R the limiting value when the temperature is extremely high when the temperature is so high that temperature is much higher than theta R then this value is approaching a value this CVM by R is approaching a value of 1 that is CVM is equal to R. However when the temperature is not high you see here when the temperature is close to 0 there is no rotation the value is mean energy rotational energy is 0 and in the intermediate temperature you see how this heat capacity rises in the region when the temperature experimental temperature is less than the rotational temperature. I hope it is clear that when the temperature is not high in that case your heat capacity at constant volume will not be equal to fully R value but it will be somewhere in between 0 and R this is how you get alright. This is rotational contribution now let us talk about the vibrational contribution. We have discussed from time to time that the energy separations are in the following order translational then rotational then vibrational then electron in the increasing order. Translational contributions as we just discussed some time ago is always there for a gas at room temperature. Rotational contribution will also be there provided the system is more than monatomic and if the system is diatomic triatomic polyatomic in addition to rotational contribution vibrational contributions are also there. Now you look at the comment it is very unusual for the vibrations to be so highly excited that equipartition is valid and it is more appropriate to use the full expression for the vibrational heat capacity by using the following expression for mean vibrational energy. What is the meaning of this comment that the vibrational energy separations are so far apart under normal conditions of temperature that equipartition will not be valid that value will not be valid and therefore it is more appropriate to actually use this direct expression mean vibrational energy expression for which we have discussed in one of the previous lectures is equal to h c nu bar divided by exponential beta h c nu bar minus 1 and we want c v m vibrational is equal to n a into del we need to take temperature derivative of this and once again the derivation or derivative of this with respect to temperature or with respect to beta whatever expression you want to use now you see here your expression is in terms of beta. So, therefore it will be a good idea to use the other form of c v in which the derivation derivative is with respect to beta. So, when you take a derivative of this with respect to temperature or with respect to beta eventually you come up with c v m is equal to r f square and in this case m represents vibration this m. In fact, a correct notation for this will be c v m vibration here this m will mean molar ok. So, therefore you can choose whatever way you want to show and what is this f f is theta v by t square exponential minus theta v by 2 t divided by 1 minus exponential minus theta v by t 2 square some cumbersome kind of equation. What is the equipartition number equipartition number is complete k t mean vibrational energy. If it is complete k t that means the value of c v m full value is going to be r. Here also in this case if you plot c v m by r versus t when the temperature is close to 0 there is no vibration. Therefore, heat capacity is approximately 0 and as the temperature increases you see it reaches towards a full value of r which is predicted by the equipartition theorem. I hope it is clear c v m vibrational is equal to r f square this is what we discussed in the previous slide here c v m is r f square and I am using this over here where f is given by this and when you plot c v m by r against t t by theta v or t you get this kind of behavior. That means when the temperature is not very high the contribution is also not very high. So, what we have done we have now discussed translational we have discussed rotational we have discussed vibrational translational and rotational. Translational the value was 3 by 2 r this is contribution to heat capacity c v m we are talking about c v m then we talked about rotational it was either r or 3 by 2 r depending upon whether the system is a linear rotor or it is a non-linear rotor then vibrational this is equal to complete r, but this is only possible if all modes are fully active by fully active means the temperature is much above the characteristic rotational temperature or characteristic vibrational temperature. Translational contribution is anyways always there for a gas. So, what about if the temperature is not high then you have to see what the temperature is and the corresponding value of c v can be taken from this kind of plots. Now there is one more contribution which one can talk about is electronic contribution to the constant volume heat capacity. Once again if the vibrational energy levels are so far apart that even under normal conditions of temperature if the vibrational contribution to heat capacity is very very small then the electronic which are further apart some up further apart from each other the electronic contribution is negligible. So, therefore in our normal discussion we will include translational contribution to constant volume heat capacity we will include rotational contribution we will include vibrational contribution. So, translational I will write again here c v m translational is 3 by 2 r fully active c v m rotational is equal to r or 3 by 2 r. C v m vibrational is equal to r for each mode and this is possible if all modes are fully active mode. So, therefore if you are dealing with a monatomic system let us say monatomic perfect gas we say that the constant volume heat capacity is 3 by 2 r why because when you have monatomic system the only degree of freedom is translation and then if the only degree of freedom is translation we stop here 3 by 2 r. And now if you are dealing with the diatomic system linear let us say then another contribution if rotation is fully active then another r adds up. So, 3 by 2 r plus r is 5 by 2 r right. So, this is how the contributions to constant volume heat capacity will increase. So, therefore now if you carefully look at this figure let us try to interpret this figure. Rotational contribution is always there and what you are doing is you are plotting c v m by r that means 3 by 2 3 by 2 r r you are consumed here. So, 3 by 2 is always there c v m by r is equal to 3 by 2 is always there. So, when you start increasing the temperature you are plotting against temperature when you start increasing the temperature rotational contribution will enter in and when the temperature is much above theta r another r from here we are talking about linear system here another r will enter in. So, 3 by 2 plus r is 5 by 2 r now when you further increase the temperature we are here at 5 by 2 r we are here when you further increase the temperature another vibrational contribution will enter in another up to another value of r. So, 5 by 2 plus r is 7 by 2 r I hope this is clear we started with 3 by 2 r because translational contribution is always there increase the temperature rotational contribution will in will come in and a full r when the temperature is much higher than rotational temperature will enter in. So, taking 3 by 2 r to 5 by 2 r and when you further increase the temperature vibrational contributions will enter in another r will add up and you have 7 by 2 r and when you capture all these in the form of an equation we can write this form of equation that C v if I further modify C v m is equal to 1 by 2 into 3 plus nu r star plus 2 nu v star into r. How to read this? This star which is represented here is star means fully active mode that means this equation when the temperature depending upon the temperature you can decide if all the modes are fully active then nu r star can be 2 or 3 let me write down nu r star 2 4. Linear and 3 for non-linear rotor and nu v star is equal to 1 if the mode is fully active. So, since you are having 2 here and 2 outside. So, it is actually giving you a contribution of r therefore, this equation is a good equation to estimate the value of C v m I am not saying to calculate the value of C v m there is a difference between estimation and calculation estimate is an approximation. Why I am saying approximation because you can use this expression only if you know whether nu r star is equal to 1 or 0 if you know nu v star is equal to 1 or 0 because there is no intermediate value which you can substitute here from a given temperature. So, what you need to do is you need to compare T and theta and decide here you can see theta r theta v it is beyond theta v or theta r that full r contribution enters into heat capacity at constant volume all right. So, what we have discussed you start with 3 by 2 1 r adds up for a linear rotor. So, 5 by 2 r another r adds up for vibration if the temperature is high enough 7 by 2 r now you keep on heating you have a diatomic molecule you keep on heating vibrations are fully set up and what happens when you further increase the temperature the thermal energy is. So, high that the bond breaks and that is dissociation when the dissociation takes place it is like a phase transition like the heat capacity at phase transition is infinite. So, when the dissociation takes place the temperature is used up in dissociating the molecule rather than increasing the temperature therefore, heat capacity becomes infinite what you have you have x 2 a diatomic molecule is dissociating into 2 x that means, what you are getting here in the end is 2 times atoms and atom will only have translational degree of freedom translational degree of freedom means only 3 by 2 r that is what is shown over here 3 by 2 into 2 which is 3 r that right. So, this is 3 r twice into contribution due to atoms to atoms translational contribution is 3 by 2 r. So, 2 into 3 by 2 r is equal to 3 r. So, this figure explains how constant volume heat capacity changes as a function of temperature for a system which is more than a monatomic system. The expression which is mentioned over here as I mentioned can be approximately used to estimate the value of c v m that means, we have to know whether the rotational modes are fully active or they are not active. We have to know whether the vibrational modes are fully active or not active and why I said that it is a you know nu v star is for one normal mode. Why I said so because if there are more than one atoms of course, there has to be more than one atom for a molecule to vibrate if it is linear or non-linear then you have to use 3 n minus 5 or 3 n minus 6 formula to calculate the total number of normal modes of vibration and then substitute the value over here. This requires a little bit more discussion, but what I have discussed is give have given you an exposure to how to treat with the different modes of motion while discussing constant volume heat capacity. We have to consider translation, we have to consider rotation, we have to consider vibration and if the temperature is even higher much higher then we have to even consider dissociation. We will be discussing more about heat capacities by taking suitable numerical problems, but that we will discuss in the next lecture. Thank you very much.