 Welcome back let us continue with the problem solving. In the previous lecture we took some examples where the developed equations could be applied in solving some numerical problems. Let us now extend our discussion to another type of problem. The question is calculate the standard molar entropy of N 2 gas at 298 Kelvin from its rotational constant is given B is equal to 1.9987 centimeter inverse and its vibrational wave number is given. The problem that we discussed in the previous lecture was based upon moment of inertia ok. Remember B is equal to h cross by 4 pi c i. So, if I know a moment of inertia I can get the rotational constant and towards the vibration side there we talked about frequency and here we are talking about the wave number. So, h nu is equal to h c nu bar. So, these should be easily inter convertible. The question that we need to address here is the standard molar entropy of nitrogen gas. When I say standard molar we are talking about the standard state standard states that we represent by the symbol this and we put pressure here is equal to 1 bar and molar means I will talk about S m naught. This is what basically I mean for the sake of simplification I will not write m and naught I will simply write S, but you take it as standard molar entropy. The question says that you evaluate you calculate the standard molar entropy from the given values of rotational constants and vibrational wave number. So, therefore we need to find out such a connection. Obviously, we need to connect entropy with the molecular partition function. So, we know that entropy is equal to u minus u 0 by t plus k log q that we know. Now, since we are dealing with nitrogen gas when you are dealing with nitrogen gas the gas molecules are indistinguishable. When the molecules are indistinguishable then canonical partition function is equal to q raise to the power n by n factorial. So, what I have now is S is equal to now u minus u 0 by t plus k log q raise to the power n over n factorial which is now I can write u minus u 0 by t plus k log q raise to the power n minus k log n factorial. Let me rewrite it S is equal to u minus u 0 by t plus n k log q minus k log n factorial is n log n minus n ok. Next S is equal to u minus u 0 by t n k is equal to n r we have discussed many time log q minus n r log n plus n r right. We are doing this k times n is equal to k n is equal to n times n a and k into n is r which is equal to n r this is what we have done. Let me write down this on the next slide what I have is S is equal to u minus u 0 by t this is 1 then we had n r log q plus n r log q minus n r log n minus n r log n plus n r. Now, we need to also connect this u minus u 0 by t with the molecular partition function and that also we know u minus u 0 is equal to minus n by q del q by del beta at constant volume which I can write as minus n del log q del beta at constant volume d q by q can be written as del log q. Now, sometimes it becomes easier if you differentiate with respect to temperature then when you differentiate with respect to beta beta is equal to 1 by k t. So, therefore, temperature derivative can also be done with respect to temperature or it can be with respect to beta. Now, you consider d by d t let us I write d by d t I can write as d beta by d t into d by d beta mathematically that is allowed and remember beta is equal to 1 over k t. Therefore, d d beta therefore, when you take its derivative it will come minus 1 over k t square this is d beta by d t into d d beta. So, that means, d d beta I can write as minus k t square d by d t this is mathematically allowed I will now use this information over here. That means, now I have u minus u 0 is equal to minus n and minus sign will go away. So, I will have it is square minus minus plus n k t square into del log q by del t at constant volume. What I have done is simply change the differentiation from d d beta to d d d. Now, let us substitute this information into this expression what I have now s is equal to u minus u 0 by t. So, therefore, 1 t with t square 1 t will get cancelled. So, I will have n k t n k t del log q del v del t at constant volume plus n r log q minus n r log n plus n r we have reached up to this this t square becomes t because u minus u 0 by t is used. Now, what I will do next is to make the things little easier I will combine these two terms. Let us say if I combine these two terms I can write n times k del t log q del t at constant volume plus n r minus n r log n right this n k also I can write n r. So, therefore, now I will write s is equal to n r del t log q del t at constant volume plus n r minus n r log n. Let me write this I hope this this is clear to you that these two can be combined to write this derivative if you open up this derivative you will get this expression. For example, when you act upon del t log q by del t then it is derivative of a into b. So, t into del log q by del t at constant volume this first term then plus log q plus log q this term n r is anyway common. Let us write this expression on the next page for further solving. So, I have s is equal to n r del t log q del t at constant volume plus n r minus n r log n. The given question is on nitrogen nitrogen gas nitrogen gas is going to have translational degree of freedom is going to have vibrational degree of freedom is going to have rotational all translational rotational vibrational degrees of freedom are there. So, therefore, I need to calculate the entropy due to translational contribution rotational contribution vibrational contribution. First of all let us talk about translational when I talk about translational then q translational is equal to V upon lambda q lambda is equal to h upon root 2 pi m k t. So, that means this is equal to V upon lambda cube upon h cube 2 pi m k t raised to the power 3 by 2 this is what I am going to get V upon lambda cube. We are asked to solve this at t equal to 298 Kelvin 25 degree centigrade and we are given standard state conditions right. I said I am not writing not, but assume that this is not that means I can get V naught V P V m naught is equal to R t where P is also P naught which is 1 bar by using this you can get V m naught equal to R t by P naught temperature is given pressure is given which is 1 bar you can convert into atmosphere and use the you know appropriate units this comes out to 24.78 liter which is equal to 24.78 into 10 raise to the power minus 3 meter cube. When you actually substitute this q t into here listen to me carefully when you put this q t into here and write an expression for translational contribution go back to Saku-Tetrode equation. We have included n factorial factor and come up with this expression including nr minus nr log n. Therefore, this expression when you apply for translational contribution will eventually take the form of Saku-Tetrode equation and what is Saku-Tetrode equation nr log I will write e raise to the power 5 by 2 2 pi m k t raise to the power 3 by 2 into V I will put V m naught right I am not writing standard state on that side, but consider this as a standard state this is equal to or I will have an a into h cube I am not deriving this because we have derived it earlier. In fact, when you put q t is equal to V by h cube 2 pi m k t raise to the power 3 by 2 here in this expression you are going to get this Saku-Tetrode equation. We know the value of V m naught m we know for nitrogen 28 divided by Avogadro constant into 10 raise to the power minus 3. We know Boltzmann constant we know Planck's constant we know Avogadro constant and when you substitute the values this is coming to be equal to 150.4 joules per Kelvin per mole. We have got now the translational contribution we still have to get the rotational contribution and vibrational contribution. Let us now work towards obtaining rotational contribution. If we go back or we can just take a look at this expression full translational contribution we have already included the n factorial expansion remember we used q is equal to q raise to the power n by n factorial ok. This molecular partition function is the product of translational rotational vibrational etcetera but n factorial is not coming with each q n factorial is coming only one that we are including here along with the translational contribution you can include with rotational also vibrational whatever because this factor is going to come only once. We have included here in Saku-Tetrode equation therefore we need not use it when we discuss rotational contribution or we discuss vibrational contribution. Let us now switch over to rotational contribution. So rotational for rotational contribution what I will do is I will use same thing n r del t log q by del t constant volume and I am not going to use the rest of the term because those have already been used with the translational contribution. Now q rotational is equal to 1 upon sigma h c beta b that we know I can write this as k t over sigma h c beta b that is equal to 1 over k t alright. So what I will do now is I will substitute here s r is equal to n r into let us see what do we get if I keep t by q del q del t this one will come plus log q will come into del t by del t 1 this all is going to be at constant volume. Let us see what we get t by q t by q q is k t into sigma h c beta b that is equal to h c b into del q by del t when I take derivative of the partition function with respect to t this is going to be k over sigma h c b that is it n plus log q this is the bracketed term and you already have here n r. So what I have now s r is equal to n r into k t sigma h c b and k t sigma h c b cancel. So I have 1 plus log q where q here is q r only or what I have n r into log q is what q is k t over sigma h c b this is log q plus 1 this is what I have. But remember that we are using the high temperature result because we are using q r is equal to k t by sigma h c b which is only possible if the temperature rotational temperature is higher than the actual temperature. So for that what you will do is you will use k theta r is equal to h c b use this you put in the value you will get theta r equal to 2.87 Kelvin right and your given temperature is equal to 298 Kelvin which is much higher than 2.87 Kelvin. So therefore it is justified to use q r is equal to k t over sigma h c b. Now you substitute the values and what you will get is 41.14 joules per Kelvin per mole k is Boltzmann constant t is the temperature which is given to you sigma a is equal to in this case it is a nitrogen. So therefore sigma is equal to 2 and h is Planck's constant c speed of speed of light and b is the rotational constant. We have done for rotational contribution also and now let us talk about the vibrational contribution. So S v is equal to n r del t log q q here is q v over del beta del t at constant volume. So what I have now is let us do it q vibrational is 1 over 1 minus exponential minus h c nu bar by k t I am writing in terms of temperature. Therefore my S v becomes n r it will be easier if I talk in terms of log q I am not writing v here for simplicity. So this is minus log 1 minus exponential minus h c nu bar by k t this is what I have. Now let us take the step forward and expand this differentiation what I have now t into derivative of log q derivative of this which is going to be minus 1 divided by 1 minus exponential minus h c nu bar by k t into 1 minus exponential minus h c nu bar by k t into next derivative is equal to h c nu bar by k t square with a plus sign this is what will come then plus log q v let us combine now. S is equal to what it will come n r into h c nu bar by k t that is coming from this term and this term and then I have into exponential minus h c nu bar by k t divided by 1 minus exponential minus h c nu bar by k t and log q v already we have it is minus log 1 minus exponential minus h c nu bar by k t right quite complicated term substitute the value h c nu bar by k t all numbers are given to you nu bar is given to you when you calculate this comes to 11.38. And when you use this into this expression what you will find that S v vibrational contribution is nearly 0 approximately 0 number is 0. Numbers are going to be very small. So, now we have S total let us say this will be S translational plus S rotational plus S vibrational we have S total is equal to S translational was 150.4 S rotational was 41.14 and this was 0. So, many joules per Kelvin per mole. So, therefore your entropy comes out to 191.54 joules per Kelvin per mole. Quite a lengthy numerical problem, but at least it discusses how the different contributions must enter in when you evaluate the overall thermodynamic quantity. Here we have shown for the case of entropy, but suppose if the question is for Gibbs function or for Helmholtz function or enthalpy then you must include the translational contribution rotational contribution or vibrational contribution whichever contribution whichever mode of motion is fully active. In this particular case you noticed that there is a sizable number coming up from translational contribution and rotational contribution and vibrational contribution was nearly 0. So, I hope that this particular example has brought in clarity on how to calculate a thermodynamic quantity from the molecular partition function which has different contributions such as translational, rotational, vibrational and if the temperature is so high then the electron. We will discuss a few more problems, but in the next lecture. Thank you very much.