 Today, we are going to discuss applications of the concepts which we have developed so far. We have derived some expressions for example, expressions for writing the partition function, expressions for fractional population of any i th state. And in today's discussion, we will solve some numerical problems and then we will try to understand that what these results represent. So, let us get started with the first numerical problem. The first problem the first question that I want to discuss with you today is that in broad sense how do the concepts of statistical thermodynamics differ from chemical thermodynamics. Explain by taking an example, this is just a discussion based problem, numerical problems we will discuss ahead, but the question that is being asked over here is that in a broad sense in a general sense how does statistical thermodynamics differ from chemical thermodynamics. Going back to our previous discussion, in chemical thermodynamics we deal with the average properties, we deal with the bulk properties and in statistical thermodynamics we deal with the molecular properties. We also discussed that statistical thermodynamics provides a link between molecular properties and bulk thermodynamic properties. Let us take an example, let us take an example of pressure of a gas pressure of a gas let me represent by P. What is P? P is force per unit area consider a gas contained in a cylinder. How do we define pressure of the gas over there? It is the force exerted by the molecules per unit area. Remember that in a container the gas molecules are colliding with each other. So, therefore, whatever is the energy possessed by the molecules that is being exchanged it is they are not just colliding with each others, but the energy is also being distributed between different modes of motion. We have discussed this earlier. Now, you see how my discussion is changing from an average quantity that is the pressure which is the force per unit area it is an average quantity we are talking about. And then I am talking about that there are different molecules they are colliding with each other, the energy is getting distributed amongst themselves not only getting distributed amongst themselves, but also between between different modes of motion. So, this is how the concepts of chemical thermodynamics that differ from statistical thermodynamics. In chemical thermodynamics we talk about bulk properties average properties. Whereas, in statistical thermodynamics we start talking about individual molecules we talk about the population of different energy levels and then we connect with the thermodynamic quantities. So, therefore, once again re-emphasizing what is statistical thermodynamics is a link between molecular properties and bulk thermodynamic quantities or properties. So, this is how in a broad sense the concepts of statistical thermodynamics differ from chemical thermodynamics. Now, let us move to another question read the question or try to understand the question carefully. The question is consider a system of 20 non interacting molecules means I am talking about independent molecules with a uniform ladder of energy levels each separated by 100 centimeter inverse. How many out of the following instantaneous configurations are simultaneously possible justify your answer with suitable calculations? Let us try to understand the question consider a system of 20 non interacting molecules with a uniform ladder of energy levels and each separated by 100 centimeter inverse. So, what I have is I have a uniform ladder of energy level and each one is separated by 100 centimeter inverse. Let me represent that 100 centimeter inverse by e. So, each one for example, if this is e 0 this is e this is 2 e this is 3 e this is 4 e and so on and this e separation is 100 centimeter inverse and then we are asked how many out of the following instantaneous configuration. What are those instantaneous configurations? First one is 20 0 0 0 how do we read this? 20 molecules are in the ground state that is with the energy equal to 0 and there is no molecule in the upper state. The second instantaneous configuration is you have 10 molecules in the ground state no molecule in the first excited state 5 molecules in the second excited state 2 in the third excited state 0 3 etcetera. And we already have learned how to read an instantaneous configuration and there are other instantaneous configurations given over here like 2 10 3 5 0 0 or 5 5 5 etcetera then 6 10 3 5 0 0 0 0 etcetera. The question is how many out of the following instantaneous configurations are simultaneously possible? There can be several instantaneous configurations, but then if you remember we discussed that all the instantaneous configurations may not be permissible because certain constraints are there certain laws certain rules have to be followed. What are those constraints? The first constraint is that the total number of molecules have to remain constant that is summation i n i is equal to n the molecules which are present in different energy state the sum has to be equal to 20 because there are total 20 non interacting molecules. And the second constraint is that the total energy of the molecules in each state has to be equal to the sum of that has to be equal to the total energy of the system. So, these 2 constraints have to be followed one is summation n i is equal to capital N that is the total number of molecules and summation n i E i is equal to the total energy. So, therefore, let us now start working out for each system. Let us take the first instantaneous configuration the first one is 20 0 0 0 etcetera. First of all let us see that in each instantaneous configuration is this constraint followed that is the total number has to be 20 in this instantaneous configuration it is 20 plus 0 0 0 0 total is 20 fine. In the second one 10 plus 5 is 15 plus 2 is 17 plus 3 is 20 that is also fine then 2 plus 10 12 12 plus 3 is 15 plus 5 is 20 that is also possible 5 times 4 is 20 fine here 6 plus 10 is 16 plus 3 is 19 plus 5 is 24 that does not fit in because that is not saying that summation n i is equal to N. So, therefore, this configuration is straight away ruled out. Now, we have to worry about the 4 instantaneous configurations this constraint is followed by these 4 instantaneous configurations first constraint. Now, we have to check about the second constraint let us start working about this. So, how much is the energy here it will be let us start writing down the energy in this case is going to be 20 times 0 plus 0 plus etcetera etcetera right 20 into because they are all in the ground state 20 times 0 0 times whatever is the energy etcetera then total value is equal to 0. Now, let us take the second one second one is 10 0 5 2 0 3 this one 10 0 5 2 0 3 let us write the energy for this this is equal to 10 times 0 plus 0 times E plus 5 times 2 E then plus 2 times 0 E 2 E 3 E plus 0 times 4 E plus 3 times 5 E right 0 E 2 E 3 E 4 E 5 let us see how much it comes to this is coming to 10 plus 6 is 16 plus 15 is equal to 31 times E where E is 100 centimeter inverse. Now, let us look at the third one the third one is 2 10 3 5 0 0 etcetera let us write the total energy for this is equal to 2 times 0 plus 10 times E plus 3 times 2 E plus 5 times 3 E plus remaining all are 0 because 0 0 0. So, how much it is comes to 10 plus 16 plus 15 is equal to 31 E. Now, let us look at the last one we can utilize this space here we have 5 5 5 5. So, energy is equal to 5 times 0 plus 5 times E plus 5 times 2 E plus 5 times 3 E and remaining are 0. So, how much this comes to 5 plus 10 15 plus 15 30. Now, the answer is very clear there are only 2 instantaneous configurations where the total energy is remaining same in other instantaneous configuration for example, here it is 0 and here it is 30 times E E is 100 centimeter inverse. So, therefore, which instantaneous configurations are simultaneously possible one is this because this instantaneous configuration has a total energy of 31 E and the second instantaneous configuration which is simultaneously possible is this one this also has a total energy of 31 E that is 31 100 centimeter inverse. And we already discussed that all these four instantaneous configuration 1 2 3 4 their total number summation they are following this summation n i is equal to n 20, but the total energy constrained is only followed by this configuration and this configuration. So, only 2 out of the given instantaneous configurations can exist simultaneously. So, remember that whenever you need to find out that which different instantaneous configurations are simultaneously possible then you need to follow these 2 constraints, but you remember our previous discussion there are several instantaneous configurations are possible, but the system is most likely going to be found out in an instantaneous configuration which has maximum weight. In this problem we have only talked about which are the different instantaneous configuration. Now, you can always talk about the weight and then try to find out the system is most likely going to show the property of that instantaneous configuration which has maximum weight ok. Let us now switch over to another type of problem let us read this problem carefully for a 3 level system in which different non degenerate energy states are equally spaced by E which out of the following expressions is correct for fractional population of second excited state show your derivation. So, we have to consider a 3 level system in which different non degenerate energy states are equally spaced. So, therefore, let us first draw we draw a 3 level system 1 2 and 3 these are equally spaced that means, this is 0 this is E this is 2 E and we need to find out the expression for fractional population of second excited state. So, let us recall what was the expression for writing fractional population the expression for writing fractional population of ith state was n i upon n is equal to exponential minus beta E i upon q ok. We will use this expression the question that is asked is corresponding to second excited state. So, this means second excited state we are talking about this one here for this E i is equal to 2 E ok and q partition function is equal to ground state contribution is 1 second exponential minus beta E plus exponential minus 2 beta E. I have expanded q is equal to summation j exponential minus beta E j there is no question of degeneracy over here. So, for g factor I have not. So, therefore, n i upon n which is p i is equal to exponential minus beta E i E i is equal to 2 E 2 beta E and q is 1 plus exponential minus beta E plus exponential minus 2 beta E. Let us take the logarithm on both sides. So, logarithm of p i is equal to log A by B is minus 2 beta E minus log this is 1 plus exponential minus beta E plus exponential minus 2 beta E beta is equal to 1 over k t this is equal to minus 2 E upon k t minus log 1 plus exponential minus beta E plus exponential minus 2 beta E let me write here. So, therefore, what we have got is logarithm of fractional population is minus 2 E upon k t minus log of 1 plus exponential minus beta E plus exponential minus 2 beta E. So, therefore, the first one is the correct expression. The way to solve this kind of problems is first of all we need to understand that how different energy levels or different energy states are arranged organized. It is possible that some of the states may have same energy and if different states have the same energy those states will form a level. Therefore, a particular energy level may be G fold degenerate we have been expressing degeneracy in terms of G. And suppose if there are more than states more more than one state which have the same energy then the fractional population of that level is going to be G fold right. Suppose, if I consider this expression and if i th state any i th it can be first state, second state, third state. If i th state is two fold degenerate that means, population of that energy level is going to be twice of that of a one state then in that case that G factor needs to appear here and that kind of you know will modify this kind of expression, but that we are going to discuss in the next set of numerical problems. So, the take home lesson from the discussion on these questions is that we should be able to write expression for the partition function. Expression for partition function if there are discrete energy levels which are given to us like uniform ladder of energy level or some sequence of energy levels we should be able to write expression for partition function and then we should be able to write expression for fractional population. Once we are able to write these kind of expression you will see in the lectures which we are going to now discuss ahead it will be easier to connect the partition function with different thermodynamic quantities. Thank you very much.