 Let us now talk about those constants you know alpha and beta which we used in Lagrange's method of undetermined multipliers. Earlier we have talked about alpha in terms of exponential alpha in one of the lectures and today we are going to express beta in terms of temperature. We have been discussing that beta is equal to 1 over k t, but today we will show we will prove that beta is equal to 1 over k t. We will need to bring in discussion on internal energy to derive that. Let us consider only the particle or molecule is able to move or free to move only in three dimensions that is translational partition function. We are not allowing rotation vibration that means we are talking the case of atoms. Atoms will only have translational degree of freedom, electronic will be there, but let us not consider electronic at this moment. Recall the discussion when you were learning equipartition theorem, a perfect gas perfect monotomic gas consider only translational. The equipartition expression for the internal energy of a perfect gas you remember that this was equal to 3 by 2 n r t, half n r t in one dimension and 3 by 2 n r t in three dimension. So, we are going to use the result of equipartition theorem and by using the concepts of statistical thermodynamics. Now, I will derive that u minus u 0 is equal to 3 n by 2 beta. I am going to derive that where n is the number of molecules and beta is what we are looking for we want to find out an expression for beta. So, once I have derived this what I will then do is I will compare these two and then equate 3 n by 2 beta equal to 3 by 2 n r t and then I will obtain an expression for beta ok. We have already talked about internal energy u is equal to u 0 minus n by q del q by del beta constant value. We are talking about translational contribution. So, therefore, the translational partition function that we know is equal to v upon lambda q we are going to use that and we are also going to use an expression for lambda which is equal to beta h square over 2 pi m square. We are going to use both these expressions fine. So, u minus u 0 is equal to minus n by q minus n by q by q is v into lambda q this is q into del q by del b beta at constant volume that means del v upon lambda q del beta at constant volume. So, what I am going to get from here now this is equal to minus n lambda q divided by v and constant volume means v can come out into volume into del del beta of 1 by lambda q at constant volume this is the expression now we are looking at. Let us work on this now. So, that means u minus u 0 is equal to volume and volume can cancel out. So, we have minus n lambda q this is lambda raise to the power minus 3. So, derivative is minus 3 over lambda 4 into del lambda del beta at constant volume. So, what are we getting now u minus u 0 is equal to 3 n 3 n divided by lambda lambda q by lambda 4 into del lambda by del beta at constant volume. Let us now act on this. This is going to be 1 by 2 into beta h square over 2 pi m minus 1 by 2 and another term that will come is h square over 2 pi m fine. So, I used this definition of lambda and I took the derivative of lambda with respect to beta. So, I have this expression u minus u 0 is equal to 3 n by lambda half beta h square 2 pi m minus raise to the power minus 1 by 2 and into I have h square over 2 pi m. Now, I need to further simplify this expression easy way to simplify is you multiply by beta and divide by beta no harm done. Now, you recognize that these two are same terms. So, what I have now my u minus u 0 is equal to 3 n by 2 beta I am combining this beta over here into 1 by lambda and this product is nothing, but lambda it is beta h square over 2 pi m square root into lambda. So, therefore, I get this expression u is equal to u 0 plus 3 n by 2 beta what we have done is we have used the equipartition theorem and internal energy is 3 by 2 n r t and then exclusively we use the translational contribution we arrived at an expression that u minus u 0 or u is equal to u 0 plus 3 n by 2 beta let us proceed further. So, we derived u is equal to u 0 plus 3 n by 2 beta and we know that u is equal to u 0 plus 3 by 2 n r t this is from equipartition theorem. Now, if you compare these two this and this are equal and that is what is equated over here 3 by 2 n r t is equal to 3 n by 2 beta. So, let us further work on this you have 3 by 2 n r t this is equal to 3 by 2 instead of n I can write number of moles into Avogadro constant that is equal to n capital N number of molecules divided by 2 beta. So, your 3 by 2 3 by 2 n n they cancel and beta is equal to n a Avogadro constant divided by r into 1 by t. Remember that gas constant r is equal to k times n a where n a is Avogadro constant k is Boltzmann constant. So, therefore, n a upon r is equal to 1 upon k. So, beta is equal to then 1 over k t we are using this r is equal to k times n. So, we have the expression beta equal to 1 over k t this is an expression that we have been using earlier, but now we have shown by using the translational partition function and equipartition theorem result that beta is equal to 1 over k t. So, this completes our discussion on the undetermined multipliers. You remember that we wanted to put one of the term equal to 0 the term which was actually not independent, but we made independent in d log w that term. So, that you know we can set that equal to 0 for a certain value of alpha and for a certain value of beta which were actually undetermined multipliers. Exponential alpha or alpha we have discussed earlier and beta we prove here that beta is equal to 1 over k t and as I have repeatedly said that sometimes you know people work in terms of beta sometimes people work in terms of just temperature. So, therefore, do not get confused all right let us move forward. Once we have the knowledge of internal energy we can calculate constant volume heat capacity. Heat capacity is a very important thermodynamic quantity because it allows determination of temperature dependent thermodynamic quantities. For example, if I know the enthalpy change for a reaction at one temperature I can calculate the change in enthalpy at another temperature if I have the knowledge of heat capacity. We can use the Kirchhoff's law any thermodynamic quantity if you determine at one temperature and if you want to connect with the same or if you want to calculate the same thermodynamic quantity at another temperature the connector is heat capacity. Therefore, heat capacity is a very important thermodynamic quantity. Usually you talk about heat capacity is under two constraints one is constant volume and other is constant pressure. When you keep constant volume we use the symbol C v and that definition the mathematical definition of C v is del u by del t at constant volume. U we have just derived an expression that u is equal to u 0 plus 3 n by 2 beta right. Now, let us write let us see what is this 3 n by 2 beta 3 n n is n times Avogadro constant and divided by 2 and beta is 1 over k t ok. So, that is what is being done over here that instead of n I am writing n times n a and for beta you are writing k t and then the second that I am using R gas constant is equal to k times n a. So, k times n a becomes R. So, it basically becomes 3 by 2 n r t that is what was the result from equipartition theorem that u is equal to u 0 plus 3 by 2 n r t. Now, you can take the derivative C v is equal to del u by del t at constant volume this is a constant. So, its derivative is equal to 0 and at constant volume derivative with respect to temperature is simply going to be 3 by 2 n r. Remember that this is the actual heat capacity at constant volume for a given size of the system. You can always express these thermodynamic quantities in terms of molar properties that is let us say I write here C v I take n on the other side is equal to 3 by 2 r and per mole C v m molar heat capacity is equal to 3 by 2 r. We should always look at whether we are talking about C v or we are talking about C v m C v heat capacity at constant volume it depends upon the size of the system how many grams how many moles because it involves n and when you talk about the molar property that does not depend upon the size because you are calculating or determining heat capacity at constant volume for 1 mole of a substance. So, therefore, do not get confused when you come across the terms C v and C v m. Since we are talking about heat capacity over here and in future also we are going to talk more about heat capacity it makes a sense here to discuss how to experimentally measure heat capacity. If you look into this definition C v is equal to del u by del t at constant volume. In fact, I can introduce right here although we will solve the numerical problems later on on that C p heat capacity at constant pressure is del h by del t at constant pressure. So, we talk about constant pressure constraint we talk about constant volume constraint and we discussed that experimental measurements of C v requires the measurements of change in internal energy that means a bomb calorimeter will be useful. And if experimentally I were to measure C p then I require a calorimeter which can work under constant pressure conditions. A literal meaning of C v and C p let us try to examine C v or C p both delta h delta u remember that q at constant volume is equal to delta u and q at constant pressure is equal to delta h this is from your previous knowledge. So, both delta u or delta h if we are talking about infinitesimally small quantities then I will write d q at constant volume is equal to d u and d q at constant pressure is equal to d h. So, whether we are talking about d u or delta u or we are talking about d h or delta h we are essentially talking about the heats in one case it is constant volume and in the other case it is constant pressure. So, experimental measurements of heat capacities require the amount of heat required to change the temperature by one kelvin either under constant volume conditions if we are talking about C v or under constant pressure conditions if we are talking about constant pressure by applying the concepts of statistical thermodynamics we connected internal energy with partition function and from that we came up with this kind of expression that is u is equal to u 0 plus 3 and by 2 beta and that partition function was expressed in terms of degeneracy and energy levels. So, that means, you can also get the information on C p and C v from the respective spectroscopic data because when you are talking about an atom a monatomic gas for example, there here if you look into the question statement it talked about monatomic gas that means, you are only talking about translational degree of freedom a monatomic system an atom cannot have vibrational or rotational degree of freedom. You also know from your previous knowledge that the value of C p and C v will be temperature dependent because the different contributions set in at different temperatures here we are only talking about monatomic gas. So, we are only including the translational contribution. So, C v is equal to 3 by 2 n r and C v m is equal to 3 by 2 r and as I just mentioned C p and C v are very important thermodynamic quantities because they are connectors of thermodynamic property at one temperature to thermodynamic property at another temperature and also if we are going through the literal meaning the amount of heat required to raise the temperature by one Kelvin. The amount of heat required to raise the temperature by one Kelvin is going to also depend upon how strong the system is what is the type of bonding which keeps different molecules together you know in the material. Right now we are not talking about beyond independent molecules we will talk about solids and liquids later on, but I am just trying to give you a literal meaning of C p and C v since it is the amount of heat required to raise the temperature by one Kelvin it also depends upon how strong the system is. Here we have discussed C v and when we derive an expression between partition function and enthalpy then we will talk about C p also. Remember that C p and C v are also connected with each other C p minus C v is equal to n r that is for the ideal gases whereas for the non-ideal gases you remember from the concepts of chemical thermodynamics if the systems are interacting there may be other terms in it. So take home lesson from this lecture is that by using the result of equipartition theorem and considering the translational partition function that is for a system which is free to move only you know in 3 dimensions having translational degree of freedom we could establish that beta is equal to 1 over k t. With this knowledge now in the next lectures we will start connecting partition function with other thermodynamic quantities and provide an interpretation to those thermodynamic quantities. Thank you very much.