 Welcome back by now we have connected entropy with canonical partition function and also we have discussed how to recover molecular partition function from canonical partition function. That means, now we have means of connecting entropy with molecular partition function when we consider concept of ensembles. That means, after having described entropy in terms of canonical partition function, then we can further decide that depending upon whether the molecules are distinguishable or indistinguishable how this equation can be further modified. So, today we will further discuss more on entropy, but let us take first an easy example that of a monatomic gas. So, when we say about monatomic gas it is understood that we are talking about perfect gases we are talking about ideal situations. So, therefore, in this discussion whenever required we will use ideal gas equation. So, in the previous lecture we have talked about the Secure-Tetrode equation which is basically an expression for entropy of a monatomic gas and this Secure-Tetrode equation which connects entropy with volume and the thermal wavelength or entropy with pressure thermal wavelength temperature of course, temperature is hidden in thermal wavelength. These two forms of Secure-Tetrode equation permit us to evaluate entropy under constant volume conditions under constant pressure conditions or under constant temperature conditions. So, depending upon the situation we can use this equation and we also last time discussed that if the molar mass is higher then thermal wavelength is lower if thermal wavelength is lower then entropy is higher and that is what is captured in this comment that Secure-Tetrode equation implies that the molar entropy of a perfect gas of high molar mass is greater than one of the low molar mass. That means, if low molar mass lambda will be high if lambda is high then entropy will be less. Let us further now discuss about entropy. Let us start now discussing the applications of the formulae that we have so far derived. The first example that we will take is calculate standard molar entropy of a gaseous argon at 25 degree centigrade and the molar mass of argon is given over here argon atom. Now the things to notice in the given statement you know before we start attempting solving a problem it is first of all wise to look at the problem statement very carefully and see what all information is given what do they want. The question is calculate standard molar entropy standard that means they want us to calculate the entropy under standard state conditions. So, standard state condition what is a standard state condition standard state means substance should be pure temperature can be any, but pressure should be 1 bar keep this definition of standard state in mind pressure should be 1 bar substance should be pure and temperature can be n. So, therefore, in the Secure-Tetrode equation look at the symbols that are notations that I have used over here S m naught m represents molar this naught is used for standard state condition and the pressure also we write p naught what is the value of p naught p naught is 1 bar that is what I was saying that when you try attempting solving a problem first of all look at the problem statement try to understand what all the information is given. Since they want us to calculate standard molar entropy so automatically we will set the pressure equal to 1 bar alright. So, in this formula of molar entropy what we have the temperature is given 25 degree centigrade. So, on absolute scale we will write it as 298 Kelvin or to be very precise 298.15 Kelvin p naught is given and lambda which is beta h square over 2 pi m whole square root over h over 2 pi m k t essentially you need the information on temperature and you need the information on molar mass that is given 39.95 gram per mole substitute here and remember I have already discussed here we are talking about one atom when you put m over here. So, therefore, what you will be using the number 39.95 convert into kilogram that means multiply by 10 raise to the power minus 3 and then divide by Avogadro constant alright. So, as we have discussed earlier that here you have to put mass of one particle. After substituting all the constants or other values you will get a thermal wavelength of 16 picometer 16 into 10 raise to the power minus 12 meter. So, now you have all the information to put into the formula. So, S m naught is equal to r and since we are talking about molar n is equal to 1 exponential 5 by 2 Boltzmann constant is 1.381 into 10 raise to the power minus 23 joules per Kelvin temperature is given 298.15 Kelvin and then pressure which is 1 bar and we have to use SI units here right. 1 bar is equal to 0.1 mega Pascal and then you convert that 0.1 mega Pascal into Newton per meter square then 1 bar will turn out to be 10 raise to the power 5 Newton per meter square. And then we have here the thermal wavelength which is 16 picometer which is 1 you convert into meter which is equal to 1.6 into 10 raise to the power minus 11 meter cube or 16.0 into 10 raise to the power minus 12 meter cube. After solving this we will get a molar entropy of 155 joules per Kelvin per mole. So, therefore, the standard molar entropy of gaseous argon at 25 degree centigrade is 155 joules per Kelvin per mole. Simple calculations we need to be careful about what value of P naught should be put. We need to be careful about what value of molar mass to be put in what way that is you need to put for one particle all right. So, we have obtained a molar entropy of 155 joules per Kelvin per mole from a thermal wavelength of 16 picometer. Let us take a look at some of the comments argon is 39.95 gram per mole neon molar mass is less 20.18 it is a lighter molecule. Lighter molecule lower molar mass if lower molar mass then thermal wavelength will be higher if thermal wavelength is higher the entropy will be less that is a reciprocal relation. So, neon which has a molar mass of 20.18 it is a lighter molecule and has higher thermal wavelength. Therefore, standard molar entropy of neon will be smaller than argon and calculation suggest that it is 146.3 argon was 155 neon is 146.3 this is the effect of the molar mass. Let us take one more case that is standard molar entropy of hydrogen and the comment if you carefully look at the comment over here what is written in the comment the translational contribution to the standard molar entropy translational contribution why I did not talk about only translational contribution in case of neon or argon because neon and argon which are atoms will have only translational degree of freedom will not have rotational and vibrational degrees of freedom. When it comes to hydrogen it is a diatomic molecule the other contributions can also come in. So, therefore, the translational contribution can be calculated by using saccortetrode equation alright. Here the molar mass is 2 and if it is molar mass is further less then thermal wavelength will be further higher and molar entropy will be further less and you can see now it is 118.118 joules per Kelvin per pol argon with a molar mass of 39.95 has a molar entropy of 155 neon with a molar mass of 20.18 has molar entropy of 146.3. Hydrogen with a molar mass of 2 has a translational contribution to molar entropy as 118 joules per Kelvin per mole. So, you can clearly here look at that how the molar mass of the given species affects the molar entropy. Now, let us further discuss about this saccortetrode equation what other type of equations we can further develop from this. So, saccortetrode equation this is one form S is equal to N R log exponential 5 by 2 volume N times N A lambda Q and we all know what is lambda we do not need to elaborate now. Now, let us maintain isothermal conditions isothermal conditions isothermal conditions means constant temperature if temperature is constant. Now, let us look at in the saccortetrode equation what happens if temperature is constant this exponential raise to the power 5 by 2 is a constant number of moles is a constant Avogadro constant is a constant and what about lambda lambda you see here is h by 2 pi m k t if temperature is constant and the molar mass is constant then lambda is also constant. That means, under isothermal conditions if you want to apply this expression for any changes happening in the system then essentially the lambda thermal wavelength is constant. So, therefore, everything other than volume that means, all these parameters exponential 5 by 2 N N A lambda cube this is all constant and I can combine all these constants into A right. So, I write S is equal to N R log A times V. Now, you consider a gas and let that gas expand under isothermal conditions isothermal conditions ok. If we are considering isothermal conditions I can use this formula. Let us say expansion takes place from V i to V f then your delta S is S f S final minus S initial what it will be? It will be N R log A V final minus N R log A V initial which now I can write as delta S is equal to N R log A V f over A V i is equal to N R log V final over V initial. We have an expression now to calculate change in entropy when a gas expands under isothermal conditions and we have only used the statistical thermodynamics concepts over here. So, this slide basically recaptures again what I have just discussed is that under constant temperature conditions S is equal to N R log A times V and then for any change where a gas expands from V final to V initial we have this expression delta S is equal to N R log V final over V initial. Remember that once again I am saying that we have used simply the statistical thermodynamics concepts over here. Does this result match with that obtained in classical thermodynamics discussion? Let us look at that remember that the definition of D S is equal to D Q reversible by T the subscript reversible that is specifically to be noticed all right. If D S is equal to D Q reversible by T then let us see what happens under constant temperature conditions I will use the first law of thermodynamics D U is equal to D Q plus D W. If the temperature is constant then the change in internal energy is 0 we are so far talking about ideal situations ideal gases. So, constant temperature means this D U is equal to 0 let me write D U is equal to 0 under isothermal conditions. So, that means D Q is equal to minus D W and if I invoke reversibility conditions then isn't this equal to minus minus P D V ok. So, what do I have now D Q reversible is equal to plus P D V and since P V is equal to N R T then I can write this as P is equal to N R T by V D V I am interested in D Q reversible by T. So, D S then becomes D Q reversible by T D Q reversible is N R T D V by V N divided by T which is N R D V by V and if I now integrate from V I to V F why? So, then delta S will be equal to N R log V final over V initial ok. You apply the limits V I and V F over here then delta S will be equal to N R log V F over V I and you are essentially getting the same result. Let us take a look at the comment. So, now we have shown that this expression delta S is equal to N R log V F over V I which by using the concepts of statistical thermodynamics we showed that at constant temperature this is N R log V F over V I and just now we also by using the classical thermodynamics concepts again we showed that under isothermal conditions or under constant volume conditions delta S is equal to N R log V F over V I. That is what is the comment over here that this is exactly the expression we obtained by using both classical thermodynamics and statistical thermodynamics. The classical expression is in fact a consequence of the increase in the number of accessible translational states when the volume of the container is increased. Look at this comment very carefully. The classical expression is in fact a consequence of the increase in the number of accessible translational states when the volume of the container is increased. What it means is that you consider gas contained in a volume certain volume you know container. Now you let the gas expand the volume is expanding that means your dimensions A B C or length of the container that is all going up right even if you consider one dimension if you are expanding like this then the length of the container is increasing. What is the effect of the change in length of the container on the number of accessible thermal trans this translational states that is what we need to consider. We will look at that. Now delta S is equal to N R log V F over V I. Obviously as we have been discussing that if V F is greater than V I. If V F is greater than V I that means we are talking about expansion and remember if we sort of you know for the sake of discussion if we restrict to one dimensional translational motion then remember E N was equal to N square H square over 8 m L square right and if it is allowed in three dimension then you have E N 1 N 2 N 3 is equal to N 1 square H square over 8 m L 1 square plus N 2 square H square over 8 m L 2 square and so and so on. I am not going into that details because that we have already covered but what is happening here is look at this scenario A is corresponding to this is corresponding to V I and B is corresponding to V final expansion. So, you are allowing the system to undergo expansion. Expansion means you are increasing the value of L. So, if L square increases the energy E N decreases and that is what you see over here that this you see that the spacing between the energy levels corresponding to different N that decreases and that is the comment written over here that as the width of a container is increased going from A to B width is increased. The energy levels become closer you can compare this versus this this system has the energy levels much closer to each other than this system and as a result more thermally accessible states at a given temperature. So, given scenario this given scenario this at a given temperature this will have more thermally accessible states and if there are more thermally accessible states if the the molecules can distribute in different in larger number of thermally accessible states and therefore, there will be more disorder and therefore, there will be more entropy. So, this is the statistical definition or interpretation explanation of why the entropy of a gas increases when it undergoes expansion under isothermal condition from a initial volume to final volume. Basically what we did was that we talked about what will be the population of different states isothermally when the gas is allowed to expand. We will discuss few more applications of sacroteroid equations as I mentioned earlier that sacroteroid equation is very very important because it will allow you to calculate the entropy changes when the system undergoes change under constant volume conditions under constant pressure conditions or under constant temperature conditions. So, we will discuss more in the next lecture. Thank you very much.