 In this lecture, we will discuss further on rotational partition function. By now we have had a thorough discussion on rotational partition function for a linear rotor and a non-linear rotor. Now, let us apply those learned concepts in solving some numerical problems. We have discussed that for a non-linear rotor, we need to have the knowledge of a, b and c. And for a linear rotor, we need to have a knowledge about b rotational constant. And rotational constant basically involves moment of inertia. And moment of inertia involves knowledge about the mass and the bond length. So, with this background now we will start discussing some numerical problems. Let us revisit that symmetry number is very very important because symmetry number that comes in the denominator and reduces the value of q r. Let us now consider this numerical problem. The NOF molecule is an asymmetric rotor. It is given information asymmetric rotor. You can take a look at the structure of molecular structure of NOF. It is asymmetric. One side is oxygen, other side is flowing. With the rotational constant 3.1752 centimeter inverse, 0.3951 centimeter inverse and 0.3505 centimeter inverse. So, basically these are a, b and c. And we have been asked to calculate the rotational partition function of the molecule at 100 degree Celsius. Now, once we are asked to calculate the rotational partition function, then we need to make a decision that whether to go by direct summation method or to go by high temperature approximation. In order to use high temperature approximation, we need to calculate characteristic rotational temperature. And how do we calculate characteristic rotational temperature? Is k, is k theta r is equal to h c b or you can have k theta r is equal to h c a, k theta r is equal to h c c. So, you will have the rotational temperature. The values of h c b everything is available. So, once we calculate theta r, if temperature is much, much higher than theta r or characteristic rotational temperature, then we can straight away go ahead and use the high temperature approximation result. In this case, you please calculate yourself, find out the value of theta r and you will find out you will observe that 100 degree Celsius which is 373 Kelvin is much, much higher than any of the theta r corresponding to a, b and c. And once you establish that, then you are free to use this high temperature approximation result. k is Boltzmann constant, t is temperature, h is Planck's constant, c is the speed of light, a, b, c all these three values are given to you. Another alternate form which is a simplified form comprising of k, h, c because these all are constant. So, you combine all these k, h, c also include pi into all these constants multiplied. It transforms into 1.0270 divided by symmetry number into substitute temperature in Kelvin raised to the power 3 by 2 and put a, b, c each one in centimeter inverse. So, it becomes centimeter minus 3, it is a square root. We should know this formula and this formula leads to another formula where you combine k, h, c, pi these are all constants combine them and the value is 1.0270. So, therefore, we just need to now substitute the values. So, what we will have now q r, you can use any of the result. Let us say if I use 1.0270, now sigma here if you rotate the molecule by 180 degree centigrade, fluorine will be on the other side, oxygen will be on this side. So, therefore, it is distinguishable that means, sigma is equal to 1 here. So, once I have sigma equal to 1 into temperature in Kelvin 373 raised to the power 3 by 2 and a, b, c each one in centimeter inverse you have 3.1752 into 0.3951 into 0.3505 raised to the power 1 by 2 treat this as separate and once you solve you should get a value of 1.12 into 10 to the power 4. Please note that this number is very high. So, for a non-linear rotor obviously, the value of rotational contribution to partition function is inversely proportional to a, b, c and a, b, c involves moment of inertia h cross by 4 pi c i. So, the higher the value of mass then the higher will be the value of moment of inertia and since moment of inertia comes in the denominator for a, b, c. So, higher value of moment of inertia will lower the value of a, b or c and if a and b and c are low the value of q r is going to be high. The previous example in the previous lecture we talked about oxygen the value was something around 79. Here we are talking about NOF molecule and c, but the value the temperature here is 100 degree centigrade. At 100 degree centigrade you have 1.12 into 10 raise to the power 4 thermally accessible rotational states alright. I hope this is clear when we want to address rotational partition function. We need to identify this I am saying again and again is it a linear rotor or is it a non-linear rotor and the third thing we need to know what is the value of symmetry number. The other information which you need is in this case linear rotor means you need the value of rotational constant b. Non-linear rotor you need to have the value of a, b and c. Everything else is a constant there h planks constant k Boltzmann constant c speed of light everything else is a constant. And the fourth thing which you need is mass or more precisely you need the reduced mass. Once you have this information it is very easy to obtain the value of rotational partition function alright. Let us move ahead. We will also now discuss some associated numerical problems. We have earlier discussed that pressure is given by k t del log Q by del b. Q is the canonical partition function. Canonical partition function can be written in terms of molecular partition function that is either Q is equal to Q raise to the power n or Q is equal to Q raise to the power n by n factorial. The problem here which is being discussed says calculate the equation of state associated with the partition function Q which is given by 1 by n factorial 2 pi m over beta h square raise to the power 3 n by 2 into v minus n b raise to the power n into beta a n square by v. This n v b these are the constraints n v beta beta is 1 by k t. So, essentially it is n v t that means it is a canonical partition function. And if you look at this expression carefully it involves b and a. So, that reminds you of Wenderwals equation of state where the attraction and repulsion terms are included. The question is to derive or calculate the equation of state. What is an equation of state? An equation of state is an equation which connects p v and t. For ideal gas p v is equal to n r t. For non-ideal gas we should get an expression from the given expression. Because if we the given expression involves b and a which are repulsive and attractive parameters. So, since we want to derive an expression between p v and t it is wise to start with the expression for pressure. And pressure is equal to k t times del log q del v at constant temperature. We need partial derivative of q or log q with respect to volume at constant temperature. So, let us now write an expression for log q because we will need to take derivative of log q. Let us write log q. Log q is equal to let me write first v minus n b term that will be n log v minus n b plus beta a n square by v log e is 1 plus you have terms containing m beta n and some other constants. Why I wrote specifically only two terms? Because is that I need to take partial derivative with respect to volume at constant temperature. And this first part of this equation n is constant pi is constant for a given system m is constant h is constant beta t temperature is constant. Therefore, everything is a constant I separate this constant term because derivative of this constant term with respect to volume is going to be 0. So, let us now take derivative of log q with respect to volume at constant temperature because this is what I need over here. So, therefore, let us take the derivative del log q with respect to volume at constant temperature is going to be n over v minus n b. This is for the first term for the second term beta a n are constant. So, 1 over v that means, minus beta a n square over v square and all other terms the derivative is 0. This is del log q by del v at constant temperature and this is equal to p over k t from this. If I substitute this over here then I get this is equal to p over k t. So, let us continue what we have now is p over k t is equal to n over v minus n b n over v minus n b minus beta a n square by v square minus beta a n square by v square minus beta a n square by v square this is what it is all right. So, therefore, p is equal to n k t over v minus n b minus beta is 1 over k t beta is 1 over k t into k t by v square. So, p into a n square over v square this k t and this k t cancel. So, p plus a n square by v square is equal to n k t over v minus n b that is what we get and when you rearrange this if you put this in one bracket. So, this into v minus n b is equal to n k t in fact, n k t is equal to n is equal to small n number of moles into Avogadro constant into Boltzmann constant into temperature and that k times n a is equal to r is equal to n r t I can write this is equal to n r t and I am sure that you recognize this expression this is Wendervals equation of state for ideal gas p v is equal to n r t because you ignore a and b for a real gas or a non-ideal gas where attractive and repulsive interactions can be present the equation is modified to p plus a n square by v square into v minus n b is equal to n r t this is the Wendervals equation of state and that is what the question was that derive an equation of state for the given canonical partition function. So, by statistical means we have now arrived at the same equation which you have studied in theory of ideal and non-ideal gases in other words we have derived the expression for a non-ideal gas which obeys p plus a n square by v square into v minus n b is equal to n r t that is Wendervals equation of state alright. Let us try one more numerical problem or theoretical problem the question is derive an equation for energy average energy for a simple system of a bare proton in a magnetic field BZ. So, we need to derive an equation for the energy total energy if I express in terms of internal energy total energy is equal to u minus u 0. So, therefore, the problem that we need to solve is for obtaining an expression for u minus u 0 and the other information which is important and given to us is they have defined they have fixed the magnetic field at a constant magnetic field. So, therefore, we can safely put the magnetic field constant in any case when we write an expression for any thermodynamic quantity in terms of canonical partition function we need to have the information about its energy levels and the energy levels or energy for a bare proton is given by E plus minus half is equal to minus plus half cross gamma BZ where gamma which is magneto-zyric ratio is a product of nuclear factor g n and nuclear magneton beta n. So, there the energy is given by both the terms with plus sign and with the minus sign representing two different energy levels. Once you have these two different energy levels then I can write q is equal to go back to your definition q is equal to summation j exponential minus beta E j if the degeneracy is not there. So, you expand this and once you expand this you include the energy corresponding to plus beta term is also there and energy corresponding to minus beta term is also there. The internal energy is given by u minus u 0 is equal to minus del log q by del beta at constant BZ which is also equal to minus 1 by q del q by del beta at constant BZ. We have the expression for q once we have the expression for q q then this q can be used over here take the partial derivative of q with respect to beta at constant PZ you work out yourself eventually you are going to get this expression this is what we were asked for that is to derive an expression for energy for simple system of bare proton in magnetic field BZ and it turns out that this energy expression is given in terms of this expression where we need information about gamma BZ and temperature. Once we have then we can calculate the value of energy for a simple system of bare proton which is in a fixed magnetic field BZ. So, in this lecture we have calculated or derived some expressions which were relevant to rotational contribution to partition function and also in general further applications of canonical partition function. Keep in mind that whenever you convert canonical partition function to molecular partition function that molecular partition function can be a product of different contribution to the partition function. For example, of translational rotational vibrational and electron and when you have to derive expressions for different thermodynamic quantities for example, in this case for internal energy we should be able to write in terms of canonical partition function and then take appropriate derivatives to arrive at the desired equation. We will further solve more numerical problems and also discuss the vibrational contribution to partition function, but those things we will discuss in the next lectures. Thank you very much.