Proof that b = 1/kT

The pressure in the state j is given by pj = - (Ej/V). \

The average energy is

The average pressure is

According the Gibbs postulate the average energy, average pressure and other average mechanical properties calculated using the partition function are equal to their thermodynamic counterparts.

Note that here McQuarrie uses E bar and p bar for the average quantities and elsewhere the angle bracket notation is used. These are equivalent notations.

If we differentiate the expression for the average energy we can treat the denominator, Q as a function of V as well since it represents a sum over e-bEj(N,V). Since Ej appears both in the exponent and as a function multiplying the exponent we have

Here we used the quotient rule to take the derivative.

This can written compactly as

We can differentiate expression (II) above with respect to b to obtain

The two derivative expressions can be combined to give 

This can be compared to the thermodynamic equation of state

 

This can be derived as follows from dE = TdS – PdV. First take the derivative of both sides with respect to V at constant T.

Now, note that

This is known as a Maxwell relation. It is obtained from

dA = SdT + PdV (III)

From the fact that A is a state function (the Helmholtz free energy) we know that the second cross derivatives must be equal. That is:

And from inspection of (III) we see that

Finally, using the relation

Showing that this is true is a little tricky. For example, we can define F = 1/T. Then (F/T) = -1/T2 and (F/F) = 1. So we can write

(F/T) = -1/T2(F/F)

(/T) = -1/T2(/F) = -1/T2(/(1/T))

which gives (IV) when both sides are multiplied by T.

The comparison with the above equation shows that b µ 1/T.

This proves that b = constant/T. The constant turns out to be kB or Boltzmann’s constant by comparison with expressions for the average energy or average pressure with known thermodynamic equations.