Entropy of Ideal Gas

A vessel of volume V₁ contains N molecules of an ideal gas held at temperature τ and pressure P₁. The energy of a molecule may be written in the form

where ε_k denotes the energy levels corresponding to the internal states of the molecules of the gas.

a) Evaluate the free energy F. Explicitly display the dependence on the volume V₁.

Now consider another vessel, also at temperature τ, containing the same number of molecules of the identical gas held at pressure P₂.

b) Give an expression for the total entropy of the two gases in terms of P₁, P₂, τ, N.

c) The vessels are then connected to permit the gases to mix without doing work. Evaluate explicitly the change in entropy of the system. Check whether your answer makes sense by considering the special case V₁ = V₂ (P₁ = P₂).

SOLUTION

a) For an ideal gas the partition function factors; however, we must take the sum of N identical molecules divided by the number of interchanges N! to account for the fact that one microscopic quantum state corresponds to a number of different points in phase space. So

(1)

Now, the Helmholtz free energy, F, is given by

(2)

Using Stirling’s formula, ln we obtain

(3)

Using the explicit expression for the molecular energy E_k, we can rewrite (S.4.38.3) in the form

(4)

Here we used the fact that the sum depends only on temperature, so we

can define f(τ):

(5)

b) Now we can calculate the total entropy S of the two gases (it is important that the gases be identical so that f(τ) is the same for both vessels):

(6)

where F is defined by (4).

(7)

We have for total entropy

(8)

c) After the vessels are connected their volume becomes V = V₁ + V₂, the number of particles becomes 2N, and the temperature remains the same (no work is done in mixing the two gases). So now

(9)

It can be easily seen that the pressure becomes  so

(10)

and

(11)

Let us show that ∆S is always nonnegative. This is equivalent to the condition

(12)

which is always true. At P₁ = P₂ (V₁ = V₂), ∆S = 0, which makes perfect sense.