See attached file.
I've included a detailed derivation below. If we skip all the detailed subtle things that are explained there we can proceed as follows. In a volume V in configuration space and in a volume Vp in momentum space the number of quantum states is:
This means that the partition function for a single particle in a volume V is:
Z1 = V/h^3 Integral over momentum space of d^3p exp[-beta p^2/(2m)] = (exploit spherical symmetry of the integrand)
V/h^3 Integral from |p| = 0 to infinity of 4 pi |p|^2 exp[-beta |p|^2/(2m)] d|p|
This integral can then be evaluated as explained after Eq. 7 in the detailed explanation below. The result is:
Z1 = V (2 pi m k T/h^2)^(3/2)
The partition function for a dilute gas consisting of N identical particles is:
Z = Z1^N/N!
During an adiabatic expansion, the entropy stays constant, so let's find the entropy of the gas. We can find the entropy from the free energy.
F = - k T Log(Z)
and the internal energy:
E = -d Log(Z)/d beta
by using that F = E - T S.
Log(Z) = N Log(Z1) - Log(N!) = N Log[V (2 pi m k T/h^2)^(3/2)] - Log(N!)
d Log(Z)/d beta = N d Log(Z1)/d beta = N d d/beta [-3/2 Log(beta)] = -3/2 N/beta = -3/2 N k T
The entropy can be expressed as:
S = (E - F)/T = 3/2 N k + N k Log[V (2 pi m k T/h^2)^(3/2)] - k Log(N!) =
N k [5/2 + Log[V/N (2 pi m k T/h^2)^(3/2)] ]
In the last line we used Stirling's formula Log(N!) = N Log(N) - N which is valid for large N. Note that the expression for the entropy is extensive: If we have two identical systems both at the same temperature and density then the combined system should have twice the entropy. If the two systems are separated initially and that separation is removed then we can consider the system as a single system, which should have twice the entropy. You can see that the above formula for the entropy is indeed consistent with this demand. The factor N in front of the brackets becomes twice as large while inside the logarithm both V and N become twice as large. If we hadn't divided the partition function by N!, then the 1/N factor in the logarithm would have been missing and we would be led to the conclusion that the entropy is not extensive. The fact that the particles are indistinguishable is the key thing here.
In case of adiabatic expansion, the entropy stays constant. From the above formula for S we can then conclude that
V (2 pi m k T/h^2)^(3/2) = constant
because N stays constant.
We can drop the constants in here:
V T^(3/2) = constant
Using the ideal gas law P V = N k T we can put T = P V up to a constant factor. This gives:
V^(5/2) P^(3/2) = constant --------->
V^5 P^3 = constant ...
A detailed solution is derived from first principles.