Question

Derivation of the Ideal Gas Law from Statistical Thermodynamics

Answer 1

This post is a kind of self-experiment, but I'm going to derive the ideal gas law from 3 fundamental theories about the world:
1. Thermodynamics: The first and second law of thermodynamics.
2. statistical thermodynamics: Rewrite the equation of state equations into statistical form.
3. Quantum mechanics: The quantum mechanical expression of energies in a container.

I'm gonna start my derivation using the full differential for the Helmholtz free energy:
\[\Large dA = -S ~ dT - p ~ dV\]This means we get the following expression:\[\Large \left( \frac{ \partial A }{ \partial V } \right)_{T}=-p\]We can from here go into statistical thermodynamics and use a relation between the Helmholtz free energy and the partition function:\[\Large A-A (0)=-kT ~ \ln(Q)\]
We can from here calculate the pressure, \(p\), from statistical thermodynamics:\[\Large p=kT \left( \frac{ \partial \ln(Q) }{ \partial V } \right)_{V}\]For an ideal gas, indistinguishable independent particles, to which we can not see difference in, we can use that:\[\Large Q=\frac{ q ^{n} }{ N! }\]We thereby get: \[\Large p=kT \left( \frac{ \partial }{ \partial V } \ln \frac{ q ^{N} }{ N! } \right)_{T}=kT \left[ \frac{ \partial }{ \partial V }\left( \ln(q ^{N})-\ln(N!) \right) \right]_{T}\]\[\Large NkT \left( \frac{\partial }{ \partial V }\ln(q) \right)_{T}=nRT \left( \frac{ \partial }{ \partial V }\ln(q) \right)_{T}\]The number of molecules \( N = nN_{A}\) does not depend of the volume \(V\). The molecular partition function is a product of partition of the various types of movement, vibration, rotation, translation, and the movement of the electrons, \[\Large q=q ^{V}q ^{R}q ^{T}q ^{E}\]Only translational movement depend of the volume \(V\). The translational movement partition contribution is given by:\[\Large p=nRT \left( \frac{ \partial }{ \partial V }\ln \left( q ^{V}q ^{R}q ^{T}q ^{E} \right) \right)_{T}=nRT \left( \frac{ \partial }{ \partial V }\ln \left( q ^{T} \right) \right)_{T}\]\[\Large q ^{T}=\frac{ V }{ \Lambda ^{3} }\]Where \(\Lambda\) is the thermal wavelength, which does not depend on the volume either. So all we gotta do is to differentiate and remember that \(\large ln(x)\) is \(\large \frac{1}{x}\), which give us:\[\Large p=\frac{ nRT }{ V }\]Which is nothing else than the ideal gas equation.
We usually think of ideal gas equation as something you have measured your self forward to, but we have been derive using 3 theories.

Answer 2

nice !

Answer 3

Ok.

Answer 4

COOL WORK. Why not from kinetic theory of gas?

Answer 5

I came to the idea while trying to model a 2D particle system. So while reading my lecture notes I came to realize the relationship. I haven't tried for Gibbs free energy yet, however I think you can derive the ideal gas law from that too using same method.

How would you do so using kinetic theory of gases?