# Derivation of ideal gas law by Hamiltonian mechanics

1. Jun 6, 2013

### Hausdorff

Hi!

I am trying to understand the statistical mechanics derivation of the ideal gas law shown at: http://en.wikipedia.org/wiki/Ideal_gas_law inder "Derivations".

First of all, the statement "Then the time average momentum of the particle is:
$\langle \mathbf{q} \cdot \mathbf{F} \rangle= ... =-3k_BT.$". Isn't this wrong? For sure, the dimension of $\langle \mathbf{q} \cdot \mathbf{F}$ is energy, as is the dimension of $-3k_BT$ and not momentum.

Secondly, I do not understand the equation
$-\langle \sum_{i=1}^{N}\mathbf{q}_k \cdot \mathbf{F}_k \rangle = P \oint_{\mathrm{surface}} \mathbf{q} \cdot d\mathbf{S}$.
I dot not understand what $\mathbf{q}$ is, since all position vectors have been subsricpted. Nor do I understand the physical interpretation of $\mathbf{q} \cdot d\mathbf{S}$: the scalar product of $\mathbf{q}$ (a position vector?) and the vector area element.

If anybody could shed some light on this, I would be very grateful.

2. Jun 7, 2013

### Andrew Mason

I agree with you there. It should be "time average kinetic energy of the particle". You should post a talk comment in the Wikipedia article.

Pressure is the average force/unit area of the gas. So the time average of the force created by a single particle on the container is F = PΔS where ΔS is the area of the container wall that is interacting with the particle (ie. in a collision between the particle and the wall). So $\oint PdS$ is the total force on all the particles of the gas.

Where it gets a little unclear is setting the total energy to P∫qds. This does not explain the physics. A better explanation can be found on the Wikipedia page on Kinetic theory.

AM

3. Jun 8, 2013

### Hausdorff

I have been discussing this on another forum as well. There it was explained to me that q and F arise from approximation of the discrete model with a continuous one, so that the derivation with additional steps, would look something like
$-\langle \sum_{k=1}^{N} \mathbf{q}_k \cdot \mathbf{F}_k \rangle \approx \int_{\mathrm{volume}}\mathbf{q}\cdot \mathbf{F}dV= \oint_{\mathrm{surface}}\mathbf{q}\cdot (Pd\mathbf{S})=P\oint_{\mathrm{surface}}\mathbf{q} \cdot d\mathbf{S}...$,
where the second is valid to due F vanishing everywhere except for at the surface.

I have looked at the more intuitive derivation on the wiki page on kinetic theory. However, I wanted to understand the more rigorous statistical mechanics proof as well.

4. Jun 9, 2013

### Andrew Mason

I don't follow this. It does not appear to be dimensionally correct. This has dimensions of Force x distance x volume.

AM

5. Jun 9, 2013

### vanhees71

This is somewhat misleading since the equipartition theorem holds only true for quadratic forms of the canonical variables, i.e., for the harmonic oscillator. The classical grand-canonical partition function is given by
$$Z=\sum_{N=0}^{\infty} \exp(\alpha N) \frac{z^N}{N!} = \exp(z \exp \alpha),$$
with the single-particle partion function
$$z=\int \mathrm{d}^3 \vec{x} \mathrm{d}^3 \vec{p} \exp \left [-\beta \left (\frac{\vec{p}^2}{2m}+\frac{\omega^2}{2} \vec{x}^2 \right ) \right ]=\frac{8 \pi m^{3/2}}{\beta^3 \omega^3}.$$
Then you get
$$\langle N \rangle=\frac{\partial \ln Z}{\partial \alpha}=\frac{8 \pi^3 m^{3/2} \exp \alpha}{\beta^3 \omega^3}$$
and
$$\langle H \rangle = -\frac{\partial \ln Z}{\partial \beta} = 3 \frac{\langle N \rangle}{\beta}.$$
As you see only in a harmonic oscillator potential you get the mean energy in the form given in Wikipedia.