# Ideal gas subject to central potential [Statistical Mech]

1. Mar 13, 2014

### Telemachus

The problem says: Consider an ideal gas of N particles in a spherical vessel of radius R. A force acts directly over the molecules and is directly proportional to the distance to the center of the sphere $V(r)=\alpha r$. Calculate the pressure of the gas, and the density of particles at the surface.

So I set the Hamiltonian: $\displaystyle H= \sum_i^N \frac{p_i^2}{2m}+\sum_i^N \alpha r_i$

Then, as the potential is independent of the momentum, I can take $Z=\frac{1}{N!}Z_1^N$, where $Z_1$ represents the partition function for only one molecule.

Then:

$\displaystyle Z_1=\frac{1}{h^3}\int_{-\infty}^{+\infty}e^{-\beta \frac{p^2}{2m}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R} dr d \phi d \theta r^2 \sin \theta e^{-\beta \alpha r}$

Now, the pressure in the canonical ensemble is proportional to the derivative with respect to the volume of the natural logarithm of the partition function. And the thing is I don't have any volume anywhere in there (I've done the integral, and I don't get the volume), so I get a pressure equal to zero. Something is wrong with what I've done.

2. Mar 13, 2014

### BvU

Can you show how the $\int_0^R r^2 e^{-\beta\alpha r}$ does not leave some power of R in Z ?

3. Mar 13, 2014

### tman12321

BvU is correct. You should end up with terms like exp(-R) and polynomials in R. Since it's a sphere you know that R is proportional to the cube root of the volume.

4. Mar 14, 2014

### Telemachus

The integral gives: $\displaystyle 4\pi \frac{e^{-\beta \alpha R}}{(\beta \alpha)^3} \left [ (\beta \alpha R)^2+ 2 \beta \alpha R + 2\right ]$

Now, the volume of the sphere is $V=\frac{4\pi}{3}R^3$ I could arrange this to get the volume in the polynomials of R:

$\displaystyle Z_1=\left ( \frac{2\pi m}{\beta h^2} \right ) \frac{\exp\left [-\beta \alpha \frac{3}{4 \pi R^2}V\right ]}{(\beta \alpha)^3}\displaystyle \left [ (\beta \alpha)^2\frac{3}{R}V+ 2 \beta \alpha \frac{3}{R^2}V + 2\right ]$

I've used that $R=\frac{4\pi}{3}R^3 \frac{3}{4 \pi R^3}R=V\frac{3}{4 \pi R^2}$

And then the pŕessure $\displaystyle p=k_B T \frac{\partial \ln Z}{\partial V}$

Is that right?

Last edited: Mar 14, 2014
5. Mar 14, 2014

### tman12321

You would really want to eliminate R--in favor of a term like V^(1/3)--from the expression entirely. It's also probably easier to use the chain rule here. Take a derivative with respect to R and then multiply by dR/dV.

6. Mar 14, 2014

### Telemachus

Yes. I thought it was a bad idea to have all those R all around, but I didn't want to introduce the cube root neither :P

Thanks.