Work done by particle in a box in expanding the box?

[fordhamdining]

Homework Statement

I'm given that the energy of a particle in a rectangular box is the following:

$$E =\frac{\hbar \pi^2}{2m}(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2})$$

I'm to show that if the length of the box is increased adiabatically and quasistatically from L_x to 8L_x, the work done by the particle is 3/4 of the initial mean energy.

Homework Equations

The mean pressure exerted on the walls is
$$p=\frac{2E}{3V}$$
where V is the volume of the box. The force on the wall parallel to the x-axis is given by
[tex] \frac{-\partial E}{\partial L_x} = \hbar \pi^2n_x^2/mL_x^3[/tex][/B]

The Attempt at a Solution

I tried calculating the work by integrating pdV from 1 to 8 (since increasing one dimension by a factor of 8 increases the volume by the same) but I did not get the 3/4 I needed. There was a factor of ln(8) that could not possibly simplify to 3/4. Then I figured since the expansion is only happening in the x-direction, I can integrate F_xdL_x from 1 to 8 but I still didn't get the right answer. Am I missing something?

 

[TSny]

Welcome to PF!

If you are allowed to start with $p = \frac{2E}{3V}$, then I think the result follows without needing the formula you gave in section 1 for the energy levels of a particle in a box.

The wording of the question is a little unclear to me. Are you meant to assume that the particle starts out in a definite energy eigenstate corresponding to some particular (but unspecified) values of $n_x, n_y$ and $n_z$? If so, then I don't understand the phrase "3/4 of the initial mean energy". If the particle is initially in a specific energy eigenstate, then it seems to me that there is no need to use the word "mean" here.