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

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data

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

    [tex]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})[/tex]

    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.

    2. Relevant equations

    The mean pressure exerted on the walls is
    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]

    3. 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?
  2. jcsd
  3. Jan 29, 2017 #2


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    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.
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