The situation is a thermally insulated container divided into two compartments, which are separated by a diathermic piston. Initially, the left compartment has a volume (V1) of 2L, and a temperature of 400K, and the right has a volume (V2) of 4L and a temperature of 200K. Both contain one mole of ideal gas each. The piston is first clamped, and then released. The question states that work associated with the movement of the piston may be neglected. The question asks to describe the graph of volume and pressure versus time for this spontaneous change, and these were the correct solutions (pictures below).

I calculated that the final volumes are 3L, and the final pressures are 100 atm in both compartments. (using the fact that that temperatures equalise to 300K while the piston cannot move, final pressures of both compartments must be equal, and number of moles in each compartment must remain the same)

What I do not understand is the manner in which this final 3L is reached. Specifically, I do not understand why V1 rises to a peak and then reduces and why V2 decreases to a minimum and then increases to the final value.

I've researched any other similar piston questions, but could not find an explanation for it. Could anyone please help with this? (I am in 12th grade, and this question is from an exam meant to be given by 12th graders)

image of a volume versus time curve. At t = 0, V1 has a value of 2L. the graph of V1 has a steep positive slope, and it reaches a maximum of 4L, before decreasing over a significantly longer time to a final volume of 3L. At t = 0, V2 has a value of 4. The graph of V2 has a steep negative slope, and it reaches a minimum of 2;, before increasing over a significantly longer time to a final volume of 3L

  • $\begingroup$ That seems an unnecessary complication added to the problem. It looks like the response of a damped oscillator (see for instance en.wikipedia.org/wiki/…). What happens is that the hotter and compressed compartment initially expands while losing heat to the second one. The piston moves into the second compartment with some inertia which results in the volume of the first compartment expanding past the equilibrium (halfway) point. The details are less important than the final condition of $V_1=V_2$. $\endgroup$
    – Buck Thorn
    Commented Jul 21, 2021 at 12:55
  • $\begingroup$ @BuckThorn thank you so much for taking the time to reply, I understand what's happening now! these were only 2 of the options, the other one was a graph with a shape similar to the pressure one but for volume, with the same condition of V1 = V2, hence my confusion. I have a follow up question, if that's okay: why does't the pressure oscillate like the volume? and what causes the damping here? as I understand it, damping usually involves some form of energy loss, but I can't figure out where energy might be lost here. $\endgroup$
    – froog
    Commented Jul 21, 2021 at 13:07
  • $\begingroup$ They must also give you data on the thermal conductivity of the piston, right? $\endgroup$ Commented Jul 21, 2021 at 18:18
  • $\begingroup$ @ChetMiller it’s a diathermic piston! Sorry, I didn’t have the space to include that in the main question $\endgroup$
    – froog
    Commented Jul 21, 2021 at 18:24
  • $\begingroup$ They just say it is diathermic, or do they give you more information about the piston? $\endgroup$ Commented Jul 21, 2021 at 18:33


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