Let's say I have an adiabatic piston at $P_1$, $V_1$, and $T_1$. Then I throw a weight on it that doubles the original pressure: $P_2=2P_1$.

After some time the piston reaches a new equilibrium state at $P_2$, $V_2$, and $T_2$.

Then we remove the weight. My question: Does the piston return to its original state of $P_1$, $V_1$, and $T_1$?


1 Answer 1


My guess? No. Here is why. System is adiabatic so only work is being done ON to the system in the compression phase and only work is being done BY the system during the expansion phase. We can calculate the constant pressure work using: $$w=-P_{ext}\Delta V$$

Compression phase: $w= -P_2(V_2-V_1)=-2P_1(V_2-V_1)$

Expansion phase:$w= -P_1(V_3-V_2)$

If we were to assume that the expansion phase does the same magnitude of work as the compression phase did (which it probably doesn't), then at the very least:

$$\lvert w\rvert\ = -P_1(V_3-V_2)=-2P_1(V_2-V_1) \Rightarrow \lvert (V_3-V_2) \rvert\ =2\lvert (V_2-V_1) \rvert\ $$

Because $P_2$ is greater than $P_1$, the system gains more energy during the compression and has to expend less of that energy to fight a smaller external pressure during expansion.

Therefore, I think the last state reached will have $P_1$, $V_3$ (where $V_3>V_1$) , and $T_3$ (where $T_3>T_1$).

Also, since $T_3>T_1$, and since this gas is ideal, we know that not all of the work that was absorbed by the system during phase 1 (compression) was used again as work during expansion. The remainder of that energy was used to increase the internal energy of the system.

The the most important conclusion we can make here is this:

The internal energy of the system before the compression phase is LESS than the internal energy of the system after the expansion phase.

  • $\begingroup$ You cannot use work done=$P_{ext}$ multiplied by change in volume, because the pressure of the system does not remain constant throughout the process. $\endgroup$
    – Tejas P
    Mar 15, 2017 at 3:07
  • $\begingroup$ @TEJAS I think you are confusing reversible vs irreversible processes. $\endgroup$
    – Nova
    Mar 15, 2017 at 3:15
  • $\begingroup$ I think that relation is valid only for an isobaric process. Here, the gas is getting compressed irreversibly, the gas goes through a sequence of different pressures before it reaches pressure $P_2$ $\endgroup$
    – Tejas P
    Mar 15, 2017 at 3:34
  • $\begingroup$ @TEJAS P Both the compression and expansion are isobaric processes though. The external pressure isn't changing throughout the process. $\endgroup$
    – Nova
    Mar 15, 2017 at 14:11

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