Say I have an ideal gas that has a known $P_1$, $P_2$, $T_1$, and $T_2$ undergoing a reversible adiabatic process. I want to find the work done so I must use $PV = RT$ to get the change in $V$, so that's $$\frac {P_1V_1}{T_1} = \frac {P_2V_2}{T_2}$$, and also $$W = \int PdV = P(V_2 - V_1)$$.

My problem is, these are only 2 equations and 3 unknowns. How do I solve $V_1$, $V_2$, and $W$? Also, in the second equation, is the value of $P$ equal to $P_1$ or $P_2$?

  • 2
    $\begingroup$ $\int P(V) dV = P(V_2 - V_1)$ only if the pressure is constant. If the pressure is changing then you need to actually do the integral, which means knowing the pressure as a function of volume. $\endgroup$
    – Michael
    Mar 27, 2013 at 2:35
  • $\begingroup$ Ah. Unfortunately I'm not given that function. Is there another way for me to get the specific work? $\endgroup$ Mar 27, 2013 at 2:40
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    $\begingroup$ @markovchain: the process is given as adiabatic, and the gas obeys the ideal gas equation of state... what does that tell you about the pressure? $\endgroup$ Mar 27, 2013 at 2:43
  • $\begingroup$ I think you're implying the pressure is constant... which would make sense as the set up is a piston cylinder. But I've been given values for $P_1$ and $P_2$. I'm really sorry for not knowing, but I honestly don't know. $\endgroup$ Mar 27, 2013 at 2:52
  • $\begingroup$ Oh, wait, do you mean that the function of increase is linear? $\endgroup$ Mar 27, 2013 at 2:58

1 Answer 1


You were onto it in the comments, so I might be late to offer anything new here. The pressure is irrelevant in this problem; it's a trick, I guess. A reversible adiabatic process is one in which there's no heat flow in or out of the gas, so all of the work done in the expansion/compression goes into the temperature change. Just calculate the change in energy like you were saying ($U_2-U_1$) and that's the work done (on it, not by it)! Good job figuring it out!

Also, as with all physics problems, make sure that the negative signs match your intuition. I can't always keep $-W$ and the like straight on paper, especially since I often mix up whether $W$ represents the work done by the gas or the work done on the gas. I'm pretty sure it's the former. So, if the gas expands, it does work, and $W>0$. If the gas contracts, work is done on it, so the work the gas does is negative.

  • $\begingroup$ I did figure it out. But it's still nice to have an answer to accept. I realized that $U_2 - U_1$ only depends on $T$ and not $P$ but only after I searched my formulas. I dont quite remember all of them yet. But still, thanks! $\endgroup$ Mar 27, 2013 at 4:27
  • $\begingroup$ No problem! It's much better that you figured it out on your own. I suppose the answer stands for future inquirers with a similar question, since that's part of what Stack Exchange is for. $\endgroup$
    – krs013
    Mar 27, 2013 at 4:44

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