In the problem there is a cylinder with a massless and frictionless piston, initially at rest. A weight is placed on the piston and the gas undergoes an adiabatic compression. The problem specifies the atmospheric pressure $p_0$. The task is to find the final temperature of the gas. In the solution, they formulate the following statement:

The mechanical work done by the gas is equal in magnitude and opposite in sign to the mechanical work done by the weight of the object.

I would argue that the atmospheric pressure also does some work on the piston, so the correct equation would be $$W_\text{gas}=-(Mg+p_0S)\Delta h$$

They use $p_0$ only to calculate the initial temperature $T=\frac{p_0V}{\nu R}$.

Is the atmospheric pressure work negligible just because $S$ and $\Delta h$ are probably too small for $p_0$ to have an effect, or is there something else to it?

  • $\begingroup$ Can we assume $S$ is the piston cross section area? In computing work you are correct that atmospheric pressure needs to be included. Since the equilibrium pressure in the cylinder prior to compression is absolute pressure, so must be the pressure doing the work on gas be absolute. $\endgroup$
    – Bob D
    Commented Jun 21, 2023 at 19:14
  • $\begingroup$ Back in the old days, and is also how we currently think of things, the objects are some heavy bars, and it is easy to see that atmospheric pressure is negligible compared to them. You are definitely correct that pressure ought to be considered too, if you want really accurate results. $\endgroup$ Commented Jun 22, 2023 at 4:39
  • $\begingroup$ Thank you but gentlemen, I find that the bigger issue is how to actually aolve the problem. When I first wrote my solution I used the equilibrium of the piston $p=p_0+\frac{Mg}{S}$ and $pV^\gamma=C$. What I do not agree about the internal energy approach is that, theoretically, for a frictionless piston, when it reaches the equilibrium position (null acceleration), it would still have some momentum which is kinetic energy. Thus we cannot say that work done by external forces fully converts to internal energy. These two approaches lead to different results. $\endgroup$
    – Neox
    Commented Jun 22, 2023 at 5:48
  • $\begingroup$ @BobD Yes, $S$ is the cross section area. $\endgroup$
    – Neox
    Commented Jun 22, 2023 at 5:50
  • 1
    $\begingroup$ @Neox $pV^{\gamma}=C$ applies to an ideal gas reversible adiabatic process. Your compression process is not reversible if the weight is applied all at once $\endgroup$
    – Bob D
    Commented Jun 22, 2023 at 11:58


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