Suppose a massless, frictionless piston assembly initially has a higher pressure than the external (atmospheric) pressure, and it is pinned so that the piston does not move. Once the pin is removed, the piston would expand until the pressure inside the piston becomes the atmospheric pressure. During the process, the work done by the gas inside the piston is
$$W_{\text{piston}}=\int_{V_1}^{V_2} P_{\text{gas}}\cdot \mathrm{d}V$$
and the work done by the surrounding is,
$$W_{\text{ext}}=\int_{V_1}^{V_2}P_{\text{ext}}\cdot \mathrm{d}V = P_{\text{ext}} \left(V_2 – V_1 \right)
\,.$$
We can pull out the external pressure from the integral because it is constant as an atmospheric pressure.
My question is, the work done by the piston is not the same with the work done by the surrounding because $\mathrm{d}V$ is the same, but ${P}_{\text{gas}}$ is greater than ${P}_{\text{ext}}$ during the process, so the work done by the piston is larger than that by the surrounding. Shouldn't they be the same?
Best Answer
If you do a free body diagram on your massless, frictionless piston, you can only conclude that the force exerted by the gas on the inside face of then piston is equal to the external force exerted on the outside face of the piston. That means that the work is the same. So, what gives?
Well, first of all, in a rapid irreversible expansion like this, the pressure of the gas within the cylinder is not uniform (spatially). It is lower at the piston face than on average within the cylinder. Secondly, a gas experiencing a rapid deformation does not obey the ideal gas law even locally; there are viscous stresses in the gas (proportional to the rate of deformation) which contribute to the force at the inner piston face. So, without solving the partial differential equations for aerodynamics, we have very little knowledge of the force on the inner piston face.
Also, as safesphere points out, if there is a gas present (like the atmosphere) external to the piston and cylinder, analogous effects can occur within this gas. However, we typically assume that we have better control over what is happening with the external load (such as through feedback control systems, or by having a vacuum outside and using a piston with mass). So, whereas, the knowledge of the internal force on the piston requires solution of the partial differential equations for aerodynamics, for textbook thermodynamics problems, we usually assume that we can impose the external load precisely. We are thereby able to more easily calculate the amount of work done by the gas on the surroundings.
Of course, if the expansion is done reversibly (say, by imposing a very gradually varying external force on the gas within the cylinder), the work can be calculated using the ideal gas law to determine the force on the inner piston face, since, in that case, viscous stresses and non-uniformities within the gas are negligible.