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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?

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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.

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  • $\begingroup$ But, the conditions he mentioned in the question seems to resemble the adiabatic expansion process(because in adiabatic expansion,the only way for expansion is that the initial pressure should be high since heat cannot be added else the piston should be pulled).I have seen the ideal gas equations being applied to these processes.Is that an approximation? $\endgroup$ – user196272 Sep 10 '18 at 15:29
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    $\begingroup$ The ideal gas equation can only be used for the very initial state and the very final state of an irreversible expansion. For the variations between these end states, the ideal gas law does not give the correct relationship. This is because, the ideal gas law only applies to thermodynamic equilibrium states. For a reversible expansion, the gas passes through continuous sequence of thermodynamic equilibrium states, so it can be applied to all these states. In my judgment, everything you have said in this comment is incorrect. $\endgroup$ – Chet Miller Sep 10 '18 at 15:58
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The pressure on both sides of the unpinned weightless piston is always the same. In case of any difference, the piston would quickly move pressuring the external gas locally above the atmospheric pressure. Your setup is not static and only can be solved by aerodynamics. However, in the limited scope of your question, the answer is that the pressure is always the same on both sides.

In reality, the piston is not weightless, so the initial pressure difference would be compensated by the inertial force of the piston acceleration untill the dynamic pressure is the same on both sides.

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  • $\begingroup$ Are you saying that when the pin is removed, the pressure inside the piston reaches external pressure instantly, and then the expansion happens thereafter? If this is the case, what force causes the piston to expand? If the pressure inside the piston is the same with the external pressure, the net force is zero. $\endgroup$ – Jeong Won Kim Oct 13 '17 at 4:59
  • $\begingroup$ No. In your case the initial internal pressure is higher. When you remove the pin, the weightless piston quickly moves creating a compression in front of it (thus making the local external pressure higher) and decompression behind it (thus making the local internal pressure lower). The force on the piston is $(P_i-P_o)\cdot S=ma$ where $m=0$ therefore $P_i=P_o$ where $P_i$ and $P_o$ are the local dynamic inside and outside pressure, $S$ is the area of the piston, $m$ is its mass, and $a$ is its acceleration. Think of the piston as imaginary. It is the internal gas pushing the external gas. $\endgroup$ – safesphere Oct 13 '17 at 5:34
  • $\begingroup$ There would be a smooth gradient of pressure that translates to a zero difference in pressure on two sides of a thin membrane. If the piston is not massless, then there would be a step in the gradient according to the formula above for $m\ne 0$. $\endgroup$ – safesphere Oct 13 '17 at 5:41
  • $\begingroup$ See also if this helps: physics.stackexchange.com/questions/45653/… - The inside and outside pressure act on different objects. The inside pressure puts a force (through the weightless oiston) on the mass of the nearby external gas and vice versa. $\endgroup$ – safesphere Oct 13 '17 at 6:39
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The only problem consists in the fact that you consider that the atmospheric pressure will remain constant. In fact it increases. Let me explain:

When your piston expands it reduces the atmosphere volume as a whole from $V_{0\mathrm{atm}}$ to $$V_{1\mathrm{atm}}=V_{0\mathrm{atm}}- \Delta V \,.$$This reduction in volume results in an increase of the atmospheric pressure from $p_{0\text{ext}}$ to $$p_{1\text{ext}}= \frac{p_{0\mathrm{atm}}}{1- \frac{\Delta V}{V_{0\text{atm}}}} \,.$$Even thought the increase in pressure is tiny when we consider $\Delta V \ll V_{0\mathrm{atm}}$ (as it always is) to the point that one could consider $p_{0\text{ext}}$ constant, the missing energy will be dispersed throughout the planet as potential energy, just as much as you store energy in a pressurized bottle.

In this model I’ve consider that the atmosphere was closed at it's top. Still, Even if you consider the open-top scenario, instead of just an increase in atmospheric pressure, your system’s expansion would push up the top of atmosphere by a certain amount. Anyway, an increase in atmospheric height would result in an increase of its gravitational potential energy and therefore work-energy balance holds.

To make this clear, consider what would happen if the increase in volume of your piston would not be small. Let’s say, it occupied half of the planet. All atmosphere would be displaced to the other part of the planet and atmospheric pressure would double there. The work performed by your system would be stored as pressurized air in the double atmosphere part of the planet.

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The work done varies because the piston will be accelerated at a higher rate.In case of contant pressure expansion(both internal and external pressures are same at all instances) the piston moves slowly.Because the force will be just enough to move the piston.But in your case the force inside will be sufficiently higher so the piston moves faster(i.e. accelerated).I had the same doubt and still doubtfull about it.

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