# Work on ideal gas by piston

Imagine a thermally insulated cylinder containing a ideal gas closed at one end by a piston.

If the piston is moved rapidly, so the gas expands from $V_i$ to $V_f$.

The expanding gas will do work on the piston, how much work is done if the piston is moved away at a speed $w$ ?

• To work this out I think you will need to use the fact that the gas has a Maxwell–Boltzmann. Sep 13, 2012 at 17:06
• and determine the distribution of the molecules that collide with the piston. Then you work out energy lost after the collision. You then integrate over the distribution to find the answer. How do you do this? Sep 13, 2012 at 17:13
• What part of this problem is giving you trouble? Do you know the definition of work? Do you understand the sign conventions in use in this system? Do you know the equation of state of an ideal gas? Can you deduce how much heat might be gained or lost during this process? Sep 13, 2012 at 19:39
• So there is no heat gained or lost. The aim is to calculate the work done, I dont now what else to do Sep 13, 2012 at 23:04

The question is not well posed. Work is done only when there is an effect on the outside world. If the outside of the piston is vacuum, NO work is done. If the outside of the piston is under a pressure, then work is done lifting the outside atmosphere.

In any case the basic definition of work in this case is:

$$w = -\int_{V_1}^{V_2} p_{ext} dV$$

where $w$ is the work, $V_1$ is the initial volume of the system, and $V_2$ is the final volume of the system. Last, $p_{ext}$ is the external pressure.

In order to do this integral one must know how the external pressure changes as the internal volume changes. Thus it is easy if $p_{ext}$ is constant, but it is hard in general.

The speed of the expansion makes no difference at all, at least under the usual assumptions of thermodynamics which here are that the piston massless and frictionless. The speed of the expansion will affect the power generated by the process, but that's not part of thermodynamics.

• I agree with your answer but would like to add that the speed matters if we are thinking of non-equilibrium thermodynamics. For most practical purposes we can get away by assuming it to be quasi-equilibrium so that we can use the equilibrium state relations. Sep 14, 2012 at 5:44
• @mythealias second viscosity for normal gases is zero, no dissipation will have place, except for the boundary layer at walls. That's just like sounds, I bet you know the derivation of the sound speed in gases, the process is rapid, but you use equilibrium relations and assume no dissipation. Of course if the piston is not so rapid to cause shock waves. Sep 14, 2012 at 9:44