# Work done by an ideal gas: generalising a cylinder to an arbitrary shape

In thermodynamics, we usually find the work done by/on an ideal gas by considering a cylinder with a moveable piston as one of its flat ends. We consider the work done by moving the piston end a small amount, and find the well known result that

$$dW = P dV$$ (up to sign convention and inexact differential notation).

In books I've used (Finn, Zemansky), it's then asserted that this relationship generalises to systems of arbitrary shapes. However, this seems highly non-obvious to me, and they don't provide a sketch of why this result is generalisable.

Could anyone shed some light on this? I'm not necessarily looking for an ironclad argument -- more, just an indication on why it is sensible to generalise this. Thanks!

• I would ask you the opposite, where in the derivation of dW=PdV it is assumed that the shape is a cylinder?
– user65081
Jun 14, 2020 at 18:43
• Basically it adapts to every shape because $dV$ can shape anything with the correct parametrizations Jun 14, 2020 at 18:52
• I see your point. While $PdV$ does adapt to every shape, all the pictures and examples in text books seem to be for the easy cylinder and piston example. It is easier to calculate $dW$ with a force holding a piston back than, say, a stretchable spherical balloon. You can calculate the work on the balloon, but you have to account for the energy used to stretch it. Even though it works out, it distracts from the main point about the ideal gas. So a piston is a better example to start with. Jun 14, 2020 at 19:04
• Who says it has to be an ideal gas for the equation to apply? Jun 14, 2020 at 21:38

If S is the surface surrounding the gas at any time, then the differential force vector acting on a differential element of area of this surface dA is $$\mathbf{dF}=p\mathbf{n}dA$$where p is the gas pressure and $$\mathbf{n}$$ is the unit normal to the surface. If $$\mathbf{v}$$ is the velocity of the gas at the surface, the rate of gas doing work on the differential element of surface area is $$\frac{dW_{dA}}{dt}=\mathbf{v}\cdot \mathbf{dF}=p(\mathbf{v}\cdot \mathbf{n})dA$$If we integrate this over the entire surface S surrounding the gas at time t, we obtain the total rate at which the gas does work on its surroundings: $$\frac{dW}{dt}=\int_S{p(\mathbf{v}\cdot \mathbf{n})dA}=p\int_S{(\mathbf{v}\cdot \mathbf{n})dA}$$But, kinematically, the rate of change of gas volume is just $$\frac{dV}{dt}=\int_S{(\mathbf{v}\cdot \mathbf{n})dA}$$Therefore, $$\frac{dW}{dt}=p\frac{dV}{dt}$$