[Physics] The pressure-volume work in thermodynamics


The change of pressure-volume work is given by:
$$\delta W = P\mathrm d V$$
Where $P$ denotes the pressure inside the system [1].

I wonder why the work is not given by:
$$\delta W = (P-P_{ext})\mathrm d V$$
For me this last expression is more logical, for instance if $P = P_{ext}$ then there is no mechanical work.

Can you explain please?

Best Answer

With your idea the $piston$ would be accelerating all of the time and some of the external work would be used to increase the kinetic energy of the "piston".

This idea crops up a lot in Physics.
You have a piston with a force of magnitude $F$ acting on a "piston" area $A$ due the gas pressure $P$.
To keep the piston from moving you need to apply an external force of magnitude $F'$ in the opposite direction to force $F$ and also make $F'=F$.

You now want to evaluate the work done by the external force in moving the piston.

This you can look at in a variety of ways.

You apply a slightly larger force $F'' >F$ on the piston just to get the piston moving (and give the piston some kinetic energy) and then once the piston is moving apply a force $F'$ on the piston which will now keep moving at constant velocity as the net force on it is zero.
All this time force $F'$ is doing work compressing the gas.
Just before the end point you reduce force the force on the piston to $F'''<F$ so that there is now a net force slowing the piston down and just as you reach the end point the piston stops moving having lost its kinetic energy and you apply a force $F'$ on the piston.
The total work done on the gas and piston is the work done by forces $F', F''$ and $F'''$ but there is a net zero amount of work done on the piston because it starts and finishes at rest.
However the net work done on the piston is zero because the extra bit of work which was done on the piston starting it was "given back" when the piston was slowing down at the end.
So the total work done on the gas was $F'\Delta x=F\Delta x=PA\Delta x= P \Delta V$

Or you could have the piston moving very, very very slowly from the start and evaluate the work done on the gas by the constant external force $F'$.

Or you could just evaluate the work done on the gas a $P\Delta V$ and assume an infinitely slow change of volume..