Assumed value of pressure during quasistatic compression

[beefbrisket]

My text (Ian Ford - Statistical physics) describes an ideal gas system in a piston being quasistatically compressed by a piston head of area $A$ under external force $f$. It assumes the system has a uniform pressure $p$. All good so far. Then it says: "the force $pA$ equals the applied external force $f$" on its way to showing $\text{d}W = -p\text{d}V$. I have some feeling that this seems sound but I cannot explain why. Is this just a good approximation we make due to the quasistatic nature of the compression or some other reason? I think they cannot exactly equal otherwise there would be no work done due to zero net force.

 

[Chestermiller]

My text (Ian Ford - Statistical physics) describes an ideal gas system in a piston being quasistatically compressed by a piston head of area $A$ under external force $f$. It assumes the system has a uniform pressure $p$. All good so far. Then it says: "the force $pA$ equals the applied external force $f$" on its way to showing $\text{d}W = -p\text{d}V$. I have some feeling that this seems sound but I cannot explain why. Is this just a good approximation we make due to the quasistatic nature of the compression or some other reason? I think they cannot exactly equal otherwise there would be no work done due to zero net force.

If the compression is very slow (quasi static), the force per unit area exerted by the gas on the piston will only be differentially lower than f. So any negative displacement -dx of the piston will result in work -fdx being applied to the gas. But this is just -pdV.