- #51

Chestermiller

Mentor

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f is different in the frictionless case from the case with friction. In the frictionless case, $$f=PA$$ and $$nC_vdT=-PdV$$In the case with friction, $$f=PA-F$$ and $$nC_vdT=-PdV+\frac{F}{A}dV$$So, for the same change in volume, the temperature change is different.I'll make another attempt, this time with equations.

I'll treat only the ideal gas as system and assume a quasistatic, adiabatic expansion.

I'll assume that the cylinder and piston have negligible mass and heat capacity and that the thermal insulation is such that no friction heat can pass into the environment.

I'll discuss two cases, one without friction and one with friction between cylinder and piston.

I'll use the same letters for the forces as in post #8.

The first law for the gas (the system) alone is

dU = dQ - dW

Frictionless case:

dQ = 0

dW = f/A dV

Therefore: dU = - f/A dV

Case with friction:

Here all the friction work is converted to heat that goes into the gas!

dQ = F/A dV

dW = F/A dV + f/A dV

Therefore: dU = - f/A dV

If I assume that the initial state of the system was the same in both cases, then the final state must also be the same, since U changed by the same amount. Therefore dS for the system is the same in both cases.

In the frictionless case dS = 0 for both system and environment.

In the case with friction dS = 0 for the system because of the reasons given above and for the surroundings it's also 0 because no heat flows into them.

I conclude that the case with friction is also reversible.