How does the general displacement work expression arise for a piston?

Force F = PA times displacement dx gives dW = P dV

Work = mass times acceleration times dx

Work = V dP, since pressure changes during motion

Work = temperature change times pressure

What does W12 = integral of P dV represent, and what does it depend on?

Area under the P-V curve; it depends on the path

A state property that depends only on end states

Area under the T-S curve; it is path independent

Area to the left of the curve, independent of path

For an ideal gas at constant T, why does the integral give a logarithm?

P = mRT/V so the integral of dV/V is ln(V2/V1)

P is constant so W = P(V2 - V1)

PV^n = C with n = 1.4 gives a power law

Because dV = 0 for constant temperature

2068 Magh · 4 marks

General displacement work; constant temperature

Short answer

For a reversible piston-cylinder device, displacement work is W = ∫ P dV. For an ideal gas undergoing constant temperature process, work is given by mRT ln(V2/V1) = P1V1 ln(P1/P2).

Derive a general expression for the displacement work transfer for a piston cylinder device. Also reduce it for an ideal gas undergoing constant temperature process.

General expression

Let a gas at pressure P be in a frictionless piston-cylinder of area A. The force on the piston is F = PA. For a small displacement dx:

δW = F dx = P A dx = P dV

For a reversible (quasi-static) process from state 1 to 2:
W₁₂ = ∫₁² P dV

This equals the area under the curve on the P-V diagram; it is positive in expansion and negative in compression, and depends on the path.

Isothermal process of an ideal gas

PV = mRT = constant, so P = mRT/V.

W₁₂ = mRT ∫₁² dV/V = mRT ln(V₂/V₁)

Since P₁V₁ = P₂V₂ = mRT:

W₁₂ = P₁V₁ ln(V₂/V₁) = P₁V₁ ln(P₁/P₂) = mRT ln(V₂/V₁)

✓ Answer checked by Claude Opus 5.5 (high effort)

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