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Gas work: constant P, constant T, PV^1.35

Short answer

Constant pressure: W = P(V₂-V₁) = 9.0 kJ. Isothermal: W = P₁V₁ ln(V₂/V₁) = 5.50 kJ. Polytropic (n=1.35): P₂ = P₁(V₁/V₂)^1.35, then W = (P₂V₂ - P₁V₁)/(1-n) = 4.70 kJ.

A gas is contained in a piston cylinder device initially at a pressure of 150 kPa and a volume of 0.04 m³. Calculate the work done by the gas when it undergoes the following processes to a final volume of 0.1 m³, (i) Constant-Pressure (ii) Constant temperature (iii) PV^1.35 = constant.

Given: P₁ = 150 kPa, V₁ = 0.04 m³, V₂ = 0.1 m³. Using W = ∫P dV (kPa × m³ = kJ).

(i) Constant pressure

W = P(V₂ − V₁) = 150 × (0.1 − 0.04) = 9.0 kJ

(ii) Constant temperature (PV = C)

W = P₁V₁ ln(V₂/V₁) = 150 × 0.04 × ln(0.1/0.04)
= 6.0 × ln(2.5) = 6.0 × 0.9163 = 5.50 kJ

(iii) PV^1.35 = constant

P₂ = P₁ (V₁/V₂)^1.35 = 150 × (0.4)^1.35 = 150 × 0.2903 = 43.54 kPa

W = (P₂V₂ − P₁V₁)/(1 − n) = (43.54 × 0.1 − 150 × 0.04)/(1 − 1.35)
= (4.354 − 6.0)/(−0.35) = 4.70 kJ

Answer: W = 9.0 kJ (isobaric), 5.50 kJ (isothermal), 4.70 kJ (polytropic, n = 1.35). The areas under the curves on the P-V diagram decrease in this order, since the pressure falls faster for larger n.

✓ Answer checked by Claude Opus 5.5 (high effort)

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