Answer to Question #156509 in Mechanical Engineering for Rodel

Question

A certain fluid in a closed system initially at a pressure of 200 psia and having a volume of 3 ft³ undergoes a frictionless process during which the product pV^1.3 = C. If p₂ = 40 psia, compute the work done by or on the system.

Answer 1

pV^1.3 = c

p₁ = 200 psia = 1.38 x 10⁶ Pa

V₁ = 3 ft³ = 0.085 m³

p₂ = 40 psia = 2.76 x 10⁵ Pa

W = $\smallint$ pdV = p₁V₁ⁿ$\smallint$ dV/Vⁿ = (p₂V₂ - p₁V₁) / (1 - n) = (p₁V₁ - p₂V₂) / (n - 1)

c = p₁V₁^1.3 = 1.38 x 10⁶ Pa x (0.085 m³)^1.3 = 847.5

V₂ = (c/p₂)^1/1.3 = (847.5 / 2.76 x 10⁵)^1/1.3 = 0.011 m³

W = (1.38 x 10⁶ x 0.085 - 2.76 x 10⁵ x 0.011) / (1.3 - 1) = 380 880 J = 381 kJ