Question

air that initially occupies 0.140 m^3 at a gauge pressure of 103.0 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and the atmospheric pressure).

Answer 1

Given data:
volume = 0.140m^3
gauge pressure of 103.0 kP
pressure of 101.3 kPa
Work done by the air is given by the integral
W = ∫ p dV from initial final state

In first step we expand at constant temperature. i.e.
p∙V = n∙R∙T = constant
hence:
p = p₁∙V₁/V
and
W₁ = ∫ p₁∙V₁/V dV from V₁ to V₂
= p₁∙V₁∙ln( V₂/V₁)

because p₁∙V₁ = p₂∙V₂ <=> V₂/V₁ = p₁/p₂

W₁ = p₁∙V₁∙ln(p₁/p₂)

Add standard atmospheric pressure to initial gauge pressure:
p₁ = 103 kPa + 101.3kPa = 204.3kPa
W₁ = 204300Pa ∙ 0.140m³ ∙ ln(204.3/101.3) = 20064.4J

In second step we operate at constant pressure
W₂ = ∫ p dV from V₂ to V₁
= p₂ ∫ dV from V₂ to V₁
= p₂ ∙ (V₁ - V₂ )

with p₁∙V₁ = p₂∙V₂ => V₂ = (p₁/p₂∙)∙V₁

W₂ = p₂∙V₁∙ (1 - (p₁/p₂∙)) = V₁∙ (p₂ - p₁)
= 0.140m³∙ (101300Pa - 204300Pa)
= -14420J
So the net work done by the gas is
W_tot = W₁ + W₂
= 20064.4J - 14420J
= 5644J