A gas at a pressure of 500 kPa and volume of 0.75 m^{3} is contained in a cylinder-piston assembly. When the piston moves slowly in the cylinder, the pressure inside the cylinder varies as V^{–1.2}. If the final volume of gas becomes doubled, then the work done by the gas, in kJ, is _______

This question was previously asked in

GATE PI 2014 Official Paper

**Concept:**

**\(Work \ done\ in\ closed \ system \ (W) = \int Pdv\)**

__Calculation:__

__Given:__

Initial pressure (P_{1}) = 500 kPa, Initial volume (V_{1}) = 0.75 m^{3}, final volume V_{2} = 1.5 m^{3}

∵ Pressure P ∝ V-1.2

PV^{1.2} = C

\(P_1V_1^{1.2} = P_2V^{1.2}_2\)

500 × 0.75^{1.2} = P_{2} × 1.5^{1.2}

P_{2 } = 217.6376 kPa

\(W = \int Pdv\)

\(W = \frac{P_1V_1\ -\ P_2V_2}{n\ -\ 1}\)

\(W = \frac{500\ \times \ 0.75\ -\ 217.64\ \times 1.5}{1.2\ -\ 1}\)

**W = 242.7 kJ**