Question

During a reversible adiabatic expansion of an ideal gas, the pressure 

and volume at any moment are related by PVγ = c where c and γ are 

constants. Show that the work done by the gas in expanding from a 

state (P1, V1) to a state (P2, V2) is

W PV P V = −

−

1 1 2 2

γ 1

Answer 1

$pV^{\gamma} = \text{const} \; \Rightarrow p_1V_1^{\gamma} = p_2V_2^{\gamma}$

For the ideal gas $pV = \nu R T.$

The work done by the gas is

$w = \int\limits_{V_1}^{V_2}p\,dV = \int\limits_{V_1}^{V_2}\dfrac{\text{const}}{V^{\gamma}}\,dV =- \text{const}\cdot\dfrac{1}{\gamma-1}\cdot \dfrac{1}{V^{\gamma-1}}\Big |_{V_1}^{V_2} = \text{const}\cdot\dfrac{1}{1-\gamma}\cdot \left(\dfrac{1}{V_2^{\gamma-1}} - \dfrac{1}{V_1^{\gamma-1}} \right) = \Big| \text{const} = p_1V_1^{\gamma} = p_2V_2^{\gamma} \Big| = \dfrac{1}{1-\gamma}\cdot \left(\dfrac{p_2V_2^{\gamma}}{V_2^{\gamma-1}} - \dfrac{p_1V_1^{\gamma}}{V_1^{\gamma-1}} \right) = \dfrac{1}{1-\gamma}\cdot \left(p_2V_2 - p_1V_1 \right)$