The pressure and volume of an ideal gas are related as PV^(3/2)=K (Constant). The work done when the gas is taken from state A(P₁, V₁, T₁) to state B(P₂, V₂, T₂) is :

Solution

Physics sir · 3 steps

Write P as KV^(-3/2), integrate P dV, then put K = PV^(3/2) back in at each end.

1. Write P using V. Work done by the gas is W = ∫_(V₁)^(V₂) P dV. From the given relation, P = K V^(-3/2), so the integral has only V in it.

   W = ∫_(V₁)^(V₂) K V^(-3/2) dV = K[-2V^(-1/2)]_(V₁)^(V₂) = 2(K/√V₁ - K/√V₂)

2. Remove K. At state A, K = P₁ V₁^(3/2), so K/√V₁ = P₁ V₁. In the same way, at state B, K/√V₂ = P₂ V₂.
3. So. W = 2(P₁V₁ - P₂V₂), which is option (d). This matches the polytropic formula (P₁V₁ - P₂V₂)/(n-1) with n = 3/2.

The idea to keep. For any process PVⁿ = constant with n ≠ 1, the work done by the gas is W = (P₁V₁ - P₂V₂)/(n-1). Always check that your answer has the units of PV.

Answer from the NTA final answer key. Solution by PrepWiser.