Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It expands isothermally to volume 4V1. After this, the gas expands adiabatically and its volume becomes 32V1. The work done by the gas during isothermal and adiabatic expansion processes are Wiso and Wadia , respectively. If the ratio Wadia Wiso =fln2, then f is ____.
Answer: 1.78
Solution
For adiabatic process from 4V₁ to 32V₁: P₁(4V₁)^(5/3) = P₂(32V₁)^(5/3), giving P₂ = P₁/128. Work done: W_adia = (P₁V₁ - P₂×32V₁)/(γ-1) = (P₁V₁ - P₁V₁/4)/(2/3) = (9/8)P₁V₁. For isothermal: W_iso = P₁V₁ ln(4) = 2P₁V₁ ln(2). Ratio: W_iso/W_adia = 2P₁V₁ ln(2)/[(9/8)P₁V₁] = (16/9)ln(2) = f ln(2), giving f = 16/9 ≈ 1.78.