JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Heat & Thermodynamics

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Consider one mole of helium gas enclosed in a container at initial pressure and volume . It expands isothermally to volume . After this, the gas expands adiabatically and its volume becomes . The work done by the gas during isothermal and adiabatic expansion processes are and , respectively. If the ratio , then 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.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer

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