Project ForgeThermodynamics
First Law of Thermodynamics
GATE 2017
Hard
A mass m of a perfect gas at pressure p₁ and volume V₁ undergoes an isothermal process. The final pressure is p₂ and volume is V₂. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is
- Ap₁V₁ ln(V₂/V₁)
- −p₁V₁ ln(p₁/p₂)
- CRT ln(V₂/V₁)
- D−mRT ln(p₂/p₁)
Explanation
Work done BY the gas (standard) is p₁V₁ ln(V₂/V₁) = mRT ln(V₂/V₁). Under this problem's convention (work done ON the system is positive), the reported work is the negative of that: W = −p₁V₁ ln(V₂/V₁). Using the isothermal relation V₂/V₁ = p₁/p₂, this becomes W = −p₁V₁ ln(p₁/p₂), matching option B exactly.
Source: GATE ME 2017 Session 2 (ME-2) (2017)
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