JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Heat & Thermodynamics

A sample of gas at temperature T is adiabatically expanded to double its volume. Adiabatic constant for the gas is

γ = 3/2

. The work done by the gas in the process is : (

μ = 1

mole)

(A) $R T [2 - 2]$
(B) $R T [1 - 22]$
(C) $R T [22 - 1]$
(D) $R T [2 - 2]$ ✓ Correct

Answer: (D)

Solution

Adiabatic process:

T V^{γ - 1} = constant

. With

γ = 3/2

γ - 1 = 1/2

T V^{1/2} = T_{f} (2 V)^{1/2} ⟹ T_{f} = \frac{T}{2}

Work done by the gas in an adiabatic process:

W = \frac{n R ( T - T _{f} )}{γ - 1} = \frac{R T ( 1 - \frac{1}{2} )}{1/2} = 2 R T (1 - \frac{1}{2}) = R T (2 - 2)

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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