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) RT[2√(2) - 1] (b) RT[2 - √(2)] (c) RT[1 - 2√(2)] (d) RT[√(2) - 2]

Answer: (b). Adiabatic: T₂ = T · 2-1/2 = T/√(2). W = (μ R(T - T₂))/(γ - 1) = 2RT(1 - 1/√(2)) = RT(2 - √(2)) So the answer is (b) RT[2 - √(2)].

A sample of gas at temperature T is adiabatically expanded to double… | JEE Main 2024 (4 Apr, Shift 2)

Q121JEE Main 2024 · 4 APR · SHIFT 2Medium

Thermodynamic Processes

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $\gamma = 3/2$ . The work done by the gas in the process is:

$(\mu = 1$ mole $)$

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Answer: (b)

Adiabatic: $T_2 = T \cdot 2^{-1/2} = \frac{T}{\sqrt{2}}$ .
$W = \frac{\mu R(T - T_2)}{\gamma - 1} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = RT(2 - \sqrt{2})$
So the answer is (b) $RT[2 - \sqrt{2}]$ .

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