JEE Advanced 2021 Paper 2 · Q15 · Kinetic Theory of Gases
The value of $\dfrac{T_R}{T_0}$ is
Reveal answer + step-by-step solution
Correct answer:A
Solution
The right side is thermally insulated and frictionless; the partition moves slowly and reversibly, so the right gas undergoes a reversible adiabatic compression from $V_0$ to $V_0/2$. With $C_V = 2R$, $\gamma = C_P/C_V = 3R/2R = 3/2$. For adiabatic: $T V^{\gamma - 1} = \text{const}$, i.e. $T_0 V_0^{1/2} = T_R (V_0/2)^{1/2}$, giving $T_R/T_0 = \sqrt{2}$.
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