JEE Advanced 2021 Paper 2 Q16 Physics Thermal Physics Kinetic Theory of Gases Medium

JEE Advanced 2021 Paper 2 · Q16 · Kinetic Theory of Gases

The value of $\dfrac{Q}{R T_0}$ is

  1. A. $4(2\sqrt{2} + 1)$
  2. B. $4(2\sqrt{2} - 1)$
  3. C. $(5\sqrt{2} + 1)$
  4. D. $(5\sqrt{2} - 1)$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Pressure equilibrium across partition: $P_L = P_R \Rightarrow R T_L/(3V_0/2) = R T_R/(V_0/2) \Rightarrow T_L = 3 T_R = 3\sqrt{2}\, T_0$. Both chambers contain 1 mole; right is adiabatic (no heat), left receives $Q$. Right: $\Delta U_R = n C_V (T_R - T_0) = 2R(\sqrt{2} - 1)T_0$, which equals work done on right side by partition. Energy conservation overall: $Q = \Delta U_L + \Delta U_R$ (work done by left on right equals work done on right, accounted in $\Delta U_R$). $\Delta U_L = 2R(T_L - T_0) = 2R(3\sqrt{2} - 1)T_0$. So $Q = 2R T_0[(3\sqrt{2} - 1) + (\sqrt{2} - 1)] = 2RT_0(4\sqrt{2} - 2) = 4 R T_0(2\sqrt{2} - 1)$. Hence $Q/(R T_0) = 4(2\sqrt{2} - 1)$.

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