As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P_1) and a freely movable but thermally insulated piston (P_2). The partition P_1 with thermal conductivity K, cross sectional area A and width x divides the container into two sections, S_1

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As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P_1) and a freely movable but thermally insulated piston (P_2). The partition P_1 with thermal conductivity K, cross sectional area A and width x divides the container into two sections, S_1 and S_2, each containing one mole of a monoatomic gas. The piston P_2 moves freely such that the gas in S_2 is always at the atmospheric pressure. Initially, the difference between the temperatures of S_1 and S_2 is T_0. The time it takes for the temperature difference to become T_0 2 is n x R / K A, where R is the universal gas constant. The value of n is: [Given: 2 0.7]

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Correct answer: A. A

Let T_1 and T_2 be the temperatures of the gases in sections S_1 and S_2 respectively. Assume T_1 > T_2, so heat flows from S_1 to S_2. The rate of heat transfer through the partition P_1 is given by Fourier's law of heat conduction: dQ dt = KA x (T_1 - T_2) For section S_1, the volume is constant (since P_1 is immovable). Thus, the process is isochoric. The heat lost by S_1 is: dQ = -n_1 C_V dT_1 Since the gas is monoatomic and n_1 = 1 mole, C_V = 3 2 R. dT_1 = - 2 dQ 3R For section S_2, the pressure is constant (since P_2 is freely movable and exposed to atmospheric pressure). Thus, the process is isobaric. The heat gained by S_2 is: dQ = n_2 C_P dT_2 Since the gas is monoatomic and n_2 = 1 mole, C_P = 5 2 R. dT_2 = 2 dQ 5R The change in the temperature difference d( T) = d(T_1 - T_2) is: d(T_1 - T_2) = dT_1 - dT_2 = - 2 dQ 3R - 2 dQ 5R = - 16 dQ 15R Substituting the rate of heat transfer dQ dt into the equation: d(T_1 - T_2) dt = - 16 15R ( KA x (T_1 - T_2) ) Let T = T_1 - T_2. The differential equation becomes: d( T) T = - 16 KA 15 R x dt Integrating both sides from t = 0 to t, where the temperature difference goes from T_0 to T_0 2 : _ T_0 ^ T_0 / 2 d( T) T = - 16 KA 15 R x _0^t dt ( 1 2 ) = - 16 KA 15 R x t - 2 = - 16 KA 15 R x t t = 15 R x 2 16 KA Given that 2 0.7: t = 15 0.7 16 x R KA = 10.5 16 x R KA = 21 32 x R KA = 0.65625 x R KA Comparing this with the given expression t = n x R K A , we get: n = 0.65625 Answer: 0.66

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