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(C) For an ideal gas in a rigid box,the heat exchange is given by $Q = n C_v \Delta T$. Since the boxes are in thermal contact and isolated from the s

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$T_f = \frac{3}{2} T_0$

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(C) For an ideal gas in a rigid box,the heat exchange is given by $Q = n C_v \Delta T$. Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the hotter gas equals the heat gained by the colder gas.Nitrogen $(N_2)$ is a diatomic gas,so its molar heat capacity at constant volume is $C_{v1} = \frac{5}{2} R$.Helium $(He)$ is a monatomic gas,so its molar heat capacity at constant volume is $C_{v2} = \frac{3}{2} R$.Let $n_1 = 1$ mole $(N_2)$ and $n_2 = 1$ mole $(He)$.Heat lost by $He$ = Heat gained by $N_2$:$n_2 C_{v2} (T_{initial, He} - T_f) = n_1 C_{v1} (T_f - T_{initial, N2})$$1 \cdot \frac{3}{2} R \left( \frac{7}{3} T_0 - T_f \right) = 1 \cdot \frac{5}{2} R (T_f - T_0)$Multiplying both sides by $\frac{2}{R}$:$3 \left( \frac{7}{3} T_0 - T_f \right) = 5 (T_f - T_0)$$7 T_0 - 3 T_f = 5 T_f - 5 T_0$$12 T_0 = 8 T_f$$T_f = \frac{12}{8} T_0 = \frac{3}{2} T_0$.

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