Two rigid boxes containing different ideal ga | vedclass

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Question

(C) Since the boxes are rigid and thermally isolated from the surroundings,the heat lost by the hotter gas (helium) equals the heat gained by the cold

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

Vedclass · Answer

(C) Since the boxes are rigid and thermally isolated from the surroundings,the heat lost by the hotter gas (helium) equals the heat gained by the colder gas (nitrogen).For helium (monatomic gas),the molar heat capacity at constant volume is $C_{V,He} = \frac{3R}{2}$.For nitrogen (diatomic gas),the molar heat capacity at constant volume is $C_{V,N_2} = \frac{5R}{2}$.Let $T_f$ be the final equilibrium temperature.Heat lost by helium = Heat gained by nitrogen$n_{He} C_{V,He} (T_{initial,He} - T_f) = n_{N_2} C_{V,N_2} (T_f - T_{initial,N_2})$$1 	imes \frac{3R}{2} 	imes (\frac{7}{3} T_0 - T_f) = 1 	imes \frac{5R}{2} 	imes (T_f - T_0)$Canceling $\frac{R}{2}$ from both sides:$3(\frac{7}{3} T_0 - T_f) = 5(T_f - T_0)$$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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