Two rigid boxes containing different ideal ga | vedclass

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(D) Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the helium gas equals the heat gained by the nitrogen g

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the helium gas equals the heat gained by the nitrogen gas.For helium (monatomic gas),$C_v = \frac{3}{2}R$. For nitrogen (diatomic gas),$C_v = \frac{5}{2}R$.Let $n_1 = 1$ mole of He and $n_2 = 1$ mole of $N_2$.Heat lost by He = $n_1 C_{v,He} (T_{initial,He} - T_f) = 1 \cdot \frac{3}{2}R \cdot (\frac{7}{3}T_0 - T_f)$.Heat gained by $N_2$ = $n_2 C_{v,N_2} (T_f - T_{initial,N_2}) = 1 \cdot \frac{5}{2}R \cdot (T_f - T_0)$.Equating the two: $\frac{3}{2}R (\frac{7}{3}T_0 - T_f) = \frac{5}{2}R (T_f - T_0)$.Dividing by $R/2$: $3(\frac{7}{3}T_0 - T_f) = 5(T_f - T_0)$.$7T_0 - 3T_f = 5T_f - 5T_0$.$12T_0 = 8T_f$.$T_f = \frac{12}{8}T_0 = \frac{3}{2}T_0$.

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