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Two rigid boxes containing different ideal gases are placed on a table. Box $A$ contains one mole of nitrogen at temperature $T_0$, while box $B$ contains one mole of helium at temperature $(7/3)$ $T_0$. The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then, the final temperature of gases, $T_f$, in terms of $T_0$ is
${T_f} = \frac{3}{7}{T_0}$
${T_f} = \frac{7}{3}{T_0}$
${T_f} = \frac{3}{2}{T_0}$
${T_f} = \frac{5}{2}{T_0}$
Solution
The loss in energy of helium gas $=$ The gain in energy of nitrogen gas
$\Longrightarrow n C_{V_{H e}}\left(T_{1}-T_{f}\right)=n C_{V_{N}}\left(T_{f}-T_{2}\right)$
$\Longrightarrow 1 \times \frac{3 R}{2} \times\left(\frac{7}{3} T_{0}-T_{f}\right)=1 \times \frac{5 R}{2}\left(T_{f}-T_{0}\right)$
$\Longrightarrow T_{f}=\frac{3}{2} T_{0}$
