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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 $\left( {\frac{7}{3}} \right){T_0}$. 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 the 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
Heat lost by He $=$ Heat gained by
$\mathrm{N}_{2}$ $\mathrm{n}_{1} \mathrm{C}_{\mathrm{v}_{1}} \Delta \mathrm{T}_{1}=\mathrm{n}_{2} \mathrm{C}_{\mathrm{v}_{2}} \Delta \mathrm{T}_{2}$
$\frac{3}{2} \mathrm{R}\left[\frac{7}{3} \mathrm{T}_{0}-\mathrm{T}_{\mathrm{f}}\right]=\frac{5}{2} \mathrm{R}\left[\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{0}\right]$
$7 \mathrm{T}_{0}-3 \mathrm{T}_{\mathrm{f}}=5 \mathrm{T}_{\mathrm{f}}-5 \mathrm{T}_{0}$
$\Rightarrow 12 \mathrm{T}_{0}=8 \mathrm{T}_{\mathrm{f}} \Rightarrow \mathrm{T}_{\mathrm{f}}=\frac{12}{8} \mathrm{T}_{0}$
$\Rightarrow \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \mathrm{T}_{0}$
