Q147 / 150 · JEE Main 2006 · Physics
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 the gases, $T_f$, in terms of $T_0$ is
- A$\mathrm{T}_{\mathrm{f}}=\frac{5}{2} \mathrm{~T}_0$
- B$\mathrm{T}_{\mathrm{f}}=\frac{3}{7} \mathrm{~T}_0$
- C$\mathrm{T}_{\mathrm{f}}=\frac{7}{3} \mathrm{~T}_0$
- Correct answerD$T_f=\frac{3}{2} T_0$
Solution
$\Delta \mathrm{U}=0$
$\Rightarrow \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_0\right)+1 \times \frac{5}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{f}}-\frac{7}{3} \mathrm{~T}_0\right)=0$
$\mathrm{~T}_{\mathrm{f}}=\frac{3}{2} \mathrm{~T}_0$
Hence, the correct answer is option: 'D'.