Question

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 $\left(7/3\right)T_0$. 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 :

• A. $T_f=\frac{5}{2}{T}_0$
• B. $T_f=\frac{3}{7}{T}_0$
• C. $T_f=\frac{7}{3}{T}_0$
• D. $T_f=\frac{3}{2}{T}_0$

Answer 1

The correct option is D $T_f=\frac{3}{2}{T}_0$

$\text{Given:-}$ Number of moles of nitrogen $=1$ At $T_0$

Number of moles of helium $=1$ At $\frac{7}{3}{T}_0$

$\text{Solution:-}$

Here, change in internal energy of the system is zero, i.e., increase in internal energy of one is equal to decrease in internal energy of other,

Change in internal energy in box A,

$\Delta U_A=1\times \frac{5}{2}R(T_f-T_0)$

Change in internal energy in box B

$\Delta U_B=1\times \frac{3R}{2}(T_f-\frac{7}{3}{T}_0)$

Now , $\Delta U_A+\Delta U_B=0$

$\Rightarrow \frac{5R}{2}(T_f-T_0)+\frac{3R}{2}(T_f-\frac{7T_0}{3})=0$ or

$5T_f-5T_0+3T_f-7T_0=0$

$\Rightarrow 8T_f=12T_0$

$\Rightarrow T_f=\frac{12}{8}{T}_0=\frac{3}{2}{T}_0$

$\text{Hence the correct option is D}$