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 $\left(\frac{7}{3}\right)T_0$. 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 -

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

Answer 1

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

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.
$\Delta U=n{C}_v\Delta T$
where $n=$Number of mole
$C_v$ for $N_2=\frac{5}{2}$
$C_v$ for $He=\frac{3}{2}$

$\Delta U_A=1\times \frac{5R}{2}\times (T_f-T_0)$
$\Delta U_B=1\times \frac{3R}{2}(T_f-\frac{7}{3}{T}_0)$
Now, $\Delta U_A+\Delta U_B=0$
put the values,
$\frac{5R}{2}(T_f-T_0)+\frac{3R}{2}(T_f-\frac{7}{3}{T}_0)=0$
$\frac{5R}{2}{T}_f-\frac{5R{T}_0}{2}+\frac{3R{T}_f}{2}-\frac{3R\times 7}{2\times 3}{T}_0=0$
$\frac{5R}{2}{T}_f+\frac{3R{T}_f}{2}=\frac{7R}{2}{T}_0+\frac{5R{T}_0}{2}$
$\frac{8R}{2}{T}_f=\frac{12R}{2}{T}_0=6R{T}_0$
$\Rightarrow T_f=\frac{3}{2}{T}_0$