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 $(7/3)\, T_0$. 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. $T_{f} =\frac{3}{7}T_{0}$
• 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

Answer: (C)

$T_{f} =\frac{3}{2}T_{0}$

Answer 2

Here, change in internal energy of the system is ??cro. i.e., increase in internal energy of one is equal to decrease in internal energy of other. $\Delta U_{A}=1\times\frac{5R}{2}\left(T_{f} -T_{0}\right)$ $\Delta U_{B}=1\times\frac{3R}{2}\left(T_{f} -\frac{7}{3}T_{0}\right)$ Now $\Delta U_{A}+\Delta U_{B}=0$ $\frac{5R}{2}\left(T_{f} -T_{0}\right)+\frac{3R}{2}\left(T_{f} -\frac{7T_{0}}{3}\right)=0$ $5T_{f} -5T_{0}+3T_{f} -7T_{0}=0$ $\Rightarrow 8T_{f} =12T_{0} \Rightarrow T_{f} =\frac{12}{8}T_{0}=\frac{3}{2}T_{0}$