Q. 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) $T_{f} = \frac{5}{2}T_{0}$
Explanation
(c) Here, change in internal energy of the system is zero, ie, increase in internal energy of one is equal to decrease in internal energy of other.
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$\mathrm{\Delta}U_{A} = 1 \times \frac{5R}{2}(T_{f} - T_{o})$
$\mathrm{\Delta}U_{B} = 1 \times \frac{3R}{2}(T_{f} - \frac{7}{3}T_{0})$
Now, $\mathrm{\Delta}U_{A} + \mathrm{\Delta}U_{B} = 0$
$\frac{5R}{2}\left( T_{f} - T_{0} \right) + \frac{3R}{2}\left( T_{f} - \frac{7T_{0}}{3} \right) = 0$
$5T_{f} - 5T_{0} + 3T_{f} - 7T_{0} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ 8T_{f} = 12T_{0}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ T_{f} = \frac{12}{8}T_{0} = \frac{3}{2}T_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.