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) $T_{f} = \frac{7}{3}T_{0}$
B) $T_{f} = \frac{3}{2}T_{0}$
C) $T_{f} = \frac{5}{2}T_{0}$
D) $T_{f} = \frac{3}{7}T_{0}$
Explanation
(b) When two gases are mixed together then
Heat lost by the Helium gas = Heat gained by the Nitrogen gas
$\mu_{B} \times \left( C_{v} \right)_{He} \times \left( \frac{7}{3}T_{0} - T_{f} \right) = \mu_{A} \times \left( C_{v} \right)_{N_{2}} \times (T_{f} - T_{0})$
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$\Rightarrow 1 \times \frac{3}{2}R \times \left( \frac{7}{3}T_{0} - T_{f} \right) = 1 \times \frac{5}{2}R \times (T_{f} - T_{0})$
By solving we get $T_{f} = \frac{3}{2}T_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.