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Standard VI

Physics

Thermal Equilibrium

Two rigid box...

Question

Two rigid boxes containing different ideal gases are placed on table. Box A contains one mole of nitrogen at temperature $T_{0}$ , while box B contains 1 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 gases, $T_{f}$ in terms of $T_{0}$ is

\frac{2 T_{0}}{5}

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\frac{3 T_{0}}{7}

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\frac{5 T_{0}}{3}

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\frac{9 T_{0}}{7}

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Solution

The correct option is C $\frac{5 T_{0}}{3}$
The loss in energy of helium gas $=$ The gain in energy of nitrogen gas
$⟹ n C_{V_{H e}} (T_{1} - T_{f}) = n C_{V_{N}} (T_{f} - T_{2})$

$⟹ 1 \times \frac{3 R}{2} \times (\frac{7}{3} T_{0} - T_{f}) = 1 \times \frac{5 R}{2} (T_{f} - T_{0})$

$⟹ T_{f} = \frac{3}{2} T_{0}$

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The correct option is C 5T03The loss in energy of helium gas = The gain in energy of nitrogen gas⟹nCVHe(T1−Tf)=nCVN(Tf−T2)⟹1×3R2×(73T0−Tf)=1×5R2(Tf−T0)⟹Tf=32T0