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Standard XII

Chemistry

Relation between Cp and Cv

Two rigid box...

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

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

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

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

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

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Solution

The correct option is D $T_{f} = \frac{3}{2} T_{0}$
Since no work is done and no heat is supplied to the system, by first law of thermodynamics, $Δ U = 0$ . Since $C_{V} = \frac{f}{2} R$ where $f$ is degree of freedom,

Hence, $\frac{3}{2} R (T - T_{0}) + \frac{5}{2} R (T - \frac{7}{3} T_{0}) = 0$

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

Answer is option D.

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The correct option is D Tf=32T0Since no work is done and no heat is supplied to the system, by first law of thermodynamics, ΔU=0. Since CV=f2R where f is degree of freedom,Hence, 32R(T−T0)+52R(T−73T0)=0T=32T0Answer is option D.