# Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $\dpi{100} T_{0},$ while Box B contains one mole of helium at temperature (7/3) $\dpi{100} T_{0}.$ 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, $\dpi{100} T_{f},$ in terms of $\dpi{100} T_{0}$ is Option 1) $T_{f}=\frac{5}{2}T_{0}$ Option 2) $T_{f}=\frac{3}{7}T_{0}$ Option 3) $T_{f}=\frac{7}{3}T_{0}$ Option 4) $T_{f}=\frac{3}{2}T_{0}$

P Plabita

As we learnt in

Property of internal energy -

U (internal energy) is a state function.

- wherein

i.e. It depends only on initial and final state.

Change in internal energy for cyclic process -

$\Delta U= 0$

- wherein

Since in a cyclic process initial and final state is same.

$U_{f}= U_{i}$

$U{_{i}} = U_{f}$

$n_{1} \:C_{V1} \:T_{o} + n_{2} \:C_{V2}\: (\frac{7}{3} T_{o}) = (n_{1} C_{V1} + n_{2} C_{V2})\: T_{f}$

$T_{f}\: = \frac{1\times \frac{5R}{2}.\: T_{o} + 1. (\frac{3R}{2}).\frac{7}{3}\:T_{o}}{1.\: \frac{5R}{2}+\: 1.\: \frac{3R}{2}}$

$=\: \frac{6R\: T_{o}}{4R}\: = 1.5\: T_{o}$

Option 1)

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

This option is incorrect.

Option 2)

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

This option is incorrect.

Option 3)

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

This option is incorrect.

Option 4)

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

This option is correct.

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