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Temperature

Two rigid box...

Question

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of $N_{2}$ at temperature $T_{0}$ , while box B contains one mole of $H_{2}$ 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

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

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

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T_{f} = \frac{5}{3} T_{o}

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

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Solution

The correct option is C $T_{f} = \frac{5}{3} T_{o}$
According to kinetic theory of gases, the KE of an ideal gas molecule at temp $T$ is given by $\frac{1}{2} f k_{B} T$ where f is the no of degrees of freedom, $k_{B}$ is the Stefan Boltzman constant and T is the temp in Kelvin.
Both Nitrogen and Hydrogen are diatomic and the no of degrees of freedom $f = 5$ .
Total KE is $K E = \frac{1}{2} f n_{1} N_{A} k_{B} T_{1} + \frac{1}{2} f n_{2} N_{A} k_{B} T_{2} = \frac{1}{2} f N_{A} k_{B} (n_{1} T_{1} + n_{2} T_{2})$ ....(1) where $N_{A}$ is the Avogadro's number.
$T_{1} = T_{0}$ , $T_{2} = \frac{7}{3} T_{0}$
$n_{1} = n_{2} = n$
Let the final temp be $T_{f}$ after both gases are mixed.
$K E = \frac{1}{2} (n_{1} + n_{2}) f N_{A} k_{B} T$ ...(2)
Equating (1) and (2) since total energy is conserved,
$\frac{1}{2} (n_{1} + n_{2}) f N_{A} k_{B} T_{f} = \frac{1}{2} f N_{A} k_{B} (n_{1} T_{1} + n_{2} T_{2})$
$∴ T_{f} = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}} = \frac{1}{2} \times (T_{1} + T_{2}) = \frac{1}{2} \times \frac{10}{3} T_{0} = \frac{5}{3} T_{0}$

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