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Thermal Equilibrium

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 $T_{o}$ , while box $B$ contains one mole of helium at temperature $(7 / 3) T_{o}$ . 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 the terms of $T_{o}$ is:-

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

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Solution

The correct option is C $T_{f} = \frac{3}{2} T_{0}$
Change in internal energy of an ideal gas
$Δ U =^{n} C_{r} Δ T$
$=^{n} C_{u} (T_{f} - T_{i})$
Fr nitrogen for helium
$C_{V} = \frac{5}{2} R$ $C_{V} = \frac{3}{2} R$
$Δ U = 1 \times \frac{5}{2} R \times (T_{f} - T_{o})$ $Δ V = 1 \times \frac{3}{2} R \times (T_{f} - 7 \frac{T_{o}}{3})$
Since no heat is given on taken from the
surrounding $Δ U$ of the system cant change
$Δ U_{n i t r o g e n} + Δ U_{h e l i u m} = 0$
$1 \times \frac{5}{2} R \times (T_{f} - T_{o}) + 1 \times \frac{3}{2} R \times (T_{f} - 7 \frac{T_{o}}{3}) = 0$
$\frac{5 R T}{2} - \frac{5 R T_{o}}{2} + \frac{3 R T_{f}}{2} - 21 \frac{R}{6} T_{o} = 0$
$4 T_{f} - 6 T_{o} = 0$
$\Rightarrow T_{f} = \frac{3}{2} T_{o}$

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