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 contains one mole of helium at temperature ( 7 \over 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

Heat and Thermodynamics PYQ — JEE Main 2006 · Shift 0 · Q109 with Solution

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Correct answer: C

Box A: $1$ mole of nitrogen $(N_2)$ at temperature $T_0$
Box B: $1$ mole of helium $(He)$ at temperature $\dfrac{7}{3}T_0$
Both boxes are rigid, so volume remains constant.
Hence, heat exchange happens at constant volume, so we use molar heat capacities at constant volume.

For helium (monoatomic ideal gas): $C_{V,He} = \frac{3}{2}R$
For nitrogen (diatomic ideal gas, neglecting vibration): $C_{V,N_2} = \frac{5}{2}R$

Since each box contains $1$ mole,

Since the two boxes are thermally isolated from the surroundings, heat lost by hotter gas = heat gained by colder gas.

Nitrogen gains heat: $Q_{N_2} = \frac{5}{2}R(T_f - T_0)$

Helium loses heat: $Q_{He} = \frac{3}{2}R\left(T_f - \frac{7}{3}T_0\right)$

Net heat exchange is zero: $\frac{5}{2}R(T_f - T_0) + \frac{3}{2}R\left(T_f - \frac{7}{3}T_0\right)=0$

Cancel $\frac{R}{2}$ : $5(T_f - T_0) + 3\left(T_f - \frac{7}{3}T_0\right)=0$

Expand: $5T_f - 5T_0 + 3T_f - 7T_0 = 0$

$8T_f - 12T_0 = 0$

$8T_f = 12T_0$

$T_f = \frac{12}{8}T_0 = \frac{3}{2}T_0$

$\boxed{T_f = \frac{3}{2}T_0}$

So, the correct option is C.