Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0 , while box B contains one mole of helium at temperature (7/3)T0 . 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, Tf, in terms of T0 is

Question

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0 , while box B contains one mole of helium at temperature (7/3)T0 . 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, Tf, in terms of T0 is (A) Tf=(3/7)T0 (B) Tf=(7/3)T0 (C) Tf=(3/2)T0 (D) Tf=(5/2)T0

Tardigrade Admin · Accepted Answer

Here, change in internal energy of the system is zero, ie, increase in internal energy of one is equal to decrease in internal energy of other.
Solution

Δ U_{A} = 1 \times \frac{5 R}{2} (T_{f} - T_{o})

Δ U_{B} = 1 \times \frac{3 R}{2} (T_{f} - \frac{7}{3} T_{0})

Now,

Δ U_{A} + Δ U_{B} = 0

\frac{5 R}{2} (T_{f} - T_{0}) + \frac{3 R}{2} (T_{f} - \frac{7 T _{0}}{3}) = 0

5 T_{f} - 5 T_{0} + 3 T_{f} - 7 T_{0} = 0

\Rightarrow 8 T_{f} = 12 T_{0}

\Rightarrow T_{f} = \frac{12}{8} T_{0} = \frac{3}{2} T_{0}

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

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

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

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

Tf​=73​T0​

Tf​=37​T0​

Tf​=23​T0​

Tf​=25​T0​

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

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

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

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