Two rigid boxes containing different ideal gases are placed on a… | JEE Main 2006 (AIEEE)

Q41JEE Main 2006 · AIEEEDifficult

Degrees of Freedom & Internal Energy

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 $\left(\frac{7}{3}\right)T_0.$ 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

Show answer and solution

Answer: (c)

Heat lost by He = heat gained by $N_2$ (rigid boxes, so constant volume).
$\frac{5}{2}R(T_f - T_0) = \frac{3}{2}R\left(\frac{7}{3}T_0 - T_f\right)$
$5T_f - 5T_0 = 7T_0 - 3T_f \Rightarrow 8T_f = 12T_0$
So the answer is (c) $T_f = \frac{3}{2}T_0$ .

Report a mistake

Degrees of Freedom & Internal Energy question from JEE Main 2006 (AIEEE)

More questions on Degrees of Freedom & Internal Energy