JEE MAIN - Physics (2006 - No. 27)
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 $$\left( {{7 \over 3}} \right){T_0}.$$ 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
$${T_f} = {3 \over 7}{T_0}$$
$${T_f} = {7 \over 3}{T_0}$$
$${T_f} = {3 \over 2}{T_0}$$
$${T_f} = {5 \over 2}{T_0}$$
Explanation
Heat lost by He $$=$$ Heat gained by $${N_2}$$
$${n_1}C{v_1}\Delta {T_1} = {n_2}C{v_2}\Delta {T_2}$$
$${3 \over 2}R\left[ {{7 \over 3}{T_0} - {T_f}} \right]$$
$$ = {5 \over 2}R\left[ {{T_f} - {T_0}} \right] \Rightarrow {T_f}$$
$$ = {3 \over 2}{T_0}$$
$${n_1}C{v_1}\Delta {T_1} = {n_2}C{v_2}\Delta {T_2}$$
$${3 \over 2}R\left[ {{7 \over 3}{T_0} - {T_f}} \right]$$
$$ = {5 \over 2}R\left[ {{T_f} - {T_0}} \right] \Rightarrow {T_f}$$
$$ = {3 \over 2}{T_0}$$
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