Two identical containers, each of volume , are joined by a small pipe of negligible volume. The cont — States of Matter and Gaseous State Chemistry Question
Question
Two identical containers, each of volume $V_o$, are joined by a small pipe of negligible volume. The containers contain identical gases at temperature $T_o$ and pressure $P_o$. One container is heated to temperature $2T_o$ while maintaining the other at the same temperature, $T_o$. The common pressure of the gas is $P$ and $n$ is the number of moles of gas in container at temperature $2T_o$, then
💡 Solution & Explanation
Total initial moles $n_{total} = \frac{P_o V_o}{R T_o} + \frac{P_o V_o}{R T_o} = \frac{2 P_o V_o}{R T_o}$. In the final state, the total moles remain conserved: $\frac{P V_o}{R(2T_o)} + \frac{P V_o}{R T_o} = \frac{2 P_o V_o}{R T_o}$. Factoring out $\frac{V_o}{R T_o}$, we get $P(0.5 + 1) = 2 P_o \implies 1.5 P = 2 P_o \implies P = \frac{4}{3} P_o$. The moles in the container at $2T_o$ is $n = \frac{P V_o}{R (2T_o)} = \frac{(4P_o/3) V_o}{2 R T_o} = \frac{2P_o V_o}{3 R T_o}$. Therefore, correct answer is B,C.