Two closed vessels of equal volume containing | vedclass

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(A) Initially,the total number of moles $n$ in both vessels of volume $V$ is given by: $n = n_1 + n_2 = \frac{P_1 V}{R T_1} + \frac{P_1 V}{R T_1} = \f

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$\frac{2 P_1 T_2}{T_1 + T_2}$

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(A) Initially,the total number of moles $n$ in both vessels of volume $V$ is given by: $n = n_1 + n_2 = \frac{P_1 V}{R T_1} + \frac{P_1 V}{R T_1} = \frac{2 P_1 V}{R T_1}$ When the temperatures are changed to $T_1$ and $T_2$ respectively,the new pressure $P$ is the same in both vessels. The total number of moles remains constant: $n = \frac{P V}{R T_1} + \frac{P V}{R T_2} = \frac{P V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right) = \frac{P V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$ Equating the initial and final moles: $\frac{2 P_1 V}{R T_1} = \frac{P V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$ $P = \frac{2 P_1 T_2}{T_1 + T_2}$

Two closed vessels of equal volume containing air at pressure $P_1$ and temperature $T_1$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_1$ and that in the other at $T_2$,what will be the final pressure in the vessels?

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