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(A) Let the volume of each container be $V$. Initially,the total number of moles of gas in both containers is $n_{total} = n_1 + n_2 = \frac{P_0 V}{R

Vedclass · Accepted Answer

$\frac{2 P_0 T}{T+T_0}$

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(A) Let the volume of each container be $V$. Initially,the total number of moles of gas in both containers is $n_{total} = n_1 + n_2 = \frac{P_0 V}{R T_0} + \frac{P_0 V}{R T_0} = \frac{2 P_0 V}{R T_0}$.When the temperatures are changed to $T_0$ and $T$ respectively,let the new pressure be $P$. Since the total number of moles remains constant,we have:$n_{total} = \frac{P V}{R T_0} + \frac{P V}{R T} = \frac{2 P_0 V}{R T_0}$.Canceling $V/R$ from both sides:$\frac{P}{T_0} + \frac{P}{T} = \frac{2 P_0}{T_0}$.Taking $P$ as a common factor:$P \left( \frac{1}{T_0} + \frac{1}{T} ight) = \frac{2 P_0}{T_0}$.$P \left( \frac{T + T_0}{T T_0} ight) = \frac{2 P_0}{T_0}$.Solving for $P$:$P = \frac{2 P_0}{T_0} 	imes \frac{T T_0}{T + T_0} = \frac{2 P_0 T}{T + T_0}$.

Two closed containers of equal volume are filled with air at pressure $P_0$ and temperature $T_0$. Both are connected by a narrow tube. If one of the containers is maintained at temperature $T_0$ and the other at temperature $T$,then the new pressure in the containers will be:

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