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(D) The number of moles in the first container is $\mu_1 = \frac{P_1V}{RT_1}$ and the number of moles in the second container is $\mu_2 = \frac{P_2V}{

Vedclass · Accepted Answer

$\frac{P_1}{2T_1} + \frac{P_2}{2T_2}$

Vedclass · Answer

(D) The number of moles in the first container is $\mu_1 = \frac{P_1V}{RT_1}$ and the number of moles in the second container is $\mu_2 = \frac{P_2V}{RT_2}$.Since the total amount of gas remains constant when the containers are connected,the total number of moles $\mu = \mu_1 + \mu_2$.The final volume of the system is $2V$. Thus,the final state is given by $\frac{P(2V)}{RT} = \frac{P_1V}{RT_1} + \frac{P_2V}{RT_2}$.Canceling $V$ and $R$ from both sides,we get $\frac{2P}{T} = \frac{P_1}{T_1} + \frac{P_2}{T_2}$.Therefore,$\frac{P}{T} = \frac{P_1}{2T_1} + \frac{P_2}{2T_2}$.

Two identical containers contain the same gas at pressures $P_1$ and $P_2$ and temperatures $T_1$ and $T_2$,respectively. If the containers are connected,the common pressure is $P$ and the common temperature is $T$. What is the value of $P/T$?

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