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(B) The total number of moles of the gas remains conserved when the two containers are joined.The number of moles in the first container is $n_1 = \fr

Vedclass · Accepted Answer

$P = \frac{1}{2}\left( {\frac{{{P_1}}}{{{T_1}}} + \frac{{{P_2}}}{{{T_2}}}} \right)T$

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(B) The total number of moles of the gas remains conserved when the two containers are joined.The number of moles in the first container is $n_1 = \frac{P_1 V}{R T_1}$.The number of moles in the second container is $n_2 = \frac{P_2 V}{R T_2}$.After joining,the total volume becomes $2V$,and the gas reaches a common pressure $P$ and temperature $T$. The total number of moles is $n = n_1 + n_2 = \frac{P(2V)}{RT}$.Equating the total moles:$\frac{P_1 V}{R T_1} + \frac{P_2 V}{R T_2} = \frac{P(2V)}{RT}$Dividing both sides by $V/R$:$\frac{P_1}{T_1} + \frac{P_2}{T_2} = \frac{2P}{T}$Rearranging for $P$:$P = \frac{1}{2} \left( \frac{P_1}{T_1} + \frac{P_2}{T_2} \right) T$

Two containers of equal volume contain identical gases at pressures $P_1$ and $P_2$ and absolute temperatures $T_1$ and $T_2$ respectively. The vessels are joined and the gas reaches a common pressure $P$ and a common temperature $T$. Then:

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