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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A gas is encl...

Question

A gas is enclosed in a vessel of volume $V$ at temperature $T_{1}$ and pressure $P$ ; the vessel is connected to another vessel of volume $V / 2$ by a tube and a stopcock. The second vessel is initially evacuated. If the stopcock is opened, the temperature of the second vessel becomes $T_{2}$ . The first vessel is maintained at a temperature of $T_{1}$ . What is the final pressure $P_{1}$ in the apparatus?

\frac{2 P T_{2}}{2 T_{2} + T_{1}}

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\frac{2 P T_{2}}{T_{2} + T_{1}}

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\frac{P T_{2}}{2 T_{2} + T_{1}}

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\frac{2 P T_{2}}{T_{1} + T_{2}}

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Solution

The correct option is A $\frac{2 P T_{2}}{2 T_{2} + T_{1}}$
Moles of gas in $1^{s t}$ sphere $= \frac{P V}{R T_{1}}$
After the stop clock is opened,
Moles of $1^{s t}$ sphere $= \frac{P_{1} V}{R T_{1}}$
Moles in $2^{n d}$ sphere $= \frac{P_{1} V / 2}{R T_{2}}$
$\frac{P V}{R T_{1}} = \frac{P_{1} V}{R T_{1}} + \frac{P_{1} V}{2 R T_{2}}$
Hence, $\frac{P}{R T_{1}} = \frac{P_{1}}{R T_{1}} + \frac{P_{1}}{2 R T_{2}} P = (\frac{2 P T_{2}}{2 T_{2} + T_{1}})$

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The correct option is A 2PT22T2+T1Moles of gas in 1st sphere =PVRT1After the stop clock is opened, Moles of 1st sphere =P1VRT1Moles in 2nd sphere =P1V/2RT2PVRT1=P1VRT1+P1V2RT2Hence, PRT1=P1RT1+P12RT2P=(2PT22T2+T1)