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Avogadro's Law & Gay Lussac's Law

Two closed ve...

Question

Two closed vessel $A$ and $B$ of equal volume containing air at pressure $P_{1}$ and temperature $T_{1}$ are connected to each other through a narrow open tube. If the temperature of one is now maintained at $T_{1}$ and other at $T_{2}$ (where $T_{1} > T_{2}$ ) then that what will be the final pressure?

\frac{T_{1}}{2 P_{1} T_{2}}

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

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\frac{2 P_{1} T_{2}}{T_{1} - T_{2}}

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

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Solution

The correct option is B $\frac{2 P_{1} T_{2}}{T_{1} + T_{2}}$
Let the final pressure of the containers be $P$ .
For connecting vessels,
$Initial moles of vessel (A + B) = Final moles of vessel (A + B) \frac{P_{1} V}{R T_{1}} + \frac{P_{1} V}{R T_{1}} = \frac{P V}{R T_{1}} + \frac{P V}{R T_{2}} \frac{P_{1}}{T_{1}} + \frac{P_{1}}{T_{1}} = \frac{P}{T_{1}} + \frac{P}{T_{2}} \frac{2 P_{1}}{T_{1}} = P (\frac{T_{1} + T_{2}}{T_{1} T_{2}}) ∴ P = \frac{2 P_{1} T_{2}}{T_{1} + T_{2}}$

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Avogadro's Law & Gay Lussac's Law

Standard XII Chemistry

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Two closed vessel A and B of equal volume containing air at pressure P1 and temperature T1 are connected to each other through a narrow open tube. If the temperature of one is now maintained at T1 and other at T2 (where T1>T2) then that what will be the final pressure?

The correct option is B 2P1T2T1+T2Let the final pressure of the containers be P. For connecting vessels, Initial moles of vessel (A+B) = Final moles of vessel (A+B)P1VRT1+P1VRT1=PVRT1+PVRT2P1T1+P1T1=PT1+PT22P1T1=P(T1+T2T1T2)∴ P=2P1T2T1+T2