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Relative Lowering of Vapour Pressure

Two closed ve...

Question

Two closed vessels of an equal volume containing air at pressure $P_{1}$ and temperature $T_{1}$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_{1}$ and that in the other at $T_{2}$ , what will be the pressure in the vessels?

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

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\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_{1} + T_{2}}

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Solution

The correct option is A $\frac{2 P_{1} T_{2}}{T_{1} + T_{2}}$
Initial mole $=$ final mole

according to the ideal gas equation

$n = \frac{P V}{R T}$

$\frac{P_{1} V}{T_{1} R} + \frac{P_{1} V}{R T_{1}} = \frac{P V}{R T_{1}} + \frac{P V}{R T_{2}}$

$\frac{V}{R} (\frac{P_{1}}{T_{1}} + \frac{P_{1}}{T_{1}}) = \frac{V}{R} (\frac{P}{T_{1}} + \frac{P}{T_{2}})$

$\frac{2 P_{1}}{T_{1}} = P (\frac{1}{T_{1}} + \frac{1}{T_{2}})$

$\frac{2 P_{1}}{T_{1}} = P (\frac{T_{2} + T_{1}}{T_{1} T_{2}})$

$\frac{2 P_{1} T_{1} T_{2}}{T_{1}} = P (T_{2} + T_{1})$

$P = \frac{2 P_{1} T_{2}}{T_{1} + T_{2}}$ .

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Two closed vessels of an equal volume containing air at pressure P1 and temperature T1 are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at T1 and that in the other at T2, what will be the pressure in the vessels?

The correct option is A 2P1T2T1+T2Initial mole=final moleaccording to the ideal gas equationn=PVRTP1VT1R+P1VRT1=PVRT1+PVRT2VR(P1T1+P1T1)=VR(PT1+PT2)2P1T1=P(1T1+1T2)2P1T1=P(T2+T1T1T2)2P1T1T2T1=P(T2+T1)P=2P1T2T1+T2.

Two closed vessels of an equal volume containing air at pressure $P_{1}$ and temperature $T_{1}$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_{1}$ and that in the other at $T_{2}$ , what will be the pressure in the vessels?