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Standard XII

Chemistry

Liquids in Liquids

Two moles of ...

Question

Two moles of $O_{2}$ gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry $O_{2}$ gas is 32.20 bar then find the vapour pressure of water under the same conditions.

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Solution

For $O_{2}$ gas collected over water, the pressure $P = \frac{n R T}{V}$
$P = \frac{2 m o l \times 0.08206 L \cdot a t m / m o l \cdot K \times 400 K}{2 L}$
$P = 32.824 a t m$
Convert unit from atm to bar
$P = 32.824 a t m \times \frac{1.01325 b a r}{1 a t m}$
$P = 33.26 b a r$
The pressure of dry $O_{2}$ gas is 32.20 bar.
The vapour pressure of water $= 33.26 b a r - 32.20 b a r = 1.06 b a r$

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Two moles of O2 gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry O2 gas is 32.20 bar then find the vapour pressure of water under the same conditions.

For O2 gas collected over water, the pressure P=nRTVP=2 mol×0.08206 L⋅atm/mol⋅K×400 K2 LP=32.824 atmConvert unit from atm to barP=32.824 atm×1.01325 bar1 atmP=33.26 barThe pressure of dry O2 gas is 32.20 bar.The vapour pressure of water =33.26 bar−32.20 bar=1.06 bar

Two moles of $O_{2}$ gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry $O_{2}$ gas is 32.20 bar then find the vapour pressure of water under the same conditions.