A given mass of a gas occupies 919 mL volume in dry state at STP- Same mass of the gas collected over water at 15 -C 750 mm of pressure occupies a volume of 1 L- Calculate vapor pressure of water at 15 -C-

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Ideal Gas Equation

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Question

A given mass of a gas occupies 919 mL volume in dry state at STP. Same mass of the gas collected over water at $15^{\circ}$ C & 750 mm of pressure occupies a volume of 1 L. Calculate vapor pressure of water at $15^{\circ}$ C.

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Solution

Let us consider the
$P_{d r y g a s} = P_{t o t a l} = P_{w a t e r v a p o u r}$
Given that:
$(A t S T P)$
$V_{1} = 919 m L$
$V_{2} = 100 m L$
$P_{1} = 760 m m H g$
$P_{2} = ? (d r y g a s)$
$T_{1} = 273 K$
$T_{2} = 273 + 15 = 288 K$
$P_{t o t a l} = 750 m m H g$
Using gas equation
$\frac{p_{1} v_{1}}{T_{1}} = \frac{p_{2} v_{2}}{T_{2}}$
implies that
$\frac{760 \times 919}{273} = \frac{p_{2} \times 1000}{288}$
implies that
$p_{2} = 736.82 m m H g$
Hence
$P_{d r y g a s} = P_{t o t a l} - P_{w a t e r v a p o u r}$
implies that
$P_{w a t e r v a p o u r} = 750 - 736.82$
$13.18 m m$

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A given mass of a gas occupies 919 mL volume in dry state at STP. Same mass of the gas collected over water at 15∘C & 750 mm of pressure occupies a volume of 1 L. Calculate vapor pressure of water at 15∘C.

A given mass of a gas occupies 919 mL volume in dry state at STP. Same mass of the gas collected over water at $15^{\circ}$ C & 750 mm of pressure occupies a volume of 1 L. Calculate vapor pressure of water at $15^{\circ}$ C.