A certain quantity of gas occupies a volume of 0.8 L collected over water at 300K and a pressure 0.92 bar. The same gas occupies a volume of 0.08 L at STP in dry conditions. Calculate the aqueous tension at 300K.

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Solution

$P_{dry gas} = P_{o b s e r v e d} - aqueous tension$
Aqueous tension = $P_{o b s e r v e d} - P_{dry gas}$
$P_{o b s e r v e d} = 0.9 a t m$
$P_{dry gas} = ?$
According to combined gas law
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V 2}{T 2}$ [Experimental condition STP condition]
Pressure of dry gas $P 1 = \frac{P_{2} V_{2}}{T_{2}} \times \frac{T_{1}}{V_{1}} = 1 \times 0.080283 \times 2730.1 = 0.7717 a t m$
Aqueous tension = $P_{o b s e r v e d} - P_{dry gas} = 0.9 - 0.7717 a t m = 0.1283 a t m$

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