Question

A certain quantity of a gas occupies a volume of 0.1 litre . When collected over water at 288 K and 0.92 pressure. The same gas occupied a volume of 0.085 litres at S.T.P in dry condition. Calculate aqueous tension at 28 K.

Answer 1

$P_{drygas}=P_{observed}-aqueoustension$

$Aqueoustension=P_{observed}-P_{drygas}$

$P_{observed}=0.92$

$P_{drygas}=?$

According to combine gas law

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

[Experimental condition STP condition]

pressure of dry gas

$P_1=\frac{P_2{V}_2}{T_2}\times \frac{T_1}{V_1}$

=$\frac{1\times 0.085}{-245}\times \frac{273}{0.1}$

=$0.94714$

Aqueous tension=$P_{observed}-P_{drygas}$

=$0.92-0.94714$ atm= $-0.02714$