Question

a certain quantity of gas occupies a volume of 0.1 L when collected over water at 288 K and a pressure of 0.92 bar. the same volume of gas occupied a volume 0.85 L at S.T.P in dry conditions. Calculate the aqueous tension at 288K.

Answer 1

We know that

Ptotal = Pdry gas + Pwater = Pdry gas + aqueous tension [Dalton's law]

Ptotal = 0.92 bar

So

Aqueous tension = 0.92 - Pdry gas

Now moles of gas is constant

We can use ideal gas equation

PV = nRT

P1V1 / T1 = P2V2 /T2 [n and R constant]

P1 X 0.1 / 288 = 1 X 0.085 / 273 .15 = 0.896

so aqueous tension = 0.92 - 0.896 = 0.024 bar