Question

Pure water vapour is trapped in a vessel of volume 10 cm³. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

Answer 1

$\begin{array}{l}\text{Here,}\\ \text{RH=40\%}\\ V_1=10\times {10}^{-6}{\text{m}}^{\text{3}}\\ \text{RH=}\frac{VP}{SVP}=0.4\\ {\text{Let SVP = P}}_o\\ {\text{Condensation occurs when VP = P}}_o\text{.}\\ \\ \Rightarrow P_1=0.4P_o\\ \Rightarrow P_2=P_o\\ \text{Since the process is isothermal, applying Boyle's law we get}\\ P_1{V}_1=P_2{V}_2\\ \Rightarrow V_2=\frac{P_1{V}_1}{P_2}\\ \Rightarrow V_2=\frac{0.4P_o\times 10\times {10}^{-6}}{P_o}\\ \Rightarrow V_2=4.0\times {10}^{-6}\\ \Rightarrow V_2=4.0{\text{cm}}^3\\ \text{Thus water vapour condenses}\mathrm{at\; volume\; 4}{\text{.0 cm}}^3\text{.}\end{array}$