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

Water vapour condenses at volume 4 cubic cm.

Explanation:

RH = 40%

V1 = 10 * 10^-6 m^3

RH = VP / SVP = 0.4

SVP = P0; condensation occurs when VP = P0

P1 = 0.4P0

P2 = P0

The process is isothermal. According to Boyle's law we get:

P1V1 = P2V2

V2 = P1V1 / P2

 = 0.4P0 * 10 * 10^-6 / P0

 = 4 * 10^-6 m

 = 4 cu. cm.

Water vapour condenses at volume 4 cubic cm.

Answer 2

The water vapor condenses $4.0 \mathrm{cm}^{3}$ in thickness

Explanation:

Step 1:

Given data in the question  

RH(Relative humidity)=40%  

$V_{1}=10 \times 10^{-6} \mathrm{m}^{3}$

$R H(\text { Relative humidity })=\frac{\text { Vapour pressure of water vapour }}{\text { saturated vapour pressure }}$

$=\frac{40}{100}=0.4$

Step 2:

Let Saturated vapour pressure = Po

When condensation occurs VP = Po.

$P_{1}=0.4 P_{0}$

$P_{2}=P_{0}$

As the phase is isothermal, we are able to apply Boyle's law

$P_{1} V_{1}=P_{2} V_{2}$

$V_{2}=\frac{P_{1} V_{1}}{P_{2}}$

$V_{2}=\frac{0.4 P_{0} \times 10 \times 10^{-6}}{P_{0}}$

$V_{2}=4 \times 10^{-6}$

$V_{2}=4.0 \mathrm{cm}^{3}$

Thus the water vapor condenses $4.0 \mathrm{cm}^{3}$ in thickness.