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

Here , 

RH = 40%

`V_1 = 10 xx 10^-6  "m"^3`

`"RH" = "VP"/"SVP" = 0.4`

Let SVP = `P_0`

condensation occurs when VP = `P_0` .

⇒ `P_1 = 0.4P_0`

⇒ `P_2 = P_0`

Since the process is isothermal , applying Boyle's law we get 

`P_1V_1 = P_2V_2`

⇒ `V_2 = (P_1V_1)/P_2`

⇒ `V_2 = (0.4P_0 xx 10 xx 10^-6)/P_0`

⇒ `V_2 = 4.0 xx 10^-6`

⇒ `V_2 = 4.0  "cm"^3`

Thus water vapour condenses at volume `4.0  "cm"^3`