Vapor Pressure of water?
A volume Vi = 27.2 L of air at temperature T = 21 °C is compressed at constant temperature to a volume Vf = 8.4 L. The relative humidity of the air before compression is R = 60.8 %. Saturated water vapour in air at this temperature has a vapour pressure vp = 2.5 kPa and a density of ρ = 18.4 g.m−3.
What is the density the water vapour would have had following compression if it behaved like an ideal gas instead of experiencing partial condensation?
What is the density of water vapour in the air following compression?
What mass of water condenses out because of the compression?
10points for good explanation, don't really need the answers just a strategy would be fine. Thanks in advance.
- Ivan ALv 61 decade agoFavorite Answer
working on this... I don't think anyone around here but me knows how to solve this.
Got it. First you need the total mass of water. At a RH = 60% the vapor pressure is
Pv = 2.5 * RH = 1.52 kPa
from steam tables, the density of water is 0.011205 kg/m^3. You can go ahead and calculate the mass of water out of this (0.3 grs). Now if you compress the water and the air and assuming that the water stays as vapor, the density is just.
density = mass of water / Vf = 0.3 / 8.4E-3 = 36 grs/m^3
This density is larger than the saturation density which means that some vapor condensed out. In saturation, the water vapor retains the density of 18.4 gr/m^3 while the liquid water density is the standard 1E6 gr/m^3. The total density, which is still 36 gr/m^3 is the combination of vapor and liquid, that is
density = mass_vapor (18.4) + mass_liquid (1E6) = 36
and the total mass is
mass_vapor + mass_liquid = 0.3
now you have two eqs with two unknowns, you can solve for either mass and find out how much it was condensed out.