262use PV = nRT
where P = 1/2 * Psaturation
the relative humidity of air at any day is 50% and saturation vapuor pressure at 300K is 3.6 kPa. the amount of water vapours pers litre of air at 300K is?
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21Relativ humidity =50%
ie in 100g of air wt of water vapour=50g,wt of other part(unsaturated)=50g
We know VD=2*mol wt.
VD of air=14.4...
Mol. wt≈29
Mol wt of water=18
therefore no. of moles of air=50/29
no. of moles of water=50/18
partial pressure of unsat. air=sat.Pressure*Mole fraction of unsat. air=1378.7Pa
using PV=nRT
n=50/29
V=50/29*8.314*300/1378.7=3.12m3
3.12m3 air contains 50g water
1 lt air contains 50/3.12 *10-3g=16mg water
ANS:16mg water .