Relative humidity id defind as the ration of partial pressure of water in air at a given temperature to the vapour pressure of water at that that temperature Calculate the mass of water per litre air at
(a) `20^(@)C` and `45%` relative humidity
(b) `0^(@)C` and `95%` relative humidity
(c ) Discuss wether temperature or relative humidity has the greater effect on the mass of water vapour in air Given `P_(H_(2)O)^(@) =17.5` torr at `20^(@)C` and 4.58 torr `0^(@)C` .

Explanation

Relative humidity = `(P'_(H_2O))/(P_(H_2O))`
(a) `P'_(H_2O)` in air at `20^(@)C = (45)/(100) xx 17.5`
` =7.88` torr (ormm)
`:. n_(H_(2)O)` in air at `20^(@)C`
` = (PV)/(RT) = (7.88 xx 1)/(760 xx 0.0821 xx 293)` (V =1 litre)
`= 4.31 xx 10^(-4)`
`:. w_(H_(2))O` in air at `20^(@)C = 4.31 xx 10^(-4) xx 18`
`= 7.76 xx 10^(-3) g`
` =7.76 mg`
(b) `P'_(H_(2))O` in air `0^(@)C =(95)/(100) xx 4.58 =4.35` torr
`:. n_(H_(2))O` in air at `0^(@)C`
`= (PV)/(RT) = (4.35 xx 1)/(760 xx 0.0821 xx 273) = 2.55 xx 10^(-4)`
`:. w_(H_(2))O` in air at `0^(@)C =2.55 xx 10^(-4) xx 18`
`= 4.60 xx 10^(-3) g = 4.60mg`
(c) Despite the higher per cent humidity in part (b) there is less water in air because of much lower vapour pressure of water at `0^(@)C` However both play important role in determing the mass of water .