1.

Mass of H_(2)O present in air in 10 lit. closed vessel with 80% relative humidity at 1 atm and 400 K ? [Vapour pressure of water at 300 K = 0.04 atm]

Answer»

0.18 gm
0.36 gm
0.09 gm
0.9 gm

Solution :`RH = 0.8 = (P_(H_(2)O (g)))/(V.P."of" H_(2)O) = (P_(H_(2)O (g)))/(0.04)`
`P_(H_(2)O(g)) = 0.8 xx 0.04`
PV = nRT for water wapour
`0.8 xx 0.04 xx 10 = n_(H_(2)O) xx (0.08)/(10) xx 400`
`n_(H_(2)O) = 0.01`
`W_(H_(2)O) = 0.01 xx 18 = 0.18 gm`


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