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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]A. 0.18 gmB. 0.36 gmC. 0.09 gmD. 0.9 gm |
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Answer» Correct Answer - A `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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