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Calculate % of 'free volume' available in 1 mol gaseous water at 1.0 atm and 100^(@)C .Density of liquid water at 100^(@)C is 0.958 g//mol. |
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Answer» Solution :We know that PV = nRT Volume OCCUPIED by 1 mol gaseous WATER at `100^(@)C` and 1 atm pressure`= ( 1 xx 0.0821 xx 373 )/( 1) = 30.62 ` litre volume of 1 mol liquid water `= ( " mass")/( " density ") = ( 18) /( 0.958) = 1.88 xx 10^(-2) ` litre % of volume occupied by liquid water `= ( 1. 88 xx 10^(-2))/( 30.62 ) xx 100 = 0.0613 %` % of free volume = 100 - 0.0613 = 99.93% |
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