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Question
One mole of water at its boiling point vaporises against a constant external pressure of 1 atm at the same temperature. Assuming ideal behaviour and initial volume of water vapours to be zero, the work done by the system (in Joules) is :
One mole of water at its boiling point vaporises against a constant external pressure of 1 atm at the same temperature. Assuming ideal behaviour and initial volume of water vapours to be zero, the work done by the system (in Joules) is :
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Solution
The correct option is A −3102
The expression for the work done is as follows:
w=−PΔV=−P×(Vf−Vi)=−P×Vf as Vi=0
However, according to the ideal gas equation, PVf=nRT
Hence, w=−nRT=−1×8.314×373=−3102 J
The expression for the work done is as follows:
w=−PΔV=−P×(Vf−Vi)=−P×Vf as Vi=0
However, according to the ideal gas equation, PVf=nRT
Hence, w=−nRT=−1×8.314×373=−3102 J
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