calculate work done when 1.0 mole of water at 373 K evaporates against an a
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V = nRT / P= (1 mol × 0.0821 L.atm/(mol K) × 373 K) / (1 atm)= 30.6 L
Work = P (Vf - Vi)= 1 atm × (30.6 L - 0 L) ………[∵ When compared to volume of vapour, volume of water is negligible]= 30.6 L atm
∴ Work done = 30.6 L atm
[Note: In case if you want answer in Joules then convert it by using the relation1 L atm = 101.3 Joules]
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Work = P (Vf - Vi)= 1 atm × (30.6 L - 0 L) ………[∵ When compared to volume of vapour, volume of water is negligible]= 30.6 L atm
∴ Work done = 30.6 L atm
[Note: In case if you want answer in Joules then convert it by using the relation1 L atm = 101.3 Joules]
plz mark it brainliest answer
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