Question

Calculate the work done when 1 mole of water evaporates at 373K against 1 atm. Assume ideal gas behaviour.

Answer 1

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 relation
1 L atm = 101.3 Joules]

Answer 2

Answer:

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 relation

1 L atm = 101.3 Joules]