Question

Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kCal mol-1. What will be the
change in internal energy of 3 mol of liquid at same temperature and pressure
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Answer 1

Answer:

3A(l)⟶3A(g)

△n=3−0=3

△E=△H−△nRT

=3×10−3×500×0.002

=27kcal

Explanation:

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