Question

Q) Latent heat of vaporisation of a liquid at 500k and 1 atm pressure is 10.0kcal/mol. What will be the change in internal energy of 3mol of liquid at the same temperature?

Answer 1

3H2O(l)→3H2O(g)3H2O(l)→3H2O(g)

Δng=3Δng=3

ΔE=ΔH−ΔngRTΔE=ΔH−ΔngRT

=3×10−3×500×0.002=27Kcal=3×10−3×500×0.002=27Kcal

Here H is the enthalpy of vaporisation......

Answer 2

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