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
(1) 13.0 kcal
(2) -13.0 kcal
(3) 27.0 kcal
(4) -27.0 kCal​

Answer 1

Answer:

1 ) 13.0 kcal is the answer

Answer 2

hii dear

this is ur answer

3H2O(l)→3H2O(g)3H2O(l)→3H2O(g)

Δng=3Δng=3

ΔE=ΔH−ΔngRTΔE=ΔH−ΔngRT

=3×10−3×500×0.002=27Kcal

(3) 27.0 kcal is ur right answer...