the volume occupied by 1kg of water at 100C and atmospheric pressure changes from 1L in the liquid phase to 1700L in the vapor phase. evaluate the difference in internal energy of 1kg of water vapor and 1kg of liquid water at the normal boiling point:

Question

So i have the equation is
delta U=Q + W
where Q=mLv and W=p*delta V

delta U=mLv - p*delta V
and
m=1kg
Lv=2260 kJ/kg
p= 1 atm= 101325 Pa
delta v= 1699L or 1.699m^3

this is not giving me the right answer though, which is supposed to be delta U=2090kJ

what am i doing wrong?

Answer 1

2260 kJ/kg * 1 kg = 2260 KJ

p delta V = 172.15 KJ

2260 KJ - 172.15 KJ = 2088 KJ

Answer 2

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Answer 3

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Answer 4

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