Question

Ex. 6. 1 mole of ice at 0°C and 4.6 mmHg pressure is converted to water vapour
at a constant temperature and pressure. Find AH and AU if the latent heat of
fusion of ice is 80 cal/g and latent heat of vaporisation of liquid water at 0°C
is 596 cal/g and the volume of ice in comparison to that of water (vapour) is
neglected.​

Answer 1

Answer:

1 mole of ice=18g.H

2

O(s)

△H=80×18+596×18=12.168Kcal

H

2

O(s)⟶H

2

O(g)

△H=△E+△n

g

RT

here △

ng

=(gaseous moles of product)-(gaseous moles of reactant)

=1

△E=△H−Rt=12.168−2×273×10

−3

△E=11.62Kcal

Answer 2

Answer:

11623 Calories.

Step-by-step explanation:

Answer is 11.623 Kcal.