Question

1 mole of ice at 0˚C and 4.6 mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find ∆H and ∆E if the latent heat of fusion of ice is 80 Cal/gm and latent heat of vaporisation of liquid water at 0˚C is 596 Cal per gram and the volume of ice in comparison to that of water (vapour) is neglected. 

Answer 1

= 3703.07 L

∆H = ∆H_fusion + ∆H_vap + H

0ºC → 0ºC

ice water

∆H = nCp∆T

∆T = 0, ∴ ∆H_process = 0

∆H_total = (80 + 596) × 18 = 12168 cal

∆E = ∆H – (∆PV)

= ∆H – P∆V

= 12168 –4.6/760 x 3703.07

= 12168 – 536.20

= 11623 cal