ThermodynamicsEasy
Question
Standard enthalpy of vapourisation ᐃvapH- for water at 100oC is 40.66 kJmol-1. The internal energy ofvapourisation of water at 100oC (in kJ mol-1) is
Options
A.+40.66
B.+37.56
C.-43.76
D.+43.76
Solution
H2O(l) → H2O(g)
V1 of liq. H2O = 18 ml
and V2 of H2O (vap) =
= 30.6233 lit
ᐃV = (30623.3 ml - 18 ml)
= 30605.3 ml = 30.605 lit
ᐃE = ᐃH - PᐃV
= 40.66 × 103J - 1 × 30.605 × 101.325 J
= 36557.9 = 37.5589 kJ mol-1
V1 of liq. H2O = 18 ml
and V2 of H2O (vap) =
= 30.6233 litᐃV = (30623.3 ml - 18 ml)
= 30605.3 ml = 30.605 lit
ᐃE = ᐃH - PᐃV
= 40.66 × 103J - 1 × 30.605 × 101.325 J
= 36557.9 = 37.5589 kJ mol-1
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