ThermodynamicsHard
Question
Assuming that water vapour is an ideal gas, the internal energy (ᐃU) when 1 mol of water is vapourised at 1 bar pressure and 100oC, (Given: Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol-1K-1) will be
Options
A.4.100 kJ mol-1
B.3.7904 kJ mol-1
C.37.904 kJ mol-1
D.41.00 kJ mol-1
Solution
H2O(l)
H2O(g)
ᐃng = 1 - 0 = 1
ᐃH = ᐃU + ᐃngRT
ᐃU = ᐃH - ᐃngRT
= 41 - 8.3 × 10-3 × 373
= 37.9 kJ mol-1
Hence, (3) is correct.
H2O(g)ᐃng = 1 - 0 = 1
ᐃH = ᐃU + ᐃngRT
ᐃU = ᐃH - ᐃngRT
= 41 - 8.3 × 10-3 × 373
= 37.9 kJ mol-1
Hence, (3) is correct.
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