vapH for water is +40.79 kJ mol-1 at 1 bar and 100 C. Change in internal energy for this vapourisation under same condition is kJ mol-1. (Integer answer) (Given R = 8.3 JK-1 mol-1)
Answer
38
Solution
H2O(l) H2O(g), so ng = 1 and T = 373 K.
Since H = U + ng RT,
U = 40.79 - 1 8.3 3731000
= 40.79 - 3.10 = 37.69 kJ mol-1 38 kJ mol-1
Answer: 38