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24/09/2021 4:22 pm
Topic starter
What is the value of change in internal energy at 1 atm in the process?
H2O(1,323k) → H2O(g, 423 K)
Given : Cv,m (H2O, l) = 75.0 JK–1 mol–1;
Cp,m (H2O, g) = 33.314 JK–1 mol–1
ΔHvap at 373 K = 40.7 kJ/mol
(a) 42.91 kJ/mol
(b) 43086 kJ/mol
(c) 42.6 kJ/mol
(d) 49.6 kJ/mol
1 Answer
1
24/09/2021 4:30 pm
Correct answer: (c) 42.6 kJ/mol
Explanation:
Cv,m(H2O, g) = 33.314 – 8.314
= 25 J/K mol
ΔU2 = ΔH2 – Δng RT = 37.6
ΔUtotal = Δu1 + Δu2 + Δu3
= Cv,m (l) ΔT + ΔVvap + Cv,m (g) ΔT
= \(\frac{75 \times 50}{1000}\) + 37.6 + \(\frac{25 \times 50}{1000}\)
= 42.6 kJ/mol