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24/09/2021 4:04 pm
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The latent heat of vapourization of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔU) of 3 moles of liquid at the same temperature
(a) 13.0 kcal/mol
(b) -13.0 kcal/mol
(c) 27.0 kcal
(d) -7.0 kcal/mol
1 Answer
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24/09/2021 4:08 pm
Correct answer: (c) 27.0 kcal
Explanation:
3H2O(l) → 3H2O(g);
Δn = 3, ΔE = ΔH - ΔnRT
= 30 - 3 x \(\frac{2}{1000}\) x 500
= 27 k cal