One mole of a liquid is vaporized at its boiling point, 65°C and 1.00 atm. ΔH vap for the liquid is 43.8 kJ/mol at 65° C.
-Calculate w.

Name: One mole of a liquid is vaporized at its boiling point, 65°C and 1.00 atm. ΔH vap for the liquid is 43.8 kJ/mol at 65° C. -Calculate w.
Brand: Quizplus
Rating: 4.7 (55 reviews)

Question

Quizplus · Accepted Answer

The work done (w) during vaporization at constant pressure is given by w = -PΔV. Using the ideal gas law, ΔV = nRT/P. For 1 mole, w = -RT = -(8.314 J/mol·K)(338 K) = -2.81 × 10^3 J.

One Mole of a Liquid Is Vaporized at Its Boiling