A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C. ∆vapH° for water at 373 K = 40.66 kJ mol–1.

Question

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy  of vaporization at 100°C. ∆vapH° for water at 373 K = 40.66 kJ mol–1.

Shiksha Nation · Accepted Answer

The process of evaporation is: 18g H2O(l) → 18g H2O(g)Number of moles in 18g H2O =18g/18g mol−1= 1 molΔng = 1 − 0 = 1 mol∴  Δvap U° = ΔvapH° − ΔngRT = 40.66 kJ mol−1 − (1 mol)(8.314 × 10−3 kJ K−1 mol−1)(298 K)= 40.66 kJ mol−1 − 3.10 kJ mol−1 = 37.56 kJ mol−1

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C. ∆_vapH° for water at 373 K = 40.66 kJ mol^–1.

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