Explanation Given information: Mass of water = 454 g. Temperature = 373 K. The heat of vaporization = 40.7 k J / m o l . Pressure = 1 atm. Number of moles = G i v e n m a s s M o l a r m a s s The molar mass of water ( H 2 O ) = 2 × 1 + 16 = 2 + 16 = 18 g / m o l . The number of moles of water = 454 g 18 m o l = 25.22 m o l . The heat absorbed by water during vaporization is given by: q = n Δ H vap Where, n is the number of moles and Δ H vap is the heat of vaporization. Substitute, n = 25.2 mol and Δ H vap = 40.7 kJ/mol in the above expression, to get, q = 25.22 mol × 40.7 kJ mol = 1026.454 kJ ≈ 1030 kJ Thus, the value of q = 1030 kJ . As, Δ H = q at constant pressure, thus, Δ H = 1030 kJ . Now, Δ E is calculated for the vaporization of water as: H 2 O ( l ) → H 2 O ( g ) The work done is given by: w = − n R T [ as work is done by the system ] Where, n is number of moles, R is the universal gas constant and T is the change in temperature

Masteringchemistry with Pearson Etext -- Standalone Access Card -- For Chemistry

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Chapter 6, Problem 127E

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To determine: q,w,ΔE, and ΔH for evaporation of water.

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Chapter 6, Problem 126E

Chapter 6, Problem 128E

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