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Question 6.PS.5: Changes of State, ΔH, and ΔE Methanol, CH3OH, boils at 65.0 ......

Changes of State, ΔH, and ΔE

Methanol, CH_3OH, boils at 65.0 °C. When 5.0 g methanol boils at 1 atm, the volume of CH_3OH(g) is 4.32 L greater than the volume of the liquid. The heat transfer is 5865 J, and the process is endothermic. Calculate ΔH and ΔE. (The units 1 L × 1 atm = 101.3 J.)

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ΔH = 5865  J;  ΔE  = 5427  J

Strategy and Explanation The process takes place at constant pressure. By definition, ΔH = q_P. Because thermal energy is transferred to the system, the sign of ΔH must be positive. Therefore ΔH = 5865  J. Because the system expands, ΔV is positive. This makes the sign of w_{atm} negative, and

w_{atm} = – PΔV = – 1  atm × 4.32  L =  – 4.32  L  atm =  – 4.32 ×  101.3  J = – 438  J

To calculate ΔE, add the expansion work to the enthalpy change.

ΔE = q_P  +  w_{atm} =  ΔH +  w_{atm} =  5865  J –  438  J =  5427 J

Reasonable Answer Check Boiling is an endothermic process, so ΔH must be positive. Because the system did work on the surroundings, the change in internal energy must be less than the enthalpy change, and it is.

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