# 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.)

Step-by-Step
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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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