Final answer:
Work done during vaporization at constant pressure is zero, and the change in internal energy equals the enthalpy of vaporization, which is 30.7 kJ for 1 mole.
Step-by-step explanation:
To calculate the work (w) done during the vaporization of 1 mole of a liquid at its boiling point and 1.00 atm pressure, and to find the change in internal energy (ΔE), we can use the first law of thermodynamics which states that ΔE = q + w.
Here, q is the heat absorbed during vaporization, which is equivalent to the enthalpy of vaporization (ΔHvap), and w is the work done on the system by the surroundings or by the system on the surroundings during expansion.
Work done (w) is calculated using the equation w = -PΔV, where P represents the external pressure (1.00 atm) and ΔV is the change in volume.
Since the volume change during vaporization is not given, we focus on the heat aspect for ΔHvap. The ΔHvap is provided as 30.7 kJ/mol at 80℃.
Therefore, the heat absorbed (q) is 30.7 kJ for 1 mole.
Because the process occurs at constant pressure (boiling point under atmospheric pressure), the heat absorbed during vaporization is equal to the change in enthalpy (ΔHvap).
Thus, we can assume no volume change, and the work done (w) would be zero. Consequently, the change in internal energy (ΔE) for the process is equal to the heat absorbed, which is 30.7 kJ.