Calculate w and ΔE when one mole of a liquid is vaporized at...

Calculate $w$ and $\Delta E$ when one mole of a liquid is vaporized at its boiling point $\left(80 .^{\circ} \mathrm{C}\right)$ and $1.00$ atm pressure. $\Delta H_{\mathrm{vap}}$ for the liquid is $30.7 \mathrm{~kJ} / \mathrm{mol}^{-1}$ at $80 .{ }^{\circ} \mathrm{C}$.

Key Concepts

Enthalpy of Vaporization

Enthalpy of vaporization (?Hvap) is the amount of energy required per mole to convert a substance from the liquid phase to the vapor phase at its boiling point. It is a measure of the latent heat involved in the phase change, reflecting both the energy needed to overcome intermolecular attractions and the work done during expansion against the ambient pressure.

Work Done in Expansion

When a liquid vaporizes at constant pressure, the system does work on the surroundings by expanding. This work (w) can be calculated using the formula w = -P?V, where P is the pressure and ?V is the change in volume. In vaporization, the volume of a gas is significantly larger than that of a liquid, so this expansion work is a key component of the energy balance.

Relationship Between Enthalpy and Internal Energy

In a constant pressure process, the change in enthalpy (?H) is related to the change in internal energy (?E) by the work done during expansion. The relation is given by ?H = ?E + P?V. In the context of vaporization, since some of the ?Hvap is used to do work (expansion), the internal energy change (?E) is lower than the enthalpy change by the amount of work done.

Ideal Gas Law Application

The ideal gas law (PV = nRT) is used to estimate the change in volume (?V) when the liquid is vaporized. Since the liquid occupies a negligible volume compared to its vapor, the molar volume of the gas at the boiling point can be approximated using this law. This allows for the calculation of work done against the external pressure during the phase change.

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