The Clapeyron-Clausisus equation (which we will derive in...

The Clapeyron-Clausisus equation (which we will derive in Chapter 23) is $$ \ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] $$ This equation, which assumes that $\Delta H_{\text {vap }}$ does not vary with temperature, relates the change in vapor pressure and temperature to a substance's enthalpy of vaporization (or sublimation), where $R$ is the molar gas constant. Use this relationship and the fact that $\Delta H_{\text {vap }}$ of water at $25^{\circ} \mathrm{C}$ is $43.99 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$ to calculate the vapor pressure of water at $5^{\circ} \mathrm{C}, 25^{\circ} \mathrm{C}, 50^{\circ} \mathrm{C}$, and $95^{\circ} \mathrm{C}$. Compare your results to the values in Table 15.7. What might account for the slight discrepancy between the values in Table $15.7$ and your results?

The Clapeyron-Clausisus equation (which we will derive in Chapter 23) is
$$
\ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]
$$
This equation, which assumes that $\Delta H_{\text {vap }}$ does not vary with temperature, relates the change in vapor pressure and temperature to a substance's enthalpy of vaporization (or sublimation), where $R$ is the molar gas constant. Use this relationship and the fact that $\Delta H_{\text {vap }}$ of water at $25^{\circ} \mathrm{C}$ is $43.99 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$ to calculate the vapor pressure of water at $5^{\circ} \mathrm{C}, 25^{\circ} \mathrm{C}, 50^{\circ} \mathrm{C}$, and $95^{\circ} \mathrm{C}$. Compare your results to the values in Table 15.7. What might account for the slight discrepancy between the values in Table $15.7$ and your results?

Key Concepts

Clapeyron-Clausius Equation

This equation relates the vapor pressure of a substance to its temperature and enthalpy of vaporization, providing a way to predict how vapor pressure changes as temperature varies. It assumes the enthalpy of vaporization remains constant over the temperature range of interest, simplifying the relationship through the integration of the Clausius-Clapeyron differential equation.

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to convert one mole of a liquid into vapor at a constant temperature and pressure. In this context, it is assumed to be independent of temperature, which simplifies the calculation but can lead to discrepancies compared to experimental values recorded in tables.

Vapor Pressure

Vapor pressure reflects the pressure exerted by a vapor in equilibrium with its liquid or solid form and is highly sensitive to temperature changes. The Clausis-Clapeyron equation is used to calculate vapor pressure at different temperatures, and comparing these calculated values with experimental data can reveal the limitations of the assumptions used in the derivation.

Assumptions and Approximations

A key assumption in using the Clausius-Clapeyron equation is that the enthalpy of vaporization remains constant over the considered temperature range. In reality, this value can vary slightly with temperature, contributing to the discrepancies observed between calculated results and those found in experimental tables. Other factors such as non-ideal behavior and measurement uncertainties can also introduce variations.

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