Question
Derive the vapor pressure equation (Clausius-Clapeyron equation): $d p / d T=?$
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The vapor pressure of a substance is the pressure exerted by its vapor when it is in equilibrium with its liquid (or solid) phase at a given temperature. Show more…
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Use the Gibbs or chemical potential function between the liquid-gas line to derive the Clausius-Clapeyron equation dP/dT = q / (T(vv - vL)), where q is the heat of vaporization of one particle, vL and vv are the volume of one particle in liquid and vapor, respectively. Assume that the vapor is an ideal gas and its density is much smaller than that of the liquid. Show that P ≈ e^(-q/kBT), where q is independent of T.
Show that the equations $$ \frac{d \ln p}{d T}=\frac{\Delta H_{\mathrm{vap}}}{R T^{2}} \text { and } \quad \frac{d p}{d T}=\frac{p \Delta H_{\mathrm{vap}}}{R T^{2}} $$ are equivalent expressions of the Clausius-Clapeyron equation.
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?
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