Question

A plot of \(\ln \left(P_{\text {vap }}\right)\) versus \(1 / T(\mathrm{~K})\) is linear with a negative slope. Why is this the case?

Short answer

The plot of \(\ln P_{\text{vap}}\) versus \(1/T\) is linear with a negative slope because the Clausius-Clapeyron equation relates vapor pressure and temperature in such a way that the natural logarithm of vapor pressure is proportional to the inverse of temperature with a negative constant of proportionality. This constant of proportionality is derived from the enthalpy of vaporization and the ideal gas constant, both of which are positive.

Step-by-step solution

Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation is used to describe the relation between the vapor pressure of a substance and its temperature. The equation is given as follows: \[\frac{d\ln P_{\text{vap}}}{dT} = \frac{\Delta H_{\text{vap}}}{RT^2}\] where \(\Delta H_{\text{vap}}\) is the enthalpy of vaporization, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(P_{\text{vap}}\) is the vapor pressure.

Integrate the Clausius-Clapeyron Equation

To find the relationship between \(\ln P_{\text{vap}}\) and \(T\), we need to integrate the equation: \[\int\frac{d\ln P_{\text{vap}}}{dT}dT = \int\frac{\Delta H_{\text{vap}}}{RT^2}dT\] Assuming that \(\Delta H_{\text{vap}}\) is constant during the process, we can integrate with respect to \(T\): \[\ln P_{\text{vap}} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T}\right) + C\] where \(C\) is the integration constant.

Understand the Linear Relationship

From the above equation, we can see that the relationship between \(\ln P_{\text{vap}}\) and \(1/T\) is linear. The equation can be written in the form of a linear equation: \[\ln P_{\text{vap}} = m\left(\frac{1}{T}\right) + b\] where \(m = -\frac{\Delta H_{\text{vap}}}{R}\) is the negative slope, and \(b = C\) is the y-intercept. This explains why the plot of \(\ln P_{\text{vap}}\) versus \(1/T\) is linear with a negative slope. The negative slope is because the enthalpy of vaporization is positive, and the gas constant is also positive.