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Chapter 10: Problem 31

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

Expert verified
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

01

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.
02

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.
03

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.

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