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Chapter 10: Problem 35

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 \(\ln P_{vap}\) versus \(1/T\) is linear with a negative slope because the Clausius-Clapeyron equation, when integrated and expressed in these terms, results in an equation of a straight line (\(y = mx + b\)) with a negative slope (\(m = -\frac{\Delta H_{vap}}{R}\)), where \(P_{vap}\) is vapor pressure, \(T\) is temperature, \(R\) is the gas constant, and \(\Delta H_{vap}\) is the enthalpy of vaporization (a constant for a given substance).

Step by step solution

01

The Clausius-Clapeyron equation is given by: \[ \frac{d \ln P_{vap}}{dT} = \frac{\Delta H_{vap}}{RT^{2}}, \] where \(P_{vap}\) is the vapor pressure, \(T\) is temperature, \(R\) is the gas constant, and \(\Delta H_{vap}\) is the enthalpy of vaporization (a constant for a given substance). #Step 2#: Integrate the Clausius-Clapeyron equation

When we integrate the Clausius-Clapeyron equation, we get: \[ \ln P_{vap} = \frac{-\Delta H_{vap}}{RT} + C, \] where \(C\) is the integration constant. #Step 3#: Express the equation in the required form
02

Our goal is to analyze the behavior of \(\ln P_{vap}\) vs \(1/T\). To do this, we should rewrite the equation we obtained in step 2 as follows: \[ \ln P_{vap} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T}\right) + C, \] #Step 4#: Compare the equation to the equation of a line

We can see that the equation obtained in step 3 matches the equation of a line (\(y = mx + b\)), where: - \(y = \ln P_{vap}\) - \(x = \frac{1}{T}\) - \(m = -\frac{\Delta H_{vap}}{R}\) (which is a negative constant) - \(b = C\) Since the equation is in the form of a straight line with a negative slope, the plot of \(\ln P_{vap}\) versus \(\frac{1}{T}\) is linear with a negative slope.

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