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Chapter 9: Problem 33

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 \(\frac{1}{T}\) is linear with a negative slope because this relationship follows the modified Clausius-Clapeyron equation, which resembles the equation of a straight line. The slope of the line is given by the constant \(\frac{-\Delta H_{vap}}{R}\). Since ΔH_vap is always positive (energy is absorbed during vaporization) and R is also positive, their ratio is bound to be negative, resulting in a negative slope for the linear relationship between \(ln(P_{vap})\) and \(\frac{1}{T}\).

Step by step solution

01

Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is used to determine the vapor pressure of a substance at a given temperature. The equation is: \[ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2}-\frac{1}{T_1}\right)\] Where: - P1 and P2 are the vapor pressures at temperatures T1 and T2 - ΔH_vap is the enthalpy of vaporization - R is the gas constant
02

Simplify the Clausius-Clapeyron Equation

We can rearrange the Clausius-Clapeyron equation to isolate ln(P_vap) on the left side of the equation, and we will assume that the enthalpy of vaporization is constant: \[ln(P_{vap}) = \frac{-\Delta H_{vap}}{R} \cdot \frac{1}{T} + C\] Where C is a constant.
03

Compare with the equation of a straight line

Now, we compare our modified Clausius-Clapeyron equation with the general equation for a straight line, which is: \[y = mx + b\] Where: - y is the dependent variable (in our case, ln(P_vap)) - x is the independent variable (in our case, 1/T) - m is the slope of the line (in our case, -ΔH_vap/R) - b is the y-intercept (in our case, C)
04

Explain the linear relationship and negative slope

From the analysis above, it is clear that the plot of ln(P_vap) versus 1/T is linear because this relationship follows the equation of a straight line. The slope of the line is given by the constant (-ΔH_vap/R). Since ΔH_vap is always positive (energy is absorbed during vaporization) and R is also positive, their ratio is bound to be negative, resulting in a negative slope for the linear relationship between ln(P_vap) and 1/T.

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