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 $ln(P_{vap})$ 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 $\frac{-\mathrm{\Delta }{H}_{vap}}{R}$. 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

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: $$ln\left(\frac{P_2}{P_1}\right)=-\frac{\mathrm{\Delta }{H}_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ 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

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: $$ln(P_{vap})=\frac{-\mathrm{\Delta }{H}_{vap}}{R}\cdot \frac{1}{T}+C$$ Where C is a constant.

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)

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.

Key concepts

Understanding Vapor Pressure

Vapor pressure is an important concept when studying thermodynamics and phase changes. It refers to the pressure exerted by the vapor phase of a substance when it is in equilibrium with its liquid phase. This means that in a closed container, when the rate of evaporation equals the rate of condensation, the pressure exerted by the vapor molecules remains constant.
This equilibrium pressure is what we call the vapor pressure.

• Vapor pressure increases with temperature because as temperature rises, more liquid molecules have enough energy to escape into the vapor phase.
• A higher vapor pressure indicates a volatile substance or a substance that easily becomes a vapor.

Understanding vapor pressure is crucial when calculating boiling points and explaining phenomena such as evaporation and condensation.

Exploring Enthalpy of Vaporization

The enthalpy of vaporization, often represented as $\mathrm{\Delta }{H}_{vap}$, is the heat required to vaporize a given amount of liquid at constant pressure. It's an indicator of the strength of the intermolecular forces within a liquid.

• A high $\mathrm{\Delta }{H}_{vap}$ means strong intermolecular bonds, such as hydrogen bonds, are present.
• Conversely, a low $\mathrm{\Delta }{H}_{vap}$ indicates weaker interactions, which is typical for substances like alkanes.

Every substance has a characteristic enthalpy of vaporization which can be approximated as constant over a limited temperature range.
This simplification is useful when using the Clausius-Clapeyron equation to relate changes in temperature to changes in vapor pressure.

Linear Relationship in Clausius-Clapeyron Equation

A plot of $\ln (P_{vap})$ versus $1/T$ creates a straight line due to the nature of the Clausius-Clapeyron equation. This equation describes how vapor pressure changes with temperature, providing a linear form when expressed as: $$\ln (P_{vap})=\frac{-\mathrm{\Delta }{H}_{vap}}{R}\cdot \frac{1}{T}+C$$Where:

• $\ln (P_{vap})$ is analogous to the dependent variable $y$.
• $1/T$ is the independent variable $x$.
• The slope $m$ is $-\mathrm{\Delta }{H}_{vap}/R$.
• $C$ is the constant y-intercept.

The negative value of the slope arises because the enthalpy of vaporization is positive (requires energy to vaporize) while both $\mathrm{\Delta }{H}_{vap}$ and the gas constant, $R$, are positive, giving $-\mathrm{\Delta }{H}_{vap}/R$ a negative sign.
This leads to a negative slope, meaning as temperature decreases, vapor pressure also decreases, which is a typical characteristic of physical processes involving endothermic transitions like vaporization.