Suppose the vapor pressure of a substance is measured at two...

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, $P_{1}$ and $P_{2}$, and the absolute temperatures at which they were measured, $T_{1}$ and $T_{2}$ : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a major component of which is octane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$. Octane has a vapor pressure of 13.95 torr at $25^{\circ} \mathrm{C}$ and a vapor pressure of 144.78 torr at $75^{\circ} \mathrm{C}$. Use these data and the equation in part (a) to calculate the heat of vaporization of octane. (c) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.80 . (d) Calculate the vapor pressure of octane at $-30^{\circ} \mathrm{C}$.

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, $P_{1}$ and $P_{2}$, and the absolute temperatures at which they were measured, $T_{1}$ and $T_{2}$ :
$$
\ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)
$$
(b) Gasoline is a mixture of hydrocarbons, a major component of which is octane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$.
Octane has a vapor pressure of 13.95 torr at $25^{\circ} \mathrm{C}$ and a vapor pressure of 144.78 torr at $75^{\circ} \mathrm{C}$. Use these data and the equation in part (a) to calculate the heat of vaporization of octane.
(c) By using the equation in part (a) and the data given in part
(b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.80 .
(d) Calculate the vapor pressure of octane at $-30^{\circ} \mathrm{C}$.

Key Concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a fundamental relation in thermodynamics that describes how the vapor pressure of a substance changes with temperature. It provides a way to relate the enthalpy of vaporization to the variation in vapor pressure over a range of temperatures, and it can be integrated (assuming constant enthalpy of vaporization) to yield a logarithmic relationship between the ratios of vapor pressures and the inverse temperatures.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. It is a measure of a substance's tendency to evaporate, and it increases with temperature because more molecules have sufficient energy to leave the liquid phase.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit amount of a substance from its liquid phase to its vapor phase at constant temperature and pressure. It is a key parameter in the Clausius-Clapeyron equation and can be calculated from the dependence of vapor pressure on temperature.

Normal Boiling Point

The normal boiling point of a substance is defined as the temperature at which its vapor pressure equals the ambient atmospheric pressure (typically 760 torr or 1 atm) under standard conditions. It is determined by finding the temperature that satisfies this condition using the vapor pressure relation provided by the Clausius-Clapeyron equation.

Transcript

Please give Ace some feedback