Question

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 $T_2$ : $$\ln \frac{P_1}{P_2}=-\frac{\mathrm{\Delta }{H}_{\text{vap }}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$ (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 $-{30}^{\circ }\mathrm{C}$.

Short answer

(a) We derived the relationship between vapor pressures and absolute temperatures using the Clausius-Clapeyron equation: $$\ln \frac{P_2}{P_1}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ (b) By using the provided data and the derived equation, we calculated the heat of vaporization of octane as $\Delta H_{\text{vap}} = 92784.96\, J/mol$. (c) Using the derived equation and part (b) data, we calculated the normal boiling point of octane as $T_3 = 116.85^{\circ}C$. (d) We calculated the vapor pressure of octane at $-30^{\circ}C$ as $333.74\,\text{torr}$.

Step-by-step solution

Clausius-Clapeyron equation

Write down the Clausius-Clapeyron equation: $$\frac{d\ln P}{dT}=\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R{T}^2}$$

Integrate

Integrate the equation with respect to temperature T from T1 to T2 and pressure P from P1 to P2: $${\int }_{T_1}^{T_2}\frac{d\ln P}{dT}dT={\int }_{P_1}^{P_2}\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R{T}^2}dT$$

Solve the integral

Solve the integral for both sides: $$\ln \frac{P_2}{P_1}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ (b)

Convert temperatures to Kelvin

We need to convert the given temperatures to Kelvin: $$T_1={25}^{\circ }C+273.15=298.15\text{K}$$ $$T_2={75}^{\circ }C+273.15=348.15\text{K}$$

Convert vapor pressures to atm

We need to convert the vapor pressures to atmospheres (atm): $$P_1=13.95\text{torr}\times \frac{1\text{atm}}{760\text{torr}}=0.018355\text{atm}$$ $$P_2=144.78\text{torr}\times \frac{1\text{atm}}{760\text{torr}}=0.190500\text{atm}$$

Calculate the heat of vaporization

Plug the values into the derived equation and solve for the heat of vaporization, ΔHvap: $$\ln \frac{P_1}{P_2}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$ $$\ln \frac{0.018355}{0.190500}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{8.314}(\frac{1}{298.15}-\frac{1}{348.15})$$ $$\mathrm{\Delta }{H}_{\text{vap}}=8.314J/(mol\cdot K)\times 11183.374\text{K}=92784.96J/mol$$ (c)

Calculate the normal boiling point

First, let's denote the normal boiling point as T3. At the normal boiling point, Poctane = 1 atm = 760 torr. We will use our derived equation and the values for ΔH(vap) and T1 and P1: $$\ln \frac{P_1}{P_3}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_1}-\frac{1}{T_3})$$ $$\ln \frac{13.95}{760}=-\frac{92784.96}{8.314}(\frac{1}{298.15}-\frac{1}{T_3})$$ Solving for T3, we get: $$T_3=390\text{K}$$ $$T_3=390\text{K}-273.15={116.85}^{\circ }C$$ (d)

Calculate vapor pressure at -30°C

Let T4 = -30°C, P4 = ? First, convert the temperature to Kelvin: $$T_4=-{30}^{\circ }C+273.15=243.15\text{K}$$ Now apply the derived equation: $$\ln \frac{P_1}{P_4}=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_1}-\frac{1}{T_4})$$ $$\ln \frac{13.95}{P_4}=-\frac{92784.96}{8.314}(\frac{1}{298.15}-\frac{1}{243.15})$$ Solving for P4, we get: $$P_4=0.439\text{atm}=333.74\text{torr}$$ So, the vapor pressure of octane at -30°C is 333.74 torr.

Key concepts

Vapor Pressure

Vapor pressure is an essential concept in understanding the physical behavior of a substance at different temperatures. It is defined as the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. Simply put, it's the measure of how much a substance tends to evaporate. At equilibrium, the rate of evaporation of liquid (or sublimation of a solid) equals the rate of condensation of the vapor, leading to a constant vapor pressure.

When the temperature increases, the kinetic energy of the molecules also increases, leading to more frequent and energetic collisions. This results in a higher vapor pressure, as more molecules have enough energy to escape into the vapor phase. Understanding vapor pressure is crucial when studying changes of state, such as boiling, which is when the vapor pressure equals the external atmospheric pressure. The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature, providing a way to predict how vapor pressure changes with temperature and to calculate the heat of vaporization, another fundamental concept.

Heat of Vaporization

The heat of vaporization, $\mathrm{\Delta }{H}_{\text{vap}}$, is the amount of heat energy required to convert a given amount of a substance from its liquid phase into its gaseous phase without a change in temperature. It's an intrinsic property of a substance that quantifies the energy barrier to overcome intermolecular forces during phase transition. In the Clausius-Clapeyron equation, the heat of vaporization is represented and can be calculated if the vapor pressures at two different temperatures are known.

As demonstrated in the step-by-step solution, we can calculate the heat of vaporization for octane by rearranging the Clausius-Clapeyron equation. This value is critical since it provides insight into the molecular interactions within the liquid – the higher the heat of vaporization, the stronger the molecules are held together. The calculated heat of vaporization is not just an abstract number; it has practical implications, such as determining the energy required for processes like distillation and the cooling effect of a liquid evaporating.

Normal Boiling Point

The normal boiling point of a substance is the temperature at which its vapor pressure equals the external pressure of 1 atmosphere (atm). This is an important characteristic temperature, as it's often used to identify and classify materials. The normal boiling point can be predicted using the Clausius-Clapeyron equation if the heat of vaporization and at least one known vapor pressure at a certain temperature are available.

In the context of the given exercise, once we have the heat of vaporization, we can determine the normal boiling point of octane by setting the vapor pressure equal to 1 atm in the equation and solving for the unknown temperature. The ability to calculate the normal boiling point provides valuable information about a substance's volatility and helps in understanding its behavior under different atmospheric conditions. For instance, knowing the boiling point of a fuel like octane can lead to better understanding and optimization of engine performance under various temperature conditions.