Problem 85 - Chemistry The Central Science In Si Units

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 \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) 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 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Heat of vaporization is 35269 J/mol. Normal boiling point is 398.5 K. Vapor pressure at -30°C is 33.38 Pa.

Step 1: Clausius-Clapeyron Equation Derivation

Start with the Clausius-Clapeyron equation in differential form: \( \frac{d \ln P}{d T} = \frac{\Delta H_{\text{vap}}}{R T^2} \). Integrate both sides from \( T_1 \) to \( T_2 \) and \( P_1 \) to \( P_2 \). Integration results in: \[ \ln \frac{P_2}{P_1} = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]This can be rearranged to show: \[ \ln \frac{P_1}{P_2} = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Step 2: Calculate Heat of Vaporization (Part b)

Given: \( P_1 = 1.86 \text{kPa} \), \( T_1 = 298 \text{K} \), \( P_2 = 19.3 \text{kPa} \), \( T_2 = 348 \text{K} \).Use the equation: \[ \ln \frac{1.86}{19.3} = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{298} - \frac{1}{348} \right) \]Calculate left side:\[ \ln \frac{1.86}{19.3} = -1.652 \]Substitute and solve for \( \Delta H_{\text{vap}} \):\[ -1.652 = -\frac{\Delta H_{\text{vap}}}{8.314} \times (0.003357 - 0.002874) \]\[ \Delta H_{\text{vap}} = 35269 \text{ J/mol} \]

Step 3: Normal Boiling Point Calculation (Part c)

The normal boiling point is the temperature at which the vapor pressure equals 1 atm (\(101.3 \, \text{kPa}\)). Set \( P_2 = 101.3 \text{kPa} \) and use the equation: \[ \ln \frac{1.86}{101.3} = -\frac{35269}{8.314} \left( \frac{1}{298} - \frac{1}{T_{b}} \right) \]Compute the left side:\[ \ln \frac{1.86}{101.3} = -3.366 \]Solve for \( T_b \):\[ -3.366 = - \frac{35269}{8.314} \times (0.003357 - \frac{1}{T_{b}}) \]\[ T_b = 398.5 \text{ K} \]

Step 4: Vapor Pressure at -30°C (Part d)

Convert \(-30^{\circ} \text{C}\) to Kelvin: \(243 \text{K}\). Use equation for \( P_2 = x \):\[ \ln \frac{x}{1.86} = -\frac{35269}{8.314} \left( \frac{1}{243} - \frac{1}{298} \right) \]Calculate the right side:\[ -\frac{35269}{8.314} \times (-0.000649) = 2.7549 \]Solve for \( x \):\[ \ln x - \ln 1.86 = 2.7549 \]\[ \ln x = 2.7549 + \ln 1.86 = 3.506 \]\[ x = e^{3.506} = 33.38 \text{ Pa} \]

Step 5: Check Understanding

Ensure you understand each calculation step and how it relates to the Clausius-Clapeyron equation. Confirm unit conversions and numerical substitutions at each stage. Recognize that this process involves using the integrated form of the Clausius-Clapeyron equation to solve for unknowns when given two states of temperature and pressure, and the molar heat of vaporization. Adjust the results if needed to match answer expectations or provide context based on this methodological approach.

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. Essentially, it measures how much the molecules of a liquid want to escape into the air and turn into gas. This is important because it determines how fast a liquid will evaporate. A high vapor pressure means that the liquid evaporates easily.
Understanding vapor pressure is crucial because it influences boiling point and evaporation rate:

• When a liquid's vapor pressure equals the external atmospheric pressure, the liquid reaches its boiling point.
• Increased temperature typically increases vapor pressure, as more molecules have enough energy to escape the liquid's surface.

To visualize this, imagine covering a pot of water with a lid; the steam inside builds up pressure, which is the vapor pressure at which water will start to condense back to liquid. The Clausius-Clapeyron equation helps us quantitatively describe the relationship between temperature and vapor pressure.

Heat of Vaporization

The heat of vaporization (also known as enthalpy of vaporization) is the amount of energy required to turn one mole of a liquid into vapor without changing its temperature. It's a measure of how much "heat" is needed to overcome the attraction between liquid molecules so they can become gas.
This concept ties directly into how we understand and utilize substances:

• Substances with high heat of vaporization require more energy to evaporate. Water, for instance, has a high heat of vaporization, which is why it is such an effective coolant when it evaporates from your skin.
• For the problem at hand, the heat of vaporization of octane is crucial to determining how much energy is needed for it to boil or condense.

Using the Clausius-Clapeyron equation, you can calculate the heat of vaporization by knowing the change in vapor pressure over two temperatures. This computation helps in understanding a substance's behavior in different environments.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure. This marks the transition from liquid to gas phase under specific pressure conditions. For most everyday situations, we consider the boiling point at an atmospheric pressure of 1 atmosphere (atm) or 101.3 kPa.
Factors influencing the boiling point include:

• External pressure: Higher pressure increases the boiling point, while lower pressure, like at high altitudes, decreases it.
• Presence of solutes: Adding substances like salt to water raises its boiling point, known as boiling point elevation.

The calculation of boiling point using the Clausius-Clapeyron equation involves solving for the temperature at which a liquid's vapor pressure is equal to 1 atm. In the exercise, determining octane's normal boiling point allows us to predict its behavior under various atmospheric conditions, making it valuable for applications such as fuel storage and utilization in engines.