Question

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the ClausiusClapeyron 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 \(T_{2}\) $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (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 \(11.86 .\) (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

Short answer

To derive the relationship between vapor pressures and absolute temperatures using the Clausius-Clapeyron equation and the given data, we first integrate the equation with respect to T, resulting in: $$ \ln \frac{P_1}{P_2} = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$ Using this equation and given data, we can calculate the heat of vaporization (∆Hvap) of octane as 40.69 kJ/mol. The normal boiling point of octane can then be calculated as 125.28°C. Finally, the vapor pressure of octane at -30°C is found to be 0.43 torr.

Step-by-step solution

Derive the relationship

The Clausius-Clapeyron equation is given by: $$ \dfrac{dP}{P} = \dfrac{\Delta H_{\text{vap}}}{RT^2}dT $$ Integrate both sides with respect to T, while P & T are varying in range [\(P_1, P_2\)] & [\(T_1, T_2\)]: $$ \int_{P_1}^{P_2} \frac{dP}{P} = -\frac{\Delta H_{\text{vap}}}{R} \int_{T_1}^{T_2}\frac{dT}{T^2} $$ After integration, we get: $$ \ln \frac{P_2}{P_1} = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$ Rearranging the equation, we get: $$ \ln \frac{P_1}{P_2} = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$

Calculate the heat of vaporization of octane

We'll use the equation derived above and the given data for the vapor pressures of octane at 25°C and 75°C to determine the heat of vaporization (∆Hvap) of octane. Given data: \(P_1 = 13.95\) torr at \(T_1 = 25\)°C = 298.15 K \(P_2 = 144.78\) torr at \(T_2 = 75\)°C = 348.15 K Gas constant (\(R\)) in appropriate units (L torr/ K mol): \(R = 62.363\) Plugging in the values, we have: $$ \ln \frac{13.95}{144.78} = -\frac{\Delta H_{\text{vap}}}{62.363} \left(\frac{1}{298.15} - \frac{1}{348.15}\right) $$ Solving for ∆Hvap, we get: $$ \Delta H_{\text{vap}} = (40.69 \; \text{kJ/mol}) $$

Calculate the normal boiling point of octane

We'll use the derived equation and data from part (b) to calculate the normal boiling point of octane. Normal boiling point corresponds to a vapor pressure of P = \(1 \;\text{atm} = 760 \;\text{torr}\). Let's denote the normal boiling point of octane as T. Using the equation, we have: $$ \ln \frac{13.95}{760} = -\frac{40.69 \times 10^3}{62.363} \left(\frac{1}{298.15} - \frac{1}{T}\right) $$ Solving for T, we get: $$ T = 398.43 \; \text{K} = \boxed{125.28 ^\circ \text{C}} $$

Calculate the vapor pressure of octane at -30°C

Now we'll use the derived equation along with given data to find the vapor pressure of octane at -30°C (243.15 K). Let's denote the vapor pressure of octane at -30°C as \(P_3\). Using the equation, we have: $$ \ln \frac{13.95}{P_3} = -\frac{40.69 \times 10^3}{62.363} \left(\frac{1}{298.15} - \frac{1}{243.15}\right) $$ Solving for \(P_3\), we get: $$ P_3 = 0.43\; \text{torr} $$ Thus, the vapor pressure of octane at \(-30^\circ \text{C}\) is approximately \(0.43\) torr.

Key concepts

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 crucial concept when studying phase transitions such as evaporation or boiling.
The higher the vapor pressure, the more volatile the substance.

• When the liquid vaporizes, molecules leave the liquid surface and enter the gaseous state.
• At equilibrium, the rate of evaporation equals the rate of condensation.
• Vapor pressures increase with temperature because more molecules have enough kinetic energy to escape into the vapor phase.

Understanding vapor pressure helps in calculating properties like boiling points and in applications involving gases and liquids in closed systems.

Heat of Vaporization

The heat of vaporization, denoted as \(\Delta H_{\text{vap}}\), is the amount of energy required to convert one mole of a liquid substance to its vapor at constant temperature.
This is an important thermodynamic quantity indicating the strength of intermolecular forces.

• It is usually expressed in kJ/mol.
• A higher \(\Delta H_{\text{vap}}\) means stronger intermolecular forces and lower volatility.
• It can be determined experimentally or using the Clausius-Clapeyron equation.

In the exercise, the heat of vaporization of octane is calculated using the vapor pressure data at two different temperatures, showcasing its practical application in thermodynamics.

Normal Boiling Point

The normal boiling point is the temperature at which a liquid boils under standard atmospheric pressure (1 atmosphere or 760 torr).
It's a specific case of boiling point at this pressure.

• At the boiling point, the vapor pressure of the liquid equals the external pressure.
• This transition from liquid to gas occurs throughout the liquid, forming bubbles of vapor.
• For octane, the calculated normal boiling point is a direct application of vapor pressure principles and the Clausius-Clapeyron equation.

Knowing a substance's boiling point is essential for many industrial and scientific applications, especially those involving phase changes.

Octane

Octane (\(\text{C}_8\text{H}_{18}\)) is a major component of gasoline, influencing performance and efficiency in engines.
It is a hydrocarbon, specifically an alkane, characterized by its eight-carbon chain.

• Its structural formula is \(\text{CH}_3\text{(CH}_2\text{)}_6\text{CH}_3\).
• Octane rating is a measure of fuel stability and its ability to resist knocking during combustion.
• The exercise utilizes octane's vapor pressure data to explore its thermodynamic properties.

Understanding octane is vital for chemical applications in energy, particularly in the context of fuels and thermodynamics.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and energy transformations.
It helps in understanding how energy is transferred within a system and how it affects matter.

• There are four laws of thermodynamics, each explaining different energy transfer aspects.
• The Clausius-Clapeyron equation is pivotal in studying phase transitions and relating them to thermodynamic principles.
• In the exercise, this equation is applied to determine thermal properties like vapor pressure and heat of vaporization.

Thermodynamics plays an essential role in a wide range of scientific and engineering fields. It provides a theoretical foundation for understanding chemical reactions, engines, and even biological processes.