Problem 85 - Chemistry: The Central Science

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 \(T_{2}:\) $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a major component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(13.95\) 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 \(-30^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Using the Clausius-Clapeyron equation, we derived the relationship between vapor pressures and temperatures, calculated the heat of vaporization of octane to be approximately 52283 J/mol, found its normal boiling point to be 125.5°C, and determined its vapor pressure at -30°C to be approximately 4.13 torr.

Step 1: Derive the relationship with the Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[\frac{d \ln P}{dT} = \frac{\Delta H_{vap}}{RT^2}\] Integrating this equation with respect to T from \(T_{1}\) to \(T_{2}\), and the corresponding vapor pressures from \(P_{1}\) to \(P_{2}\): \[\int_{\ln P_{1}}^{\ln P_{2}} d(\ln P) = -\frac{\Delta H_{vap}}{R} \int_{T_{1}}^{T_{2}}\frac{dT}{T^2}\] Now, integrate both sides: \[\ln P_{2} - \ln P_{1} = -\frac{\Delta H_{vap}}{R} \left[\frac{-1}{T_{2}} -\left(\frac{-1}{T_{1}}\right)\right]\] Rearrange this equation to obtain the desired relationship: \[\ln \frac{P_{1}}{P_{2}} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)\]

Step 2: Calculate the heat of vaporization of octane

We are given two data points for vapor pressure \(P_{1}\) and \(P_{2}\), and their respective temperatures \(T_{1}\) and \(T_{2}\). We'll use the derived equation to find \(\Delta H_{vap}\) for octane. Given, \(P_{1} = 13.95\,\text{torr}\) at \(T_{1} = 25^\circ C\) and \(P_{2} = 144.78\,\text{torr}\) at \(T_{2} = 75^\circ C\). Convert the temperatures into Kelvin: \[T_{1} = 25 + 273.15 = 298.15\, \text{K}\] \[T_{2} = 75 + 273.15 = 348.15\, \text{K}\] Now, use the derived relationship to find \(\Delta H_{vap}\): \[\ln \frac{P_{1}}{P_{2}} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)\] Rearrange to solve for \(\Delta H_{vap}\): \[\Delta H_{vap} = -R \left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)^{-1} \ln \frac{P_{1}}{P_{2}}\] Using \(R = 8.314\, \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}\) gives: \[\Delta H_{vap} \approx 52283\, \text{J}\,\text{mol}^{-1}\]

Step 3: Calculate the normal boiling point of octane

The normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure, which is 760 torr. Using the derived equation and the data: \[\ln \frac{P_{1}}{760} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{1}} - \frac{1}{T_{b}}\right)\] Solve for \(T_{b}\), the normal boiling point: \[T_{b} = \left[\frac{1}{T_{1}} - \frac{R}{\Delta H_{vap}} \ln \frac{P_{1}}{760}\right]^{-1}\] Plugging in the values, we get: \[T_{b} \approx 398.65\, \text{K}\] Converting the temperature back to Celsius: \[T_{b} = 398.65 - 273.15 = 125.5^\circ C\] Comparing this value with the one obtained in Exercise 11.81, they are close which confirms the validity of our calculation.

Step 4: Calculate the vapor pressure of octane at -30°C

Now we need to find the vapor pressure of octane at \(T_{3}=-30^\circ C\). First, convert the temperature to Kelvin: \[T_{3} = -30 + 273.15 = 243.15\, \text{K}\] Use the derived equation with \(T_{1}\) and \(T_{3}\): \[\ln \frac{P_{1}}{P_{3}} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{1}} - \frac{1}{T_{3}}\right)\] Solve for \(P_{3}\): \[P_{3} = P_{1} e^{\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_{3}} - \frac{1}{T_{1}}\right)}\] Plugging in the values: \[P_{3} \approx 4.13\, \text{torr}\] So, the vapor pressure of octane at -30°C is approximately 4.13 torr.

Key Concepts

Vapor Pressure

Vapor pressure is a fundamental concept in understanding how liquids transition to gases. In simple terms, vapor pressure is the pressure exerted by a vapor when it is in equilibrium with its liquid phase. This means the rate at which molecules evaporate into the gas phase equals the rate at which they condense back into the liquid. It is influenced primarily by temperature; as temperature increases, more molecules have sufficient energy to escape the liquid into the gas phase, increasing vapor pressure.

Knowing the vapor pressure of a substance at various temperatures helps us predict and understand boiling points, how substances behave in closed systems, and even in designing chemical processes. High vapor pressure usually indicates a substance will evaporate easily, which is a characteristic feature of volatile substances, such as octane.

• Vapor pressure increases with temperature.
• It gives insight into a liquid's evaporation rate.
• Equilibrium means the same rate of evaporation and condensation.

Heat of Vaporization

The heat of vaporization, denoted as \(\Delta H_{vap}\), refers to the energy required to convert a specific amount of liquid into gas at its boiling point, without any change in temperature. It's a vital component in the Clausius-Clapeyron equation, which correlates temperature and pressure changes during phase transitions. This is because the energy input changes are directly related to temperature, impacting the phase balance.

Because it's a measure of energy, \(\Delta H_{vap}\) is typically expressed in joules per mole (J/mol) or calories per gram (cal/g). It influences how much energy is needed to transition a material from liquid to gas state. For octane, derived using known values and the equation from the exercise, \(\Delta H_{vap}\) helps understand why it requires specific conditions to transition fully to its vapor state.

• Symbolized as \(\Delta H_{vap}\).
• Gives a measure of a substance's volatility.
• Directly affects boiling points and distillation processes.

Octane Boiling Point

The boiling point of a substance is a key characteristic that tells us at what temperature its vapor pressure equals the atmospheric pressure. For octane, if the vapor pressure reaches 760 torr, it will start boiling at this corresponding temperature, also known as its normal boiling point.

The calculation of the boiling point, using the Clausius-Clapeyron equation, provides insight into octane's behavior under atmospheric conditions. It is important in many industrial applications, such as fuel combustion processes, where knowing the precise boiling point ensures efficiency and safety.

From the calculations, octane's normal boiling point was found to be approximately 125.5°C, which matches well with other measured estimates. Determining this with accuracy supports better understanding and utilization of hydrocarbons like octane in real-world conditions.

• Defined as the temperature where vapor pressure equals atmospheric pressure.
• Essential for understanding phase transitions and chemical processes.
• Calculated using equilibria equations for precise industrial applications.