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, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\) \(\mathrm{C} \mathrm{H}_{2} \mathrm{C} \mathrm{H}_{3} .\) 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

The heat of vaporization of octane is approximately \(39144.30\,\mathrm{J/mol}\). The normal boiling point of octane is about \(125^{\circ}\mathrm{C}\). The vapor pressure of octane at \(-30^{\circ}\mathrm{C}\) is approximately \(0.53\,\mathrm{torr}\).

Step-by-step solution

Derive the relationship between the vapor pressures and the absolute temperatures

The Clausius-Clapeyron equation is given by: $$ \frac{d\ln P}{dT}=\frac{\Delta H_{vap}}{RT^2} $$ Integrating both sides, we get: $$ \int_{P_1}^{P_2} d\ln P = -\frac{\Delta H_{vap}}{R} \int_{T_1}^{T_2} \frac{dT}{T^2} $$ After integrating, we obtain the relationship: $$ \ln \frac{P_1}{P_2}= -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) $$

Calculate the heat of vaporization of octane

Given the vapor pressures of octane, \(P_1 = 13.95\,\mathrm{torr}\) at \(T_1 = 25^{\circ}\mathrm{C}\) and \(P_2 = 144.78\,\mathrm{torr}\) at \(T_2 = 75^{\circ}\mathrm{C}\). We can use these values and the derived relationship to find \(\Delta H_{vap}\). First, convert temperatures to Kelvin: $$ T_1 = (25 + 273.15)\,\mathrm{K} = 298.15\,\mathrm{K} $$ $$ T_2 = (75 + 273.15)\,\mathrm{K} = 348.15\,\mathrm{K} $$ Plug the values into the equation and solve for \(\Delta H_{vap}\): $$ \ln \frac{13.95}{144.78}= -\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{298.15}-\frac{1}{348.15}\right) $$ $$ \Delta H_{vap}= 8.314 \times \frac{\ln\frac{13.95}{144.78}}{\frac{1}{298.15}-\frac{1}{348.15}} \approx 39144.30\,\mathrm{J/mol} $$

Calculate the normal boiling point of octane

The normal boiling point of octane occurs at a vapor pressure of \(1\,\mathrm{atm}\), or \(760\,\mathrm{torr}\). Using the relationship derived earlier, we can solve for the temperature (\(T_{nb}\)) at this vapor pressure: $$ \ln \frac{13.95}{760}= -\frac{39144.30}{8.314}\left(\frac{1}{298.15}-\frac{1}{T_{nb}}\right) $$ Solving for \(T_{nb}\), we get: $$ T_{nb}= \frac{1}{\frac{1}{298.15} - \frac{8.314}{39144.30} \times \ln\frac{13.95}{760}} \approx 398.15\,\mathrm{K} $$ The normal boiling point in Celsius is: $$ T_{nb}^{\circ}\mathrm{C} = 398.15\,\mathrm{K} - 273.15 \approx 125^{\circ}\mathrm{C} $$

Calculate the vapor pressure of octane at \(-30^{\circ}\mathrm{C}\)

To find the vapor pressure of octane at \(-30^{\circ}\mathrm{C}\), we need to determine the temperature in Kelvin: $$ T = (-30 + 273.15)\,\mathrm{K} = 243.15\,\mathrm{K} $$ Now, we use the derived relationship and the known values (e.g., \(\Delta H_{vap}\) and \(P_1\) at \(T_1=298.15\,\mathrm{K}\)) to find the vapor pressure (\(P\)) at \(T=243.15\,\mathrm{K}\): $$ \ln \frac{13.95}{P}= -\frac{39144.30}{8.314}\left(\frac{1}{298.15}-\frac{1}{243.15}\right) $$ Solving for \(P\), we get: $$ P = \frac{13.95}{\exp\left(-\frac{39144.30}{8.314}\left(\frac{1}{298.15}-\frac{1}{243.15}\right)\right)} \approx 0.53\,\mathrm{torr} $$ Therefore, the vapor pressure of octane at \(-30^{\circ}\mathrm{C}\) is approximately \(0.53\,\mathrm{torr}\).

Key concepts

Vapor Pressure

Vapor pressure is a crucial concept in understanding phases of materials. It refers to the pressure exerted by the vapor (gas phase) of a liquid substance in equilibrium with its liquid phase at a given temperature. A substance with high vapor pressure at a particular temperature is considered volatile, meaning it evaporates easily.

Let's explore how vapor pressure changes:

• As temperature increases, more molecules escape from the liquid to the gas phase, raising the vapor pressure.
• Every liquid has a unique vapor pressure-temperature relationship influenced by intermolecular forces.
• It's vital because it impacts boiling point calculation and material stability in varying conditions.

Understanding vapor pressure helps predict and control conditions under which substances will vaporize, affecting fields like chemical processing and aviation.

Heat of Vaporization

The heat of vaporization (\(\Delta H_{\text{vap}}\)) is the amount of energy required to convert a given quantity of a substance from a liquid to a gas at constant pressure. It's a key factor in determining how a substance behaves under heat.

Some important aspects include:

• It is a measure of the energy needed to overcome intermolecular forces in a liquid.
• Substances with strong intermolecular forces have high heats of vaporization.
• It influences both vapor pressure and boiling point.

In the context of the given problem, we calculate the heat of vaporization using the Clausius-Clapeyron equation, which relates temperature and vapor pressure changes to the heat of vaporization. This calculation helps understand phase changes and energy transfers in various scientific and industrial applications.

Boiling Point Calculation

Calculating the boiling point of a substance involves using the intricate relationship between temperature, vapor pressure, and heat of vaporization. The boiling point is the temperature at which a liquid's vapor pressure equals external pressure. A thorough understanding can help predict and control boiling in practical scenarios.

Here are some considerations:

• The normal boiling point refers to the boiling at a pressure of 1 atmosphere (atm).
• Using the Clausius-Clapeyron equation, we can derive the temperature at which vapor pressure becomes 760 torr (1 atm).
• This method allows accurate boiling point calculation by using known vapor pressures and heat of vaporization.

By applying this approach, we gain insights into conditions required for phase shifts and utilize these for designing heating processes in diverse industries.