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 \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (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

01

(Part a: Derive the relationship between vapor pressures and temperatures)

The Clausius-Clapeyron equation is given by: $$ \frac{d \ln P}{dT}=\frac{\Delta H_{\text{vap}}}{RT^2} $$ To derive the relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and their absolute temperatures, \(T_{1}\) and \(T_{2}\), we will integrate the Clausius-Clapeyron equation as follows: $$ \int_{T_1}^{T_2} \frac{d \ln P}{dT} dT = \int_{T_1}^{T_2} \frac{\Delta H_{\mathrm{vap}}}{RT^2} dT $$ Integrating both sides, we get: $$ \left[ \ln P \right]_{P_1}^{P_2} = -\frac{\Delta H_{\mathrm{vap}}}{R} \left[\frac{1}{T}\right]_{T_1}^{T_2} $$ Simplifying the above equation, we get: $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ Now, we will use this equation to solve parts (b), (c), and (d).

02

(Part b: Calculate the heat of vaporization of octane)

We are given the vapor pressures and temperatures for octane: $$ P_1=1.86\, \text{kPa}, \quad T_1 = 25^{\circ} \mathrm{C} + 273.15 = 298.15\,\mathrm{K} $$ $$ P_2=19.3\, \text{kPa}, \quad T_2 = 75^{\circ} \mathrm{C} + 273.15 = 348.15\,\mathrm{K} $$ And the universal gas constant, R: $$ R = 8.314\, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} $$ Using the equation we derived in part (a), we can solve for the heat of vaporization \(\Delta H_{\text{vap}}\): $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ Inserting the given values, we get: $$ \Delta H_{\mathrm{vap}}=-R \cdot \frac{\ln \frac{1.86}{19.3}}{\frac{1}{298.15}-\frac{1}{348.15}} $$ Calculating the value of \(\Delta H_{\mathrm{vap}}\): $$ \Delta H_{\text{vap}} \approx 3.85 \times 10^4\,\mathrm{J/mol} $$

03

(Part c: Calculate the normal boiling point of octane)

The normal boiling point is achieved when the vapor pressure, \(P,\) is equal to atmospheric pressure, 101.3 kPa. We can use the equation derived in part (a) and input the new values to solve for the boiling point temperature, \(T_{\text{bp}}\): $$ \ln \frac{P_{1}}{P}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{\text{bp}}}\right) $$ Rearranging the equation to solve for \(T_{\text{bp}}\): $$ \frac{1}{T_{\text{bp}}} = \frac{1}{T_1} - \frac{R}{\Delta H_{\text{vap}}} \ln \frac{P_{1}}{P} $$ Now we insert the known values: $$ \frac{1}{T_{\text{bp}}} = \frac{1}{298.15} - \frac{8.314}{3.85 \times 10^4} \ln \frac{1.86}{101.3} $$ Calculating \(T_{\text{bp}}\): $$ T_{\text{bp}} \approx 398.5\,\mathrm{K} $$ Converting to Celsius: $$ T_{\text{bp (C)}} = 398.5 - 273.15 = 125.35^{\circ} \mathrm{C} $$

04

(Part d: Calculate the vapor pressure of octane at -30°C)

We can use the equation derived in part (a) to solve for the vapor pressure of octane at -30°C. First, convert -30°C to Kelvin: $$ T_3=-30^{\circ} \mathrm{C} + 273.15 = 243.15\,\mathrm{K} $$ Plugging the known values into the derived equation: $$ \ln \frac{P_{1}}{P_3}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{243.15}\right) $$ Rearranging the equation to solve for \(P_3\): $$ P_3 = P_1 \cdot e^{\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{243.15}\right)} $$ Inserting the known values, we get: $$ P_3 = 1.86 \cdot e^{\frac{3.85 \times 10^4}{8.314}\left(\frac{1}{298.15}-\frac{1}{243.15}\right)} $$ Calculating the value of \(P_3\): $$ P_3 \approx 0.157\, \mathrm{kPa} $$ The vapor pressure of octane at \(-30^{\circ} \mathrm{C}\) is approximately \(0.157\) kPa.

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