Question

The data below give the vapor pressure of octane, a major component of gasoline. $$ \begin{array}{lllcl} \mathrm{vp}(\mathrm{mm} \mathrm{Hg}) & 10 & 40 & 100 & 400 \\ t\left({ }^{\circ} \mathrm{C}\right) & 19.2 & 45.1 & 65.7 & 104.0 \end{array} $$ Plot \(\ln (\mathrm{vp})\) versus \(1 / T\). Use your graph to estimate the heat of vaporization of octane. \(\left(\ln P=A-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T}\right),\right.\) where \(A\) is the \(y\) -intercept and \(\Delta H_{\text {vap }}\) is the slope.)

Short answer

#Short Answer# To estimate the heat of vaporization of octane, we first convert the temperatures from Celsius to Kelvin and calculate the natural log of vapor pressure (ln(vp)). Then, we plot ln(vp) versus the reciprocal of temperature (1/T) in Kelvin. We find the slope and y-intercept of the plotted graph, and finally use the slope as the heat of vaporization of octane (ΔHvap). The value of ΔHvap in J/mol represents the energy required to vaporize one mole of octane at a constant temperature.

Step-by-step solution

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, add 273.15 to each temperature in Celsius. The conversion is: \(T(K)=T(^{\circ}C)+273.15\)

Calculate the natural log of vapor pressure

Once the temperatures have been converted to Kelvin, calculate the natural log of the vapor pressure using: \(\ln(\text{vp}) = \ln(\text{vapor pressure})\)

Plot ln(vp) versus 1/T

Once you have the natural logs of vapor pressures and the reciprocals of corresponding temperatures in Kelvin, plot ln(vp) (y-axis) against 1/T (x-axis) using graph paper or software like Excel or MATLAB.

Find the slope and y-intercept of the plotted graph

After plotting the points, fit a straight line through them and determine the slope ΔHvap (m) and y-intercept A (c) using any two points from the graph or software.

Use the slope as the heat of vaporization of octane

With the slope (ΔHvap) determined from the graph, we can use it as the heat of vaporization of octane. The value of ΔHvap will be in J/mol, and it can be used to estimate the energy required to vaporize one mole of octane at constant temperature.

Key concepts

Clausius-Clapeyron equation

Understanding the Clausius-Clapeyron equation is foundational when studying the phase transitions of substances. This equation provides a method to estimate the vapor pressure of a liquid at different temperatures. The equation typically takes the form of
\(\ln(P) = A - \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T}\right)\),
where \(P\) is the vapor pressure, \(A\) is a constant related to the entropy of vaporization, \(\Delta H_{\text{vap}}\) is the heat of vaporization, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. Essentially, it illustrates how the natural logarithm of the vapor pressure of a liquid is inversely proportional to the temperature, and directly related to the heat of vaporization. This becomes particularly useful for substances like octane, where predicting vapor pressure at various temperatures is important for applications like engine performance.

Heat of vaporization

The heat of vaporization, \(\Delta H_{\text{vap}}\), represents the amount of energy needed to convert a given amount of a liquid into a gas at a constant temperature, which is an important aspect of understanding how substances behave under heat. In the context of octane, the heat of vaporization can be estimated using the slope of the graph when plotting \(\ln(vp)\) versus \(1/T\). The value obtained from this graph can be used to determine the energy needed to vaporize octane - insights crucial for designing engines and fuel systems to ensure efficiency and safety. The units of the heat of vaporization are typically in Joules per mole (J/mol), indicating the energy required to vaporize one mole of the substance.

Plotting ln(vp) vs 1/T

Graphical analysis provides a visual representation of the relationship between variables and is an invaluable tool in the sciences. When plotting the natural logarithm of vapor pressure (\(\ln(vp)\)) against the reciprocal of the temperature in Kelvin (\(1/T\)), the result should be a straight line if the Clausius-Clapeyron equation is valid for the substance in the given temperature range. By using graph paper or software for plotting, we can obtain the slope and intercept of the line which correspond to the heat of vaporization and a constant term, respectively. These plots not only allow us to estimate such thermodynamic properties but also help in visually demonstrating the consistency of the Clausius-Clapeyron equation with experimental vapor pressure data, as shown in the example with octane.

Converting Celsius to Kelvin

For many scientific calculations, temperature must be expressed in absolute terms, which is where the Kelvin scale comes in handy. The Kelvin scale is pivotal in thermodynamic equations, including the Clausius-Clapeyron equation. To convert Celsius to Kelvin, which is a straightforward process, you add 273.15 to the Celsius temperature. This conversion is essential before any vapor pressure calculations can be made, as seen in the example with octane. The Kelvin temperature provides a consistent base whereby absolute zero (0 K) represents the absence of thermal energy, and thus it allows for accurate and meaningful comparisons of temperature-related data across different systems.