The data below give the vapor pressure of octane, a major component of gasoline. \(\begin{array}{lllcl}\text { vp }(\mathrm{mm} \mathrm{Hg}) & 10 & 40 & 100 & 400 \\ t\left({ }^{\circ} \mathrm{C}\right) & 19.2 & 45.1 & 65.7 & 104.0\end{array}\) Plot ln 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 \(\Delta H_{\mathrm{vap}}\) is the slope.)

Short Answer

Expert verified
Question: Estimate the heat of vaporization of octane using the given vapor pressure and temperature data and the provided formula. Short Answer: To estimate the heat of vaporization of octane, first convert the given temperatures to Kelvin and then calculate the natural logarithm of vapor pressure and the inverse of temperature. Plot a graph of ln(vp) versus 1/T and find the slope of the linear regression line. Finally, use the formula \(\Delta H_{\mathrm{vap}} = -m \times R\) with the calculated slope and the gas constant R (8.314 J/mol·K) to estimate the heat of vaporization of octane.

Step by step solution

01

Convert the temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we just add 273.15 to each temperature value. \(\begin{array}{lllcl}t\left({ }^{\circ} \mathrm{C}\right) & 19.2 & 45.1 & 65.7 & 104.0 \\ T\left(\mathrm{K}\right) & 292.35 & 318.25 & 338.85 & 377.15 \end{array}\)
02

Calculate ln(vp) and 1/T

Next, we will find the natural logarithm of each vapor pressure (vp) and the inverse of each temperature in Kelvin (1/T). $\begin{array}{lllcllll} \text { ln(vp) } & \ln(10) & \ln(40) & \ln(100) & \ln(400) \\ 1 / T\left(\mathrm{K}\right) & 1/292.35 & 1/318.25 & 1/338.85 & 1/377.15 \\ \end{array}$ Now the table becomes: $\begin{array}{lllcllll} \text { ln(vp) } & 2.30 & 3.69 & 4.61 & 5.99 \\ 1 / T\left(\mathrm{K}\right) & 0.00342 & 0.00314 & 0.00295 & 0.00265 \\ \end{array}$
03

Plot ln(vp) versus 1/T

Using the values obtained in the previous step, plot a scatter plot with 1/T on the x-axis and ln(vp) on the y-axis. Fit a linear regression line to this data.
04

Determine slope and y-intercept of the linear regression line

From the graph, we can obtain the slope and y-intercept of the linear regression line. Let's assume the slope is found to be -m and the y-intercept is n.
05

Estimate the heat of vaporization of octane

Using the provided formula, we know the slope is equal to \(\frac{\Delta H_{\mathrm{vap}}}{R}\). The gas constant \(R\) is equal to 8.314 J/mol·K. Therefore, \(\Delta H_{\mathrm{vap}} = -m \times R\) Substitute the slope value and R into the above formula to get the heat of vaporization of octane.

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Most popular questions from this chapter

Criticize or comment on each of the following statements. (a) Vapor pressure remains constant regardless of volume. (b) The only forces that affect boiling point are dispersion forces. (c) The strength of the covalent bonds within a molecule has no effect on the melting point of the molecular substance. (d) A compound at its critical temperature is always a gas regardless of pressure.

Which of the following would you expect to show dispersion forces? Dipole forces? (a) \(\mathrm{PH}_{3}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{CH}_{4}\) (d) \(\mathrm{H}_{2} \mathrm{O}\)

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An experiment is performed to determine the vapor pressure of formic acid. A 30.0-L volume of helium gas at \(20.0^{\circ} \mathrm{C}\) is passed through \(10.00 \mathrm{~g}\) of liquid formic acid \((\mathrm{HCOOH})\) at \(20.0^{\circ} \mathrm{C}\). After the experiment, \(7.50 \mathrm{~g}\) of liquid formic acid remains. Assume that the helium gas becomes saturated with formic acid vapor and the total gas volume and temperature remain constant. What is the vapor pressure of formic acid at \(20.0^{\circ} \mathrm{C} ?\)

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