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Chapter 12: Problem 41

The normal boiling point of acetone, an important laboratory and industrial solvent, is \(56.2^{\circ} \mathrm{C}\) and its \(\Delta H_{\text {vap }}\) is \(25.5 \mathrm{kJmol}^{-1} .\) At what temperature does acetone have a vapor pressure of \(375 \mathrm{mmHg} ?\)

Short Answer

Expert verified
The temperature at which acetone has a vapor pressure of \(375 \mathrm{mmHg}\) is approximately \(67.12^{\circ} \mathrm{C}\).

Step by step solution

01

Conversion of given data

Before starting, it's necessary to convert the given data into appropriate units. \(\Delta H_{\text {vap }}\) is given as \(25.5 \mathrm{kJ mol}^{-1}\), which needs to be converted to \(J mol^{-1}\), resulting in \(\Delta H_{\text {vap }}=25500 \mathrm{Jmol}^{-1}\). Similarly, the vapor pressure of \(375 \mathrm{mmHg}\) needs to be converted to atmospheres by using the conversion \(1 \mathrm{atm} = 760 \mathrm{mmHg}\), resulting in a vapor pressure of \(375/760\approx 0.493 \mathrm{atm}\).
02

Initial Setup of the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is: \[\ln\left(\frac{P2}{P1}\right)=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T2}-\frac{1}{T1}\right)\] where \(P1\) is the initial pressure, \(P2\) is the final pressure, \(T1\) is the initial temperature, \(T2\) is the final temperature, \(\Delta H_{\text {vap }}\) is the heat of vaporization, and \(R\) is the gas constant. In this problem, \(P1\) is 1 atm (as the boiling point is defined as the temperature at which the vapor pressure of a liquid is equal to 1 atm), \(P2\) is 0.493 atm, \(T1\) is the boiling point of the acetone in Kelvin (56.2°C = 329.35 K), and \(T2\) is our unknown.
03

Solve for T2

By substituting the known values into the Clausius-Clapeyron equation, we can solve for \(T2\). We have that \(R=8.314\) as the ideal gas constant in \(J \mathrm{mol}^{-1} \mathrm{K}^{-1}\), therefore we can isolate \(T2\) to solve for it as follows: \[T2=\left(\frac{[\ln (P2/P1)\cdot R + \Delta H_{\text {vap }} / T1] }{ \Delta H_{\text {vap }}} \right)^{-1}\]
04

Calculation

Substitution the known values in the above expression to find \(T2\), it is calculated to be approximately 340.27 K. This temperature should be converted back to Celsius, yielding a final answer of \(T2 \approx 67.12^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

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