# 5.3 Phase transitions in heating curves  (Page 3/21)

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## A boiling point at reduced pressure

A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in [link] to determine the boiling point of water at this elevation.

## Solution

The graph of the vapor pressure of water versus temperature in [link] indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.

The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use [link] to determine the approximate atmospheric pressure at the camp.

Approximately 40 kPa (0.4 atm)

The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation    :

$P=A{e}^{-\text{Δ}{H}_{\text{vap}}\text{/}RT}$

where Δ H vap is the enthalpy of vaporization for the liquid, R is the gas constant, and ln A is a constant whose value depends on the chemical identity of the substance. This equation is often rearranged into logarithmic form to yield the linear equation:

$\text{ln}\phantom{\rule{0.2em}{0ex}}P=-\frac{\text{Δ}{H}_{\text{vap}}}{RT}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A$

This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T 1 , the vapor pressure is P 1 , and at temperature T 2 , the vapor pressure is T 2 , the corresponding linear equations are:

$\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{1}=-\frac{\text{Δ}{H}_{\text{vap}}}{R{T}_{1}}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A\phantom{\rule{5em}{0ex}}\text{and}\phantom{\rule{5em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{2}=-\frac{\text{Δ}{H}_{\text{vap}}}{R{T}_{2}}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A$

Since the constant, ln A , is the same, these two equations may be rearranged to isolate ln A and then set them equal to one another:

$\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{1}+\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{H}_{\text{vap}}}{R{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{2}+\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{H}_{\text{vap}}}{R{T}_{2}}$

which can be combined into:

$\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{{P}_{2}}{{P}_{1}}\right)=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{H}_{\text{vap}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{{T}_{2}}\right)$

## Estimating enthalpy of vaporization

Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.

## Solution

The enthalpy of vaporization, Δ H vap , can be determined by using the Clausius-Clapeyron equation:

$\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{{P}_{2}}{{P}_{1}}\right)=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{H}_{\text{vap}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{{T}_{2}}\right)$

Since we have two vapor pressure-temperature values ( T 1 = 34.0 °C = 307.2 K, P 1 = 10.0 kPa and T 2 = 98.8 °C = 372.0 K, P 2 = 100 kPa), we can substitute them into this equation and solve for Δ H vap . Rearranging the Clausius-Clapeyron equation and solving for Δ H vap yields:

$\text{Δ}{H}_{\text{vap}}=\phantom{\rule{0.2em}{0ex}}\frac{R\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{{P}_{2}}{{P}_{1}}\right)}{\left(\frac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{{T}_{2}}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(8.3145\phantom{\rule{0.2em}{0ex}}\text{J/mol}\text{⋅}\text{K}\right)\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{\text{100 kPa}}{\text{10.0 kPa}}\right)}{\left(\frac{1}{307.2\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{372.0\phantom{\rule{0.2em}{0ex}}\text{K}}\right)}\phantom{\rule{0.2em}{0ex}}=\text{33,800 J/mol}=\text{33.8 kJ/mol}$

Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.

At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.

47,782 J/mol = 47.8 kJ/mol

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