# 5.3 Phase transitions in heating curves  (Page 4/21)

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## Estimating temperature (or vapor pressure)

For benzene (C 6 H 6 ), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?

## Solution

If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, Δ H vap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) 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 the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value ( T 1 = 80.1 °C = 353.3 K, P 1 = 101.3 kPa, Δ H vap = 30.8 kJ/mol) and want to find the temperature ( T 2 ) that corresponds to vapor pressure P 2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T 2 . Rearranging the Clausius-Clapeyron equation and solving for T 2 yields:

${T}_{2}={\left(\frac{-R\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{{P}_{2}}{{P}_{1}}\right)}{\text{Δ}{H}_{\text{vap}}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{{T}_{1}}\right)}^{-1}={\left(\frac{-\left(8.3145\phantom{\rule{0.2em}{0ex}}\text{J/mol}\text{⋅}\text{K}\right)\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{83.4\phantom{\rule{0.2em}{0ex}}\text{kPa}}{101.3\phantom{\rule{0.2em}{0ex}}\text{kPa}}\right)}{\text{30,800 J/mol}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{353.3\phantom{\rule{0.2em}{0ex}}\text{K}}\right)}^{-1}=\text{346.9 K or}\phantom{\rule{0.2em}{0ex}}{73.8}^{\circ }\text{C}$

For acetone (CH 3 ) 2 CO, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?

30.1 kPa

## Enthalpy of vaporization

Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, Δ H vap . For example, the vaporization of water at standard temperature is represented by:

${\text{H}}_{2}\text{O(}l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O(}g\right)\phantom{\rule{5em}{0ex}}\text{Δ}{H}_{\text{vap}}=\text{44.01 kJ/mol}$

As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:

${\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O(}l\right)\phantom{\rule{5em}{0ex}}\text{Δ}{H}_{\text{con}}=\text{−Δ}{H}_{\text{vap}}=-44.01\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$

## Using enthalpy of vaporization

One way our body is cooled is by evaporation of the water in sweat ( [link] ). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); Δ H vap = 43.46 kJ/mol at 37 °C.

## Solution

We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

$1.5\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1000\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}{1\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}}}{18\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{43.46\phantom{\rule{0.2em}{0ex}}\text{kJ}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}}}\phantom{\rule{0.2em}{0ex}}=3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.

How much heat is required to evaporate 100.0 g of liquid ammonia, NH 3 , at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?

28 kJ

## Melting and freezing

When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting    . At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid ( [link] ).

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