# 11.6 Humidity, evaporation, and boiling  (Page 2/9)

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Relative humidity is related to the partial pressure of water vapor in the air. At 100% humidity, the partial pressure is equal to the vapor pressure, and no more water can enter the vapor phase. If the partial pressure is less than the vapor pressure, then evaporation will take place, as humidity is less than 100%. If the partial pressure is greater than the vapor pressure, condensation takes place. In everyday language, people sometimes refer to the capacity of air to “hold” water vapor, but this is not actually what happens. The water vapor is not held by the air. The amount of water in air is determined by the vapor pressure of water and has nothing to do with the properties of air.

Saturation vapor density of water
Temperature $\left(\text{º}\text{C}\right)$ Vapor pressure (Pa) Saturation vapor density (g/m 3 )
−50 4.0 0.039
−20 $1\text{.}\text{04}×{\text{10}}^{2}$ 0.89
−10 $2\text{.}\text{60}×{\text{10}}^{2}$ 2.36
0 $6\text{.}\text{10}×{\text{10}}^{2}$ 4.84
5 $8\text{.}\text{68}×{\text{10}}^{2}$ 6.80
10 $1\text{.}\text{19}×{\text{10}}^{3}$ 9.40
15 $1\text{.}\text{69}×{\text{10}}^{3}$ 12.8
20 $2\text{.}\text{33}×{\text{10}}^{3}$ 17.2
25 $3\text{.}\text{17}×{\text{10}}^{3}$ 23.0
30 $4\text{.}\text{24}×{\text{10}}^{3}$ 30.4
37 $6\text{.}\text{31}×{\text{10}}^{3}$ 44.0
40 $7\text{.}\text{34}×{\text{10}}^{3}$ 51.1
50 $1\text{.}\text{23}×{\text{10}}^{4}$ 82.4
60 $1\text{.}\text{99}×{\text{10}}^{4}$ 130
70 $3\text{.}\text{12}×{\text{10}}^{4}$ 197
80 $4\text{.}\text{73}×{\text{10}}^{4}$ 294
90 $7\text{.}\text{01}×{\text{10}}^{4}$ 418
95 $8\text{.}\text{59}×{\text{10}}^{4}$ 505
100 $1\text{.}\text{01}×{\text{10}}^{5}$ 598
120 $1\text{.}\text{99}×{\text{10}}^{5}$ 1095
150 $4\text{.}\text{76}×{\text{10}}^{5}$ 2430
200 $1\text{.}\text{55}×{\text{10}}^{6}$ 7090
220 $2\text{.}\text{32}×{\text{10}}^{6}$ 10,200

## Calculating density using vapor pressure

[link] gives the vapor pressure of water at $\text{20}\text{.}0\text{º}\text{C}$ as $2\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Pa}\text{.}$ Use the ideal gas law to calculate the density of water vapor in $\text{g}/{\text{m}}^{3}$ that would create a partial pressure equal to this vapor pressure. Compare the result with the saturation vapor density given in the table.

Strategy

To solve this problem, we need to break it down into a two steps. The partial pressure follows the ideal gas law,

$\text{PV}=\text{nRT,}$

where $n$ is the number of moles. If we solve this equation for $n/V$ to calculate the number of moles per cubic meter, we can then convert this quantity to grams per cubic meter as requested. To do this, we need to use the molecular mass of water, which is given in the periodic table.

Solution

1. Identify the knowns and convert them to the proper units:

1. temperature $T=\text{20}\text{º}\text{C=293 K}$
2. vapor pressure $P$ of water at $\text{20}\text{º}\text{C}$ is $2\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Pa}$
3. molecular mass of water is $\text{18}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{g/mol}$

2. Solve the ideal gas law for $n/V$ .

$\frac{n}{V}=\frac{P}{\text{RT}}$

3. Substitute known values into the equation and solve for $n/V$ .

$\frac{n}{V}=\frac{P}{\text{RT}}=\frac{2\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Pa}}{\left(8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}\right)\left(\text{293}\phantom{\rule{0.25em}{0ex}}\text{K}\right)}=0\text{.}\text{957}\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}$

4. Convert the density in moles per cubic meter to grams per cubic meter.

$\rho =\left(0\text{.}\text{957}\frac{\text{mol}}{{\text{m}}^{3}}\right)\left(\frac{\text{18}\text{.}\text{0 g}}{\text{mol}}\right)=\text{17}\text{.}2\phantom{\rule{0.25em}{0ex}}{\text{g/m}}^{3}$

Discussion

The density is obtained by assuming a pressure equal to the vapor pressure of water at $\text{20}\text{.}0\text{º}\text{C}$ . The density found is identical to the value in [link] , which means that a vapor density of $\text{17}\text{.}2\phantom{\rule{0.25em}{0ex}}{\text{g/m}}^{3}$ at $\text{20}\text{.}0\text{º}\text{C}$ creates a partial pressure of $2\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Pa,}$ equal to the vapor pressure of water at that temperature. If the partial pressure is equal to the vapor pressure, then the liquid and vapor phases are in equilibrium, and the relative humidity is 100%. Thus, there can be no more than 17.2 g of water vapor per ${\text{m}}^{3}$ at $\text{20}\text{.}0\text{º}\text{C}$ , so that this value is the saturation vapor density at that temperature. This example illustrates how water vapor behaves like an ideal gas: the pressure and density are consistent with the ideal gas law (assuming the density in the table is correct). The saturation vapor densities listed in [link] are the maximum amounts of water vapor that air can hold at various temperatures.

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I think
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