## Section27.7Water Vapor and Humidity

The vapor pressure of water plays an important role in atmospheric physics. Humidity\index{Humidity} or dampness of air is a measure of the presence of water molecules in air. Air is mostly made up of the Nitrogen gas (approx. 78% by volume) and the Oxygen gas (approx. 21% by volume). The remaining 1% by volume consists of many other gases, Argon (0.93\%), Carbon dioxide (0.03%), Neon (0.008%), Helium (0.0005%), Hydrogen (0.00005%), varying amounts of water and trace amounts of other gases, where all numbers are given as percentages by volume.

When the temperature of air increases more water is boiled off from oceans and lakes into the vapor phase which changes the ratio of water to other gases in air. Since water molecules are lighter than diatomic nitrogen and oxygen molecules, the density of air decreases when the humidity in the air increases. The humidity or relative humidity of air is defined in terms of partial pressure of water in the air compared to the saturated vapor pressure for the corresponding temperature and pressure.

Partial pressure is a useful concept for mixture of gases. If you go back to the beginning of this chapter and think about how pressure in a gas is related to the motion of molecules, you would find that total pressure of a mixture of gases will be equal to the pressure that each gas will exert if it alone occupied the entire volume at the given temperature. The net pressure of air would be equal to the sum of the partial pressures of the component gases.

\begin{equation*} p_{\text{atm}} = p_{\text{N}_2} +p_{\text{O}_2}+p_{\text{H}_2\text{O}}+\cdots \end{equation*}

Relative humidity is defined as the ratio of partial pressure $p_{\text{H}_2\text{O}}$ of water actually present in air and the saturated vapor pressure $p_{\text{H}_2\text{O}}^\text{sat}$ of water expected at that temperature from the coexistence curve of water.

\begin{equation} \text{Relative Humidity} = \frac{p_{\text{H}_2\text{O}}}{p_{\text{H}_2\text{O}}^\text{sat}} \times 100\%.\tag{27.7.1} \end{equation}

Thus, if the humidity is close to 100% then air holds the maximum amount of water it can under the equilibrium condition. Table 27.7.1 shows the saturated water vapor pressure and the corresponding densities at various temperatures.

If the relative humidity is more than 100%, then we would have more water vapor pressure than the value of the saturated vapor pressure for that condition. We say that the air is supersaturated with water. When the air is supersaturated with water, water will precipitate out of air. This is what happens at night when temperature drops but water vapor content remains the same as in the evening.

Suppose there is 50% humidity at $77^{\circ}\text{F}\text{,}$ which corresponds to a partial pressure of 0.5 $\times$ 23.76 = 11.88 mmHg from Table 27.7.1 . Now, if the temperature overnight drops to $50^{\circ}\text{F}$ without any change in partial pressure of water molecules in the air. At $50^{\circ}\text{F}\text{,}$ saturated vapor pressure is 9.21 mmHg. That means that when the temperature drops to $50^{\circ}\text{F}\text{,}$ the air will be supersaturated. The water vapor will then condense to liquid. This is the process behind dew formation at night. The dew point is the temperature at which the partial pressure of water in a given atmospheric condition will correspond to the saturated vapor pressure for that condition. In our example of 50% humidity at $77^{\circ}F\text{,}$ the dew point will be the temperature at which the saturated vapor pressure is $11.8$ mmH, which is between $55.4^{\circ}\text{F}$ and $57.2^{\circ}\text{F}$ as extrapolated from the data in Table 27.7.1.

A meteorologist reports that the humidity is $60\%$ on a day when temperature is $77^{\circ}\text{F}\text{.}$ What is the partial pressure of water in the air that day?

Hint

Use $p_\text{sat} = 23.76 \text{ mmHg}\text{.}$

$15.689\ \text{mmHg}\text{.}$

Solution

We look up the saturated vapor pressure from the table. At $77^{\circ}\text{F}$ the $p_\text{sat} = 23.76 \text{ mmHg}\text{.}$ Therefore, $60\%$ humidity will correspond to the partial pressure $p$ of water to be

\begin{equation*} 66\% = \frac{p}{23.76\ \text{mmHg}}\ 100\%. \end{equation*}

Solving for $p$ we obtain

\begin{equation*} p = 15.689\ \text{mmHg}. \end{equation*}

Humidity in a room is to be maintained at $30\%\text{.}$ On a humid day the humidity is found to be $80\%\text{.}$ If the room has a volume of 8 cubic meters, how much water vapor must be removed?

Data: Saturation pressure, $p_\text{sat} = 40\ \text{mmHg} \text{.}$

Hint

Get partial pressure of water from $p_{w} = h\, p_\text{sat}\text{,}$ where $h$ is the humidity as fraction. Then, use ideal gas law.

Solution

Let $p_\text{sat}$ be the saturation pressure corresponding to the temperature. Then, the partial pressure $p_{w}$ of water in the room is

\begin{equation*} p_{w} = 0.8 p_\text{sat}. \end{equation*}

Using ideal gas for water molecules filling the volume $V = 8\text{ m}^3$ space, the number of moles on water in the room would be

\begin{equation*} n = \frac{p_w V}{RT}. \end{equation*}

When the partial pressure is 30\% we would have

\begin{equation*} n' =\frac{p'_w V}{RT} \end{equation*}

The amount to be removed

\begin{equation*} \Delta n = \frac{p_\text{sat} V}{RT} \left(0.8-0.3 \right). \end{equation*}

Using

\begin{equation*} p_\text{sat} \sim 40\ \text{mmHg} =(40 /760)\ \text{atm},\ \ T = 300\ \text{K}, \end{equation*}

the volume in L,

\begin{equation*} V = 8\ \text{m}^3 = 8000\ \text{L}. \end{equation*}

give

\begin{equation*} \Delta n = 17\ \text{mol}, \end{equation*}

which corresponds to about $300\text{ grams}$ or $300\text{ cm}^3$ of water.