Calculating Density Using Vapor Pressure

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

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.

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 $2\text{.}\text{33}\times {\text{10}}^3\text{Pa}$
3. molecular mass of water is $\text{18}\text{.}0\text{g/mol}$

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

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

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

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 Table 13.5, which means that a vapor density of $\text{17}\text{.}2{\text{g/m}}^3$ at $\text{20}\text{.}0\text{º}\text{C}$ creates a partial pressure of $2\text{.}\text{33}\times {\text{10}}^3\text{Pa,}$ 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 Table 13.5 are the maximum amounts of water vapor that air can hold at various temperatures.