Question:On a certain day, the temperature is 25.0°C and the relative humidity is 90.0%. How many grams of water must condense out of each cubic meter of air if the temperature falls to 15.0°C? Such a drop in temperature can, thus, produce heavy dew or fog.

In this scenario, we need to find out the amount of water condensed when the temperature falls from25.0°C to 15.0°C. The relative humidity at 25.0°C is 90.0%.The relative humidity is the ratio of vapor density to the saturation vapor density expressed in percentage.

$\mathbf{Percent} \mathbf{RelativeHumidity} \mathbf{=} \frac{\mathbf{Vapor} \mathbf{Density}}{\mathbf{SaturationVapor} \mathbf{Density}}\mathbf{\times }\mathbf{100}$…………….(1)

Here, vapor density is the density of water vapor. Saturation vapor density is the maximum density of water vapor. So, the difference between saturation vapor density at 15.0°C and vapor density at 25.0°C gives the amount of condensed water.

Saturation vapor density at 25.0°C is 23.0 g/m³. Using the equation (1), vapor density can be expressed as;

$\text{Vapor} \text{Density} \text{=} \frac{\text{Percent} \text{Relative Humidity}\times \text{Saturation Vapor} \text{Density}}{100}$

Substitute the values of relative humidity and saturation vapor density.

$$\begin{gathered} \text{Vapor} \text{Density} \text{=} \frac{{\displaystyle 90\times 23.0g/m^3}}{100} \\ =20.7g/m^3 \end{gathered}$$

So, the density of water vapor at 25.0°C is 20.7 g/m³.

At 15.0 °C, the saturation vapor density is 12.8 g/m3. Since the difference between the calculated vapor density and this saturation density gives the amount of condensed water, subtract the value.

$$\begin{gathered} \text{Condensed Water}= \text{Vapor density}\left(25.0°C\right)-\text{Saturation vapor density}\left(\text{15}.0°C\right) \\ {\text{=20.7g/m}}^3\text{-12.8}g/m^3 \\ =7.9g/m^3 \end{gathered}$$

So, 7.9 g of water condensed out of each cubic meter of air.