Q61E - College Physics (Urone)

Question

Question: Late on an autumn day, the relative humidity is 45.0% and the temperature is 20.0°C. What will the relative humidity be that evening when the temperature has dropped to 10.0°C, assuming constant water vapor density?

Step-by-Step Solution

Answer

Answer

Relative humidity in the evening is 82.34%.

Step 1: Definition of relative humidity

The relative humidity is the ratio of vapor density to the saturation vapor density expressed in percentage.

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

Here, vapor density is constant and saturation vapor density decreases with the temperature. Saturation vapor density corresponding to 20.0°C is 17.2 g/m³ and 10.0°C is 9.40 g/m³.

Step 2: Find relative humidity in the evening

Relative humidity at 20.0°C is given as 45.0%. Substitute these values in the equation to find relative humidity at both temperatures.

$$\begin{gathered} 45.0\%=\frac{{\displaystyle vapor density}}{17.2g/m^2}\times 100 \\ \text{Relative} \text{humidity} \text{at 10}.0°C=\frac{{\displaystyle vapor density}}{9.40g/m^2} \end{gathered}$$
 Dividing these two equations gives;

$$\begin{gathered} \frac{45.0\%}{\text{Relative} \text{humidity} \text{at 10}.0°C}=\frac{940g/m^2}{17.2g/m^2} \\ \text{Relative} \text{humidity} \text{at 10}.0°C=\frac{17.2g/m}{940g/m^2} \\ =82.34\% \end{gathered}$$

So, the relative humidity at 10.0°C is 82.34%.