What is the relative humidity on a 25.0 °C day when the air contains 14 g/m3 of water vapor?
To determine the relative humidity on a 25.0 °C day when the air contains 14 g/m³ of water vapor, we need to know two key values:
1. Saturation vapor pressure at 25.0 °C: This refers to the maximum amount of water vapor that can exist in the air at a given temperature.
2. Actual vapor pressure: This represents the amount of water vapor present in the air at the given conditions.
To calculate relative humidity, we use the formula:
Relative Humidity = (Actual vapor pressure / Saturation vapor pressure) * 100
Let's break down the steps to find the answer:
1. Determine the saturation vapor pressure at 25.0 °C: We can look up this value on a psychrometric chart or use formulas such as the Magnus equation.
For 25.0 °C, the saturation vapor pressure is approximately 3.17 kPa.
2. Convert the given water vapor density from g/m³ to kPa:
We divide the density by the molar mass of water (18.015 g/mol) to convert it into the units of pressure (kPa).
14 g/m³ / 18.015 g/mol = 0.7775 mol/m³
To convert mol/m³ to kPa, we multiply the molar volume of an ideal gas at standard temperature and pressure (22.414 L/mol) by the gas constant (0.008314 kPa·m³/(mol·K)) and divide by 1000 to convert L to m³:
0.7775 mol/m³ * (22.414 L/mol * 0.008314 kPa·m³/(mol·K)) / 1000 = 0.1459 kPa
3. Plug the values into the relative humidity formula:
Relative Humidity = (Actual vapor pressure / Saturation vapor pressure) * 100
= (0.1459 kPa / 3.17 kPa) * 100
= 4.6%
Therefore, the relative humidity on a 25.0 °C day when the air contains 14 g/m³ of water vapor is approximately 4.6%.