Final answer:
The vapor pressure of water at 20.0°C is 17.5 mmHg, which is 2.30% of standard atmospheric pressure. If the air is at 100% relative humidity, then 2.30% of the air is water vapor.
Step-by-step explanation:
The vapor pressure of water at 20.0°C is 17.5 mmHg. To determine what percentage of atmospheric pressure this corresponds to, we use the fact that standard atmospheric pressure is 760.0 mmHg. Therefore, the vapor pressure of water at this temperature is 17.5 mmHg / 760.0 mmHg × 100% = 2.30% of atmospheric pressure. If the air has 100% relative humidity at this temperature, the percent of the air that is water vapor is 2.30%.
Although the information regarding how to calculate this in units of atmospheres (atm) and kilopascals (kPa) is provided, the actual calculations aren't directly related to the given question. However, for reference, the conversion from mmHg to atm can be done using the conversion factor 1 atm = 760 mmHg, and the conversion from atm to kPa can be done using 1 atm = 101.325 kPa.
The vapor pressure of water at different temperatures can be found using a vapor pressure curve. At 760 mmHg (which is equal to 1 atm or 100 kPa), the boiling point of water is 100°C or 373.15 K.