Nitrogen (80 kPa), oxygen (21.0 kPa), carbon dioxide (0.03 kPa), and water vapor are the usual components of air. What is the partial pressure of water in mm Hg if the total atmospheric pressure in 1.00 atm?
To find the partial pressure of water vapor in mm Hg, we need to use the given data. Let's break down the problem step by step.
1. The total atmospheric pressure is given as 1.00 atm, but we need to convert it to mm Hg since we want the partial pressure of water vapor in mm Hg.
1 atm = 760 mm Hg
Therefore, 1.00 atm = 760 mm Hg.
2. Next, we need to find the partial pressure of water vapor. We are given the partial pressures of nitrogen, oxygen, and carbon dioxide but not water vapor. However, we know that the partial pressures of all the components of air add up to the total atmospheric pressure.
Partial pressure of nitrogen + Partial pressure of oxygen + Partial pressure of carbon dioxide + Partial pressure of water vapor = Total atmospheric pressure
80 kPa + 21.0 kPa + 0.03 kPa + Partial pressure of water vapor = 1.00 atm
To convert the pressure from kPa to mm Hg, we need to use the following conversion factor:
1 kPa = 7.50062 mm Hg
Therefore, we can rewrite the equation as:
80 kPa + 21.0 kPa + 0.03 kPa + Partial pressure of water vapor = (1.00 atm)(760 mm Hg / 1 atm)
Simplifying the equation:
101.03 kPa + Partial pressure of water vapor = 760 mm Hg
3. Finally, we isolate the variable "Partial pressure of water vapor" to find its value:
Partial pressure of water vapor = 760 mm Hg - 101.03 kPa
Converting the remaining kPa value to mm Hg:
Partial pressure of water vapor = 760 mm Hg - (101.03 kPa)(7.50062 mm Hg / 1 kPa)
Partial pressure of water vapor = 760 mm Hg - 757.753 mm Hg
Partial pressure of water vapor = 2.247 mm Hg
Therefore, the partial pressure of water vapor in mm Hg is approximately 2.247 mm Hg.
how many kPa in 1 atm?
Subtract the sum of all the others from that amount. Then convert back to mm Hg.