Given a total pressure of 740 mm Hg for three gases oxygen, nitrogen, and water vapor H20(g), if the partial pressures for nitrogen and oxygen are 600 mm Hg and 100 mm Hg What is the partial pressure of the water vapor in ATM?

Answer 1

To convert from mm Hg to atm, we divide by 760 mm Hg/atm:

Partial pressure of nitrogen = 600 mm Hg / 760 mm Hg/atm = 0.789 atm
Partial pressure of oxygen = 100 mm Hg / 760 mm Hg/atm = 0.132 atm

To find the partial pressure of water vapor, we can subtract the sum of the partial pressures of nitrogen and oxygen from the total pressure:

Partial pressure of water vapor = (Total pressure - Partial pressure of nitrogen - Partial pressure of oxygen)
= (740 mm Hg / 760 mm Hg/atm) - 0.789 atm - 0.132 atm
= 0.319 atm

Therefore, the partial pressure of water vapor in ATM is 0.319 atm.