Question
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?
Answers
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.
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.
Related Questions
if the total pressure of a container filled with oxygen gas, nitrogen gas and water vapor is equal t...
Nitrogen (80 kPa), oxygen (21.0 kPa), carbon dioxide (0.03 kPa), and water vapor are the usual compo...
The partial pressure of nitrogen in the atmosphere is 0.7829 arm and the partial pressure of carbo...
A mixture of oxygen, helium, and nitrogen gases exerts a total pressure of 287 kPa. If the partial p...