The gases in a mixture have the following partial pressures at the same volume and temperature: 1,013,000 Pa nitrogen, 700. mmHg argon, 1.20 atm of helium, 790. torr oxygen, and 0.50 atm of carbon dioxide. What is the total pressure of the mixture in atm?
Convert each of the pressures to atm, then add them.
760 torr = 1 atm
1.01325 x 10^5 Pa = 1 atm.
To find the total pressure of the mixture, we need to convert all the individual partial pressures to the same unit of pressure. In this case, we will convert all the partial pressures to atm, as mentioned in the question.
Given:
Partial pressure of nitrogen = 1,013,000 Pa
Partial pressure of argon = 700. mmHg
Partial pressure of helium = 1.20 atm
Partial pressure of oxygen = 790. torr
Partial pressure of carbon dioxide = 0.50 atm
We will use the following conversions:
1 atm = 101,325 Pa
1 atm = 760 mmHg (Torr)
Converting partial pressures to atm:
1,013,000 Pa nitrogen ÷ 101,325 Pa/atm = 10 atm nitrogen
700. mmHg argon ÷ 760 mmHg/atm ≈ 0.9211 atm argon
1.20 atm helium = 1.20 atm helium (already in atm)
790. torr oxygen ÷ 760 Torr/atm ≈ 1.0395 atm oxygen
0.50 atm carbon dioxide = 0.50 atm carbon dioxide (already in atm)
Now, we can add up all the partial pressures to find the total pressure of the mixture:
Total pressure = 10 atm + 0.9211 atm + 1.20 atm + 1.0395 atm + 0.50 atm
Total pressure = 13.6606 atm
Therefore, the total pressure of the mixture is approximately 13.6606 atm.