7.9 partial pressures (dalton's law)
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 1
Partial Pressure (Dalton’s Law)
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
Dalton’s Law of Partial Pressure
states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
this empirical law was observed by John Dalton in 1801 and is related to the ideal gas laws.
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The partial pressure of a gas is the pressure of each gas in a mixture is the pressure that gas would exert if it were by
itself in the container
Partial Pressure
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Dalton’s Law of Partial Pressures indicates that pressure depends on the total number of gas
particles, not on the types of particles the total pressure exerted by gases in a mixture is the
sum of the partial pressures of those gases
or PT = P1 + P2 + P3 + .....
Dalton’s Law of Partial Pressures
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Dalton’s Law of Partial Pressures (continued)
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Guide to Solving for Partial Pressure
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Learning Check
1. A scuba tank contains O2 with a pressure of 0.450 atm and He at 855 mmHg.
What is the total pressure in mmHg in the tank?2. For a deep descent, a scuba diver uses a mixture of
helium and oxygen with a total pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium?
3. The pressure of a mixture of nitrogen, carbon dioxide, and oxygen is 150 kPa. What is the partial pressure of oxygen if the partial pressures of the nitrogen and carbon dioxide are 100 kPA and 24 kPa, respectively?
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1.STEP 1 Write the equation for partial pressures.Ptotal = PO2 + PHe
0.450 atm x 760 mmHg = 342 mmHg = PO2 1 atm
STEP 2 Solve for the unknownPtotal = PO2 + PHe
STEP 3 Substitute data and calculatePtotal = 342 mmHg + 855 mmHg
= 1197 mmHg
Solution
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2.STEP 1 Write the equation for the sum of partial
pressures. PTotal = PO2 + PHe
PTotal = 8.00 atm x 760 mmHg = 6080 mmHg 1 atm
STEP 2 Solve for the unknown partial pressure. PHe = PTotal – PO2
STEP 3 Substitute pressures and calculate unknown. PHe = 6080 mmHg – 1280 mmHg
= 4800 mmHg
Solution
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.
STEP 1 Write the equation for the sum of partial pressures. P = Pnitrogen + Pcarbon dioxide + Poxygen
STEP 2 Solve for the unknown partial pressure. P = Pnitrogen + Pcarbon dioxide + Poxygen
Poxygen = P - Pnitrogen - Pcarbon dioxide
STEP 3 Substitute pressures and calculate unknown. Poxygen = 150 kPa - 100 kPa - 24 kPa
Poxygen = 26 kPa
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 11
Gases We Breathe
The air we breathe is a gas mixture contains mostly N2 and
O2 and small amounts of other gases
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A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O2 in the air?
B. At an atmospheric pressure of 714 mmHg, what is the partial pressure (mmHg) of N2 in the air?
Learning Check
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A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O2 in the air?
2) 156 mmHg 745 mmHg (total) x 21 mmHg O2 = 156 mmHg 100 mmHg (total)
B. At an atmospheric pressure of 714 mmHg, what is the partial pressure (mmHg) N2 in the air?
1) 557 mmHg 714 mmHg (total) x 78 mmHg N2 = 557 mmHg of N2
100 mmHg (total)
Solution