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Chapter 14 The Behavior of Gases 14.3 Ideal Gases

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1 Chapter 14 The Behavior of Gases 14.3 Ideal Gases
14.1 Properties of Gases 14.2 The Gas Laws 14.3 Ideal Gases 14.4 Gases: Mixtures and Movements Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2 CHEMISTRY & YOU Why do balloons filled with helium deflate faster than balloons filled with air ? The surface of a latex balloon has tiny pores through which gas particles can pass, causing the balloon to deflate over time. The rate which the balloon deflates depends on the gas it contains. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 Dalton’s Law How is the total pressure of a gas mixture related to the partial pressures of the component gases? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

4 Dalton’s Law Gas pressure results from collisions of particles in a gas with an object. If the number if particles increases in a given volume, more collisions occur. If the average kinetic energy of the particles increases, more collisions occur. In both cases, the pressure increases. Gas pressure depends only on the number of particles in a given volume and on their average kinetic energy. Particles in a mixture of gases at the same temperature have the same average kinetic energy. So the kind of particle is not important. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

5 Composition of Dry Air Component Volume (%) Partial pressure (kPa) Nitrogen 78.08 79.11 Oxygen 20.95 21.22 Carbon dioxide 0.04 Argon and others 0.93 0.95 Total 100.0 101.32 The contribution each gas in a mixture makes to the total pressure is called partial pressure.

6 Dalton’s Law In a mixture of gases, the total pressure is the sum of the partial pressures of the gases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

7 Dalton’s Law Dalton’s law of partial pressures states that, at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. Ptotal = P1 + P2 + P3 + … Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

8 Dalton’s Law Dalton’s law holds true because each component exerts its own pressure independent of the pressure exerted by the other gases. If the percent composition of a mixture of gases does not change, the fraction of the pressure exerted by a gas does not change as the total pressure changes. This fact is important for people who must operate at high altitudes. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

9 Dalton’s Law At top of mount Everest, the total atmospheric pressure is kPa. This pressure is about one third of its value at sea level. The partial pressure of oxygen is also reduced by one third, to 7.06 kPa. In order to support respiration in humans, the partial pressure of oxygen must be kPa or higher. So, climbers of Mount Everest need an oxygen mask and a cylinder of compressed oxygen to survive ascent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

10 Using Dalton’s Law of Partial Pressure
Sample Problem 14.5 Using Dalton’s Law of Partial Pressure . Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressures of oxygen (PO2) at kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are kPa, kPa, and 0.94 kPa, respectively? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

11 Sample Problem 14.5 Analyze List the knowns and the unknown.
Use the equation for Dalton’s law of partial pressures (Ptotal = PO2 + PN2 + PCO2 + Pothers) to calculate the unknown value (PO2) KNOWNS PN2 = kPa PCO2 = kPa Pothers = 0.94 kPa Ptotal = kPa UNKNOWNS PO2 = ? kPa Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

12 Sample Problem 14.5 Calculate Solve for the unknown
PO2 = Ptotal – (PN2 + PCO2 + Pothers) = kPa – (79.10 kPa kPa kPa) = kPa Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

13 Graham’s Law How does the molar mass of a gas affect the rate at which the gas diffuses or effuses? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

14 Graham’s Law Diffusion is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout. Another process that involves the movement of molecules in a gas is effusion. During effusion, a gas escapes through a tiny hole in its container. With effusion and diffusion, the type of particle is important. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

15 Graham’s Law Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

16 Graham’s Law Thomas Graham’s Contribution
The Scottish chemist Thomas Graham studied rates of effusion during the 1840s. From his observations, he proposed a law. Graham’s law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass. This law can also be applied to the diffusion of gases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

17 Graham’s Law This law makes sense if you know the mass, velocity, and kinetic energy of a moving object are related. KE = ½ mv2 For the kinetic energy to be constant, any increase in mass must be balanced by a decrease in velocity. If two objects with different masses have the same kinetic energy, the lighter object must move faster. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

18 Graham’s Law Comparing Effusion Rates
Supposed you have two balloons, one fille with helium and the other filled with air. If the balloons are at the same temperature, the particles in each balloon have the same average kinetic energy. But helium atoms are less massive than oxygen or nitrogen molecules. So, the molecules in air move more slowly than helium atoms with the same kinetic energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

19 √ = Graham’s Law Comparing Effusion Rates
Because the rate of effusion is related only to a particle’s speed, Graham’s law can be rewritten as follows for two gases, A and B. RateA molar massB RateB molar massA = ________ ________________ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

20 √ = Graham’s Law Comparing Effusion Rates
The rates of effusion of two gases are inversely proportional to the square roots of their molar masses. = ________ ________________ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

21 Sample Problem 14.8 Comparing Effusion Rates How much faster does helium effuse than nitrogen at the same temperature? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

22 Analyze List the knowns and the unknown.
Sample Problem 14.8 Analyze List the knowns and the unknown. 1 Use Graham’s law and the molar masses of the two gases to calculate the ratio of effusion rates. KNOWNS UNKNOWN molar massHe = 4.0 g Molar massN2= 28.0 g ratio of effusion rates = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

23 √ √ = = = Calculate Solve for the unknown. 2
Sample Problem 14.5 Calculate Solve for the unknown. 2 Helium effuses nearly three times faster than nitrogen at the same temperature. RateHe g RateN g = = = 7.0 2.7 ________ ________ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

24 END OF 14.4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.


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