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Gases Chapter 13 Pg. 372-402 Goal To learn about the behavior of gases both on molecular and macroscopic levels.

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Presentation on theme: "Gases Chapter 13 Pg. 372-402 Goal To learn about the behavior of gases both on molecular and macroscopic levels."— Presentation transcript:


2 Gases Chapter 13 Pg. 372-402

3 Goal To learn about the behavior of gases both on molecular and macroscopic levels.

4 General Characteristics of Gases Uniformly fills any container. Mixes completely with any other gas. Exerts pressure on its surroundings.

5 Force = mass X acceleration To understand pressure, one has to understand force. Weight = your mass X acceleration due to gravity Mr. Yoos’ force F = 87 kg X 9.8 m/s 2 F = 852.6 kg m /s 2 = 850 N Acceleration due to gravity

6 Pressure = Force / Area If Mr. Yoos weighs 850 N on earth, and I am standing on a scale that is 0.5m X 0.5m or 0.25m 2, the pressure I exert is …

7 Let’s Say Mr. Yoos is Wearing High Heels Total area for the heels = 1X10 -4 m 2. The force is 850 N. What is the pressure? 1N/m 2 = 1 Pascal or 1 Pa

8 Air Exerts Pressure The “standard atmosphere” is equal to 101,325 Pa. 1 atm = 101,325 Pa 1 atm = 760 torr = 760 mm Hg

9 Barometer


11 The Gas Laws Boyle’s Law- the pressure exerted by a gas is inversely proportional to the volume the gas occupies if the temperature remains constant.

12 Boyle’s Law Example: A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature?

13 Boyle’s Law

14 Charles’s Law At constant pressure, the volume of a gas is directly proportional to the temperature (in Kelvins) of the gas.

15 Charles’s Law

16 Example: A gas at 30.0 0 C and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0 0 C and 1.00 atm?

17 Avogadro’s Law For a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas.

18 (at constant T, P) If you triple the number of moles of gas (at constant temperature and pressure), the volume will also triple.

19 Example: A 5.20 L sample at 18.0 0 C and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be?

20 The Ideal Gas Law P in atm V in L n in moles T in Kelvins R =


22 Keep in Mind: 1.This relationship assumes that the gas behaves ideally (conditions of low pressure and high temperature). Correction factors must be added under certain conditions. 2.Keep track of dimensions! Many ideal gas law problems are best solved using DA. 3.Always list what you are given. You may be able to simplify the problem.

23 Example: A sample containing 0.614 moles of a gas at 12.0 0 C occupies a volume of 12.9 L. What pressure does the gas exert?

24 Example: A sample of methane gas (CH 4 ) at 0.848 atm and 4.0 o C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature is increased to 11.0 o C?

25 Example: How many moles of a gas at 104 o C would occupy a volume of 6.8 L at a pressure of 270 mmHg?

26 Gas Stoichiometry Many gas law problems involve calculating the volume of a gas produced by the reaction of volume of other gases. The problem solving strategy that we have used throughout is still the same. That is, you want to relate moles of reactants to moles of products. The ideal gas law will allow you to use the following strategy:

27 Standard Temperature and Pressure “STP” P = T = The molar volume of an ideal gas is _________ at STP

28 volume of reactants  (apply the ideal gas law)  moles of reactants  (apply stoichiometry)  moles of products (apply ideal gas law)  volume of products

29 Example: A sample containing 15.0 g of dry ice (CO 2 (s)) is put into a balloon and allowed to sublime according to the following equation: CO 2 (s)  CO 2 (g) How big will the balloon be (ie, what is the volume of the balloon), at 22.0 o C and 1.04 atm, after all of the dry ice has sublimed?

30 Example: 0.500 L of H 2 (g) are reacted with 0.600 L of O 2 (g) at STP according to the equation 2H 2 (g) + O 2 (g)  2H 2 O (g) What volume will the H 2 O occupy at 1.00 atm and 350. o C?

31 Density and Molar Mass PV = nRT P = nRT/V

32 Example: A gas at 34.0 o C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (MM) of the gas.

33 Dalton’s Law of Partial Pressures For a mixture of gases in a container, the total pressure is the sum of the pressures that each gas would exert if it were alone. Because RT/V will be the same for each of the different gases in the same container.

34 Example: A volume of 2.0 L of He at 46 o C, and 1.2 atm pressure, was added to a vessel that contained 4.5 L of N 2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added?

35 The Production of Oxygen by Thermal Decomposition of KCIO 3

36 Mole Fraction The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. The mole fraction of a particular component is a mixture of ideal gases is directly related to its partial pressure

37 Kinetic Molecular Theory of Gases 1. The volume of the individual particles of a gas can be assumed to be negligible. 2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3. The particles are assumed to exert no forces on each other 4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

38 Temperature is a Measure of the Average Kinetic Energy of a Gas

39 Root Mean Square Velocity The expression dealing with the average velocity of gas particles is called the root mean square velocity. Where R = 8.3145 J/K mol = 8.3145 kg m 2 /s 2 / Kmol T = temp in Kelvins M = mass of a mole of the gas in Kilograms

40 Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at a.0.0 o C b.300. o C

41 Effusion or Diffusion? Diffusion- term used to describe the mixing of gases. Effusion- relates to the passage of a gas through an orifice into an evacuated chamber.

42 Graham’s Law of Effusion The higher the molar mass of the gas, the slower the rate of effusion through a small orifice.

43 Example: How many times faster than He would NO 2 gas effuse?

44 With Regard to Diffusion The important idea is that even though gases travel very rapidly (hundreds of meters per second), their motions are in all directions, so mixing is relatively slow. The basic structure of Graham’s Law holds.

45 Chemistry in the Atmosphere The atmosphere is composed of 78% N 2, 21% O 2, 0.9% Ar, and 0.03% CO 2 along with trace gases. The composition of the atmosphere varies as a function of distance from the earth’s surface. Heavier molecules tend to be near the surface due to gravity. Upper atmospheric chemistry is largely affected by UV, X-rays, and cosmic radiation emanating from space. The ozone layer is especially reactive to UV radiation. Manufacturing and other processes of our modern society affect the chemistry of our atmosphere. Air pollution is a direct result of such processes.

46 Air Pollution Photochemical smog reactions: 1.N 2 (g) + O 2 (g) + heat  2NO (g) 2.2NO (g) + O 2 (g)  2NO 2 (g) 3.NO 2 (g) + radiant energy  NO (g) + O (g) 4.O (g) + O 2 (g)  O 3 (g) Ozone Ozone causes lung and eye irritation and can be dangerous for people with asthma, emphysema, and other respiratory conditions.

47 Photochemical Smog

48 Ozone Layer

49 A Schematic Diagram of the Process for Scrubbing Sulfur Dioxide from Stack Gases in Power Plants

50 Acid Rain 1.2NO 2 (g) + H 2 O (l)  HNO 2 (aq) + HNO 3 (aq) 2.2SO 2 (g) + O 2 (g)  2SO 3 (g) 3.SO 3 (g) + H 2 O (l)  H 2 SO 4 (aq)

51 Acid Rain


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