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Gases and Their Properties Chapter 11. Gases Some common elements and compounds exist in the gaseous state under normal conditions of pressure and temperature.

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Presentation on theme: "Gases and Their Properties Chapter 11. Gases Some common elements and compounds exist in the gaseous state under normal conditions of pressure and temperature."— Presentation transcript:

1 Gases and Their Properties Chapter 11

2 Gases Some common elements and compounds exist in the gaseous state under normal conditions of pressure and temperature. Many common liquids can be vaporized. Our gaseous atmosphere provides one means of transferring energy and material throughout the globe. Gas behavior is reasonably simple and they are well understood. Simple mathematical models may be used.

3 11.1 Properties of Gases Gas Pressure Pressure is the force exerted by particles on an object divided by the area upon which the force is exerted. Gas pressure may be measured in millimeters of mercury (mm Hg or torr), standard atmospheres (atm), the SI unit pascal (Pa), or the bar. 1 atm = 760 mm Hg = 101.325 kPa = 1.01325 bar

4 Practice Problem Convert 575 mm Hg into atmospheres and kilopascals.

5 11.2 Gas Laws: Experimental Basis  The volume of a fixed amount of gas at a given temperature is inversely proportional to the pressure of the gas is a statement of Boyle’s Law.  Therefore, at constant temperature and number of moles… P 1 V 1 = P 2 V 2

6 Practice Problem A sample of CO 2 has a pressure of 55 mm Hg in a volume of 125 mL. The sample is compressed so that the new pressure of the gas is 78 mm Hg. What is the new volume of the gas, assuming constant temperature?

7 Gas Laws: Experimental Basis  Charles’ Law: If a given quantity of gas is held at constant pressure, its volume is directly proportional to the Kelvin temperature. V 1 /T 1 = V 2 /T 2 Remember that T must always be expressed in kelvins!

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9 Practice Problem  A balloon is inflated with helium to a volume of 45 L at room temperature (25 o C). If the balloon is cooled to -10. o C, what is the new volume of the balloon? Assume that the pressure does not change.

10 Gas Laws: Experimental Basis  Boyle’s and Charles’ Laws may be combined into one equation to give the general gas law or combined gas law. It applies to situations in which the amount of gas does not change. (P 1 V 1 )/T 1 = (P 2 V 2 )/T 2 OR P 1 V 1 T 2 = P 2 V 2 T 1

11 Practice Problem You have a 22 L cylinder of helium at a pressure of 150 atm and at 31 o C. How many balloons can you fill, each with a volume of 5.0 L, on a day when the atmospheric pressure is 755 mm Hg and the temperature is 22 o C?

12 Avogadro’s Hypothesis Avogadro proposed that equal volumes of gases under the same conditions of temperature and pressure have equal numbers of molecules.

13 Practice Problem If 22.4 L of gaseous CH 4 is burned, what volume of O 2 is required for complete combustion? What volumes of CO 2 and H 2 O are produced? Assume all gases have the same temperature and pressure. CH 4 + 2O 2  CO 2 + 2H 2 O

14 11.3 Ideal Gas Law  The four quantities that can be used to describe a gas include pressure, volume, temperature, and amount.  All three laws can be combined to determine a universal gas constant, R, that can be used to interrelate the properties of a gas. PV = nRT R = 0.082057 Latm/Kmol

15 Practice Problem  The balloon used by Jacques Charles in his historic flight in 1783 was filled with about 1300 mol of H 2. If the temperature of the gas was 23 o C, and its pressure was 750 mm Hg, what was the volume of the balloon?

16 Density of Gases  Since n = mass/molar mass, then PV = (m/M)RT  Since density is defined as d = m/V, we can rearrange the ideal gas equation to say D = (PM)/(RT) Calculate the density of dry air at 15.0 o C and 1.00 atm if its molar mass (average) is 28.96 g/mol.

17 Practice Problem  A 0.105 g sample of a gaseous compound exerts a pressure of 561 mm Hg in a volume of 125 mL at 23.0 o C. What is its molar mass?

18 Homework  After reading sections 11.1-11.3, you should be able to do the following…  P. 546 (12-30 even)

19 11.4 Gas Laws and Chemical Reactions  Gaseous ammonia is synthesized by the following reaction N 2 (g) + 3H 2 (g)  2NH 3 (g) in the presence of an iron catalyst at 500 o C. Assume that 355 L of H 2 gas at 25 o C and 542 mm Hg is combined with excess N 2 gas. What amount (mol) of NH 3 gas can be produced? If this amount of NH 3 gas is stored in a 125 L tank at 25.0 o C, what is the pressure of the gas?

20 11.5 Gas Mixtures and Partial Pressures  The pressure of each gas in a mixture of gases (such as our atmosphere) is the partial pressure.  Dalton’s Law of Partial Pressures states that a mixture of gases is the sum of the partial pressures of the different gases in the mixture. P total = P 1 + P 2 + P 3 +… P total = n total (RT/V)

21 Partial Pressures of Gases  A mole fraction, X, is defined as the number of moles of a substance divided by the total number of moles of all substances present.  P A = X A (P total )

22 Practice Problem  15.0 g of halothane (C 2 HBrClF 3 ) is mixed with 23.5 g of oxygen gas. If the mixture is placed in a 5.00 L tank at 25.0 o C, what is the total pressure (mm Hg) of the gas mixture in the tank? What are the partial pressures (mm Hg) of the gases?

23 11.6 Kinetic-Molecular Theory of Gases  The kinetic-molecular theory is a description of the behavior of gases at the molecular level.  Gases consist of particles whose separation is much greater than the size of the particles themselves.  The particles of a gas are in continual, random, and rapid motion. As they move, they collide with one another and the walls of their container.  The average kinetic energy of gas particles is proportional to the gas temperature. All gases, regardless of their molecular mass, have the same average kinetic energy at the same temperature.

24 Kinetic-Molecular Theory  Kinetic energy of a gas particle of mass m may be calculated by KE = ½ (mass)(speed) 2 = ½ mu 2 where u is the speed of that molecule.  Average kinetic energy depends only upon Kelvin temperature. Therefore, average kinetic energy for a gas sample is equal to one half the mass times the average of the mean square speed.

25 Kinetic-Molecular Theory  Since average kinetic energy is proportional to temperature, then ½ mu 2 must also be proportional to temperature. Therefore the square root of the mean square speed, the temperature, and the molar mass are related. √u 2 = √(3RT)/M where R is 8.3145 J/Kmol and 1 J = 1 kg(m 2 /s 2 ) All gases have the same average kinetic energy at the same temperature.

26 Practice Problem  Calculate the rms speed of helium atoms and N 2 molecules at 25 o C.

27 Homework  After reading sections 11.4 – 11.6, you should be able to do the following… p. 547 (31-36 all, 38-44 even)

28 11.7 Diffusion and Effusion  The mixing of molecules of two or more gases due to their motion is called diffusion. Ex: pizza, bleach  The movement of gas through a tiny hole in a container is called effusion.  Graham’s Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Rate 1 /Rate 2 = √(M 2 /M 1 )

29 Common Misconception!  A gas with a lower molar mass does not effuse more quickly because it is smaller, it does so because it is moving more quickly and would have more collisions. This would increase the probability that it would effuse through a tiny hole!

30 Practice Problem  A sample of pure methane, CH 4, is found to effuse through a porous barrier in 1.50 min. Under the same conditions, an equal number of molecules of an unknown gas effuses through the barrier in 4.73 min. What is the molar mass of the unknown gas?

31 11.9 Nonideal behavior: Real Gases  Most gases behave ideally at low pressures and high temperatures.  Assumptions are not always valid. Sometimes atoms or molecules stick to each other and may be liquified. Volume available to gas molecules may be small.  The van der Waals equation adjusts for nonideal behavior

32 van der Waals Equation (P + a[n/v] 2 )(V-bn) = nRT where “a” and “b” are experimentally determined constants  the pressure correction term accounts of intermolecular forces, and the volume correction term accounts for molecular volume

33 Practice Problem  Using both the ideal gas law and the van der Waals equation, calculate the pressure expected for 10.0 mol of helium gas in a 1.00 L container at 25 o C. For helium, a=0.034 (atmL 2 /mol 2 ) and b=0.0237 L/mol.

34 Homework  After reading sections 11.7 – 11.10, you should be able to do the following…  P. 548 (48-52)


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