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Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering.

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Presentation on theme: "Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering."— Presentation transcript:

1 Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering the air density, making the balloon buoyant. 12 The Gaseous State of Matter Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

2 12.1 Properties of Gases A. Measuring the Pressure of a Gas B. Pressure Dependence: Number of Molecules and Temperature 12.2 Boyle’s Law 12.3 Charles’ Law 12.4 Avogadro’s Law A. Mole-Mass-Volume Calculations 12.5 Combined Gas Laws 12.6 Ideal Gas Law A. Kinetic-Molecular Theory B. Real Gases 12.7 Dalton’s Law of Partial Pressures 12.8 Density of Gases 12.9 Gas Stoichiometry Chapter Outline © 2014 John Wiley & Sons, Inc. All rights reserved.

3 Properties of Gases © 2014 John Wiley & Sons, Inc. All rights reserved. Gases: i) Have indefinite volume ii) Have indefinite shape Expand to fill a container Assume the shape of a container iii) Have low densities Example d air = 1.2 g/L at 25 °C d water = 1.0 g/mL at 25 °C iv) Have high velocities and kinetic energies Volume occupied by 1 mol of H 2 O: as a liquid (18 mL) as a gas (22.4 L)

4 Measuring Pressure © 2014 John Wiley & Sons, Inc. All rights reserved. Pressure: Force per unit area Pressure = area force Pressure results from gas molecule collisions with the container walls. Pressure depends on: 1) The number of gas molecules 2) Gas temperature 3) Volume occupied by the gas SI unit of pressure is the pascal (Pa) = 1 newton/meter 2 1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 1.013 bar = 14.69 psi Unit Conversions:

5 × 101.3 kPa 760 mm Hg Convert 740. mm Hg to a) atm and b) kPa. = 0.974 atm 740. mm Hg a) 1 atm = 760 mm HgUse the conversion factor: = 98.63 kPa 740. mm Hg b) Use the conversion factor: 101.3 kPa = 760 mm Hg × 1 atm 760 mm Hg Practicing Pressure Conversions © 2014 John Wiley & Sons, Inc. All rights reserved.

6 Measuring Pressure © 2014 John Wiley & Sons, Inc. All rights reserved. Measuring Pressure 1) Invert a long tube of Hg over an open dish of Hg. Use a Barometer 2) Hg will be supported (pushed up) by the pressure of the atmosphere. 3) Height of Hg column can be used to measure pressure.

7 1) On the Number of Molecules Pressure (P ) is directly proportional to the number of gas molecules present (n ) at constant temperature (T ) and volume (V ). Increasing n creates more frequent collisions with the container walls, increasing the pressure 1 mol H 2 P = 1 atm 0.5 mol H 2 P = 0.5 atm 2 mol H 2 P = 2 atm V = 22.4 L T = 25.0 °C Pressure Dependence © 2014 John Wiley & Sons, Inc. All rights reserved.

8 Pressure Dependence © 2014 John Wiley & Sons, Inc. All rights reserved. Pressure is directly proportional to temperature when moles (n ) and volume (V ) are held constant. Increasing T causes: a) more frequent and b) higher energy collisions 0.1 mol of gas in a 1L container T = 0 °C T = 100 °C 2.24 atm3.06 atm 2) On Temperature

9 The volume of a fixed quantity of gas is inversely proportional to the pressure exerted by the gas at constant mass and temperature. PV = constant (k ) P P  V 1 or P = k × V 1 Most common form: P 1 V 1 = P 2 V 2 Graph showing inverse PV relationship Boyle’s Law © 2014 John Wiley & Sons, Inc. All rights reserved.

10 730. mm Hg 600. mm Hg × What volume will 3.5 L of a gas occupy if the pressure is changed from 730. mm Hg to 600. mm Hg? V 1 = 3.5 L P 1 = 730. mm Hg P 2 = 600. mm Hg P 1 V 1 = P 2 V 2 = 4.3 L 3.5 L V 2 = Knowns Solve For V 2 V 2 = P2P2 P1V1P1V1 Calculate Boyle’s Law Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

11 Kelvin Temperature Scale Derived from the relationship between temperature and volume of a gas. As a gas is cooled by 1 ºC increments, the gas volume decreases in increments of 1/273. All gases are expected to have zero volume if cooled to −273 ºC. V -T relationship of methane (CH 4 ) with extrapolation (-----) to absolute zero. Temperature in Gas Law Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

12 This temperature (−273 ºC) is referred to as absolute zero. Absolute zero is the temperature (0 K) when the volume of an ideal gas becomes zero. All gas law problems use the Kelvin temperature scale! T K = T °C + 273 Kelvin temperature Celsius temperature Temperature in Gas Law Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

13 The volume of a fixed quantity of gas is directly proportional to the absolute temperature of the gas at constant pressure. Most common form: V  T V = k T or V 1 V 2 T 1 T 2 = V T = k Charles’ Law © 2014 John Wiley & Sons, Inc. All rights reserved.

14 3.0 L of H 2 gas at −15 ºC is allowed to warm to 27 ºC at constant pressure. What is the gas volume at 27 ºC? V 1 = 3.0 L T 1 = −15 ºC = 258 K T 2 = 27 ºC = 300. K Knowns Solving For V 2 V 2 = T1T1 V1T2V1T2 Calculate V 1 V 2 T 1 T 2 = 300. K 258 K = 3.5 L 3.0 L= V 2 = T1T1 V1T2V1T2 × Charles’ Law Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

15 Equal volumes of different gases at constant T and P contain the same number of molecules. 1 volume unit 4 molecules 1 volume unit 4 molecules 2 volume units 8 molecules Avogadro’s Law © 2014 John Wiley & Sons, Inc. All rights reserved.

16 Given the following gas phase reaction: N 2 + 3 H 2 2 NH 3 If 12.0 L of H 2 gas are present, what volume of N 2 gas is required for complete reaction? T and P are held constant. By Avogadro’s Law, we can use the reaction stoichiometry to predict the N 2 gas needed. Knowns Solving For V N 2 Calculate V H 2 = 12.0 L 12.0 L H 2 × 1 L N 2 3 L H 2 = 4.00 L N 2 required Avogadro’s Law © 2014 John Wiley & Sons, Inc. All rights reserved.

17 At constant T and P, how many liters of O 2 are required to make 45.6 L of H 2 O? a) 11.4 L b) 45.6 L c) 22.8 L d) 91.2 L Given the following gas phase reaction: 2 H 2 + O 2 2 H 2 O 45.6 L H 2 O × 1 L O 2 2 L H 2 O = 22.8 L O 2 required Sense Check: Less moles of O 2 equal less L of O 2 ! Avogadro’s Law © 2014 John Wiley & Sons, Inc. All rights reserved.

18 A combination of Boyle’s and Charles’ Laws. Used in problems involving changes in P, T, and V with a constant amount of gas. The volume of a fixed quantity of gas depends on the temperature and pressure. It is not possible to state the volume of gas without stating the temperature and pressure. 0.00 °C (273.15 K) and 1 atm (760 torr) Standard Temperature and Pressure (STP): P 1 V 1 P 2 V 2 T1T1 T2T2 = Combined Gas Laws © 2014 John Wiley & Sons, Inc. All rights reserved.

19 A single equation relating all properties of a gas. where R is the universal gas constant Constant n and T Constant P and T Constant n and P V  1/P Boyle’s Law V  T Charles’ Law V  n Avogadro’s Law PV = nRT Ideal Gas Law © 2014 John Wiley & Sons, Inc. All rights reserved.

20 mol. K R is derived from conditions at STP. Calculate R. Knowns Solving For R Calculate R PV nT = P = 1.00 atm V = 22.4 L T = 273 K n = 1.00 mol R = P × V n × Tn × T × 22.4 L 1.00 mol × 273 K = 0.0821 L. atm 1.00 atm = PV = nRT Units are critical in ideal gas problems! Ideal Gas Constant © 2014 John Wiley & Sons, Inc. All rights reserved.

21 A general theory developed to explain the behavior and theory of gases, based on the motion of particles. Assumptions of Kinetic Molecular Theory (KMT): 1) Gases consist of tiny particles. 2) The distance between particles is large when compared to particle size. The volume occupied by a gas is mostly empty space. 3) Gas particles have no attraction for one another. 4) Gas particles move linearly in all directions, frequently colliding with the container walls or other particles. Kinetic Molecular Theory © 2014 John Wiley & Sons, Inc. All rights reserved.

22 What the Nose Knows Sensing low concentrations of chemicals is useful! Dogs use smell to detect many drugs, explosives, etc. based on trace amounts of chemical compounds in the air. For more information, see: http://www.scs.illinois.edu/suslick/smell_seeing.htmlhttp://www.scs.illinois.edu/suslick/smell_seeing.html Chemistry in Action © 2014 John Wiley & Sons, Inc. All rights reserved. Better Coffee Artificial noses could sniff out cancer or explosives! Better Science

23 1) Explain atmospheric pressure and how it is measured. 2) Be able to convert between the various units of pressure. 12.1 Properties of Gases 3) Use Boyle’s Law to calculate changes in pressure or volume of a gas at constant temperature. 12.2 Boyle’s Law 4) Use Charles’ Law to calculate changes in temperature or volume of a gas at constant pressure. 12.3 Charles’ Law Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.

24 5) Solve problems using the relationships between moles, mass, and volume of gases. 12.4 Avogadro’s Law 6) Use the combined gas law to calculate changes in pressure, volume, or temperature of a gas sample. 12.5 Combined Gas Law 7) Use the ideal gas law to solve problems involving pressure, volume, temperature, and moles of a gas. 12.6 Ideal Gas Law Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.


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