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Chapter 10 Gases GASES John Dalton Characteristics, Pressure, Laws, and Ideal-Gas Equation Sections 10.1-10.4.

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Presentation on theme: "Chapter 10 Gases GASES John Dalton Characteristics, Pressure, Laws, and Ideal-Gas Equation Sections 10.1-10.4."— Presentation transcript:

1

2 Chapter 10 Gases

3 GASES John Dalton

4 Characteristics, Pressure, Laws, and Ideal-Gas Equation Sections 10.1-10.4

5 Objectives Compare distinguishing characteristics of gases with those of liquids and solids Study gas pressure and the units that express it Examine volume, pressure, and temperature and their relationship to gases. Use the ideal-gas equation to make calculations.

6 Characteristics of Gases Earth’s Atmosphere: –N 2 (78%) –O 2 (21%) –Other gases, ex: Argon (0.9%) Diatomic gases- halogens Monatomic gases- nobles Vapors

7 Gases vs. Solids & Liquids Gases, unlike solids and liquids,: –Expand to fill their containers’ volumes –Are highly compressible –Form homogeneous mixtures with each other

8 Pressure P = F A Gases exert pressure on any surface they contact

9 Atmospheric Pressure Gases experience gravitational acceleration BUT they have tiny masses So gravity operates on atmosphere as a whole to press down on Earth –Atmospheric pressure

10 Magnitude of Atmospheric Pressure F = ma Force = mass x acceleration a= 9.8 m/s 2

11 SI Units Force: kg-m/s 2, the newton (N) Pressure: N/m 2, the pascal (Pa)

12 Barometer Early 17 th century Evangelista Torricelli, student of Galileo Proved that atmosphere had weight Height of Hg, h, measures atmospheric pressure

13 Standard Atmospheric Pressure Pressure at sea level Supports a column of Hg 760mm high –1.01325 x 10 5 Pa Defines non-SI units of pressure –Atmospheres (atm) –Millimeters of mercury (mm Hg) a.k.a. torr (for Torricelli)

14 Pressure Conversions 1 atm = 760 mm Hg = 760 torr = 1.01325 x 10 5 Pa

15 Gas Laws Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law) Quantity and Volume (Avogadro’s Law)

16 Boyle's Law British chemist Robert Boyle (1627-1691) The pressure of a gas is inversely related to the volume when T does not change PV product remains constant P x V = k (constant) when T remains constant

17 Boyle’s Law and the Breathing

18 P and V Changes P 1 = 8atm P 2 = 4atm V 1 = 2L V 2 = 4L

19 Pressure/Volume Changes P 1 V 1 = P 2 V 2 P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L

20 Charles’ Law V = 125 mL V = 250 mL T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature?

21 Charles’ Law French scientist, Jacques Charles (1746-1823) At constant pressure, the volume of a gas is directly related to its absolute (K) temperature V= constant x T V 1 = V 2 T 1 T 2

22 Learning Check Use Charles’ Law to complete the statements: 1. If final T is higher than initial T, final V is (greater, or less) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) than the initial T.

23 Solution V 1 = V 2 T 1 T 2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T.

24 Avogadro’s Law Joseph Louis Gay-Lussac (1778-1823) –At a given P and T, the V of gases react with one another in a ratio of small whole numbers Amedeo Avogadro (1776-1856) –Hypothesis: Equal volumes of gases at the same T and P contain equal numbers of moecules –Law: V of a gas maintained at constant T and P is directly proportional to the number of moles of gas. V= constant x n

25 Ideal Gas Law The four variables in the gas laws all deal with proportionality: Boyle’s: V  1/P (constant n, T) Charles’: V  T (constant n, P) Avogadro’s: V  n (constant P, T) They combine into a general gas law: V  nT P

26 Ideal Gas Equation If the proportionality constant is R, than: PV = nRT R = ideal gas constant

27 Units for Ideal-Gas Equation T must be in K n is expressed in moles P is usually atm V is typically L * * *

28 STP Standard Temperature and Pressure 0 ºC 1 atm V of 1 mole of ideal gas at STP = 22.41 L

29 Temperature Conversions ºF ºC K -45932212 -2730100 0273373 K = ºC + 273


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