Chapter 8 Gases 8.1 Gases and Kinetic Theory 8.2 Gas Pressure 8.8 Ideal Gas Law * You do not need to know Boyle s (8.3), Charles (8.4), Gay-Lussac s (8.5), Avogadro s (8.7) or the Combined gas (8.6) laws. They are all contained within the ideal gas law. We will not cover Dalton s law (8.9). 8.1 Kinetic Theory of Gases Particles of a gas Move rapidly in straight lines. Have kinetic energy that increases with an increase in temperature. Are very far apart. Have essentially no attractive (or repulsive) forces. Have very small volumes compared to the volume of the container they occupy. 1 2 Properties of Gases 8.2 Barometer Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). A barometer measures the pressure exerted by the gases in the atmosphere. The atmospheric pressure is measured as the height in mm of the mercury column. 3 4 1
A. The downward pressure of the Hg in a barometer is than/as the weight of the atmosphere. 1) greater 2) less 3) the same A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/ml) because 1) H 2 O is less dense 2) H 2 O is heavier 3) air is more dense than H 2 O B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/ml) because 1) H 2 O is less dense 5 6 Pressure Units of Pressure A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa). One atmosphere (1 atm) is 760 mm Hg. 1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr 7 8 2
A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.638 atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3) 22,300 mm Hg A. What is 475 mm Hg expressed in atm? 2) 0.638 atm 485 mm Hg x 1 atm = 0.638 atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm 9 10 8.8 Ideal Gas Law STP The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nt Rearranging this expression gives the expression called the ideal gas law. PV = nrt The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 273 K (0 C ) Standard pressure (P) 1 atm (760 mm Hg) 11 12 3
Molar Volume Molar Volume as a Conversion Factor At STP, 1 mole of a gas occupies a volume of 22.4 L. The molar volume at STP can be used to form conversion factors. The volume of one mole of a gas is called the molar volume. 22.4 L and 1 mole 1 mole 22.4 L 13 14 A. What is the volume at STP of 4.00 g of CH 4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g A. 1) 5.60 L 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He 15 16 4
Universal Gas Constant, R The universal gas constant, R, can be calculated using the molar volume of a gas at STP. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nt (1 mole) (273K) n T = 0.0821 L atm mole K Note there are four units associated with R. Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? 17 18 What is the value of R when the STP value for P is 760 mmhg? R = PV = (760 mm Hg) (22.4 L) nt (1 mole) (273K) Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N 2 O at 23 C, what is the pressure (mm Hg) in the tank? = 62.4 L mm Hg mole K 19 20 5
1. Adjust the units of the given properties to match the units of R. V = 20.0 L, T = 296 K, n = 2.8 moles, P =? 2. Rearrange the ideal gas law for P. P = nrt V P = (2.8 moles)(62.4 L mm Hg)(296 K) (20.0 L) (mole K) A cylinder contains 5.0 L of O 2 at 20.0 C and 0.85 atm. How many grams of oxygen are in the cylinder? = 2.6 x 10 3 mm Hg 21 22 Molar Mass of a Gas 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mole O 2 x 32.0 g O 2 = 5.8 g O 2 1 mole O 2 What is the molar mass of a gas if 0.250 g of the gas occupy 215 ml at 0.813 atm and 30.0 C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT (0.0821 L atm/mole K)(303K) = 0.00703 mole 2. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mole mole 0.00703 mole 23 24 6
Gases in Equations Gases in Equations (continued) The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. Problem: What volume (L) of Cl 2 gas at 1.2 atm and 27 C is needed to completely react with 1.5 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl 2 = 0.083 mole Cl 2 27.0 g Al 2 moles Al 2. Place the moles Cl 2 in the ideal gas equation. V = nrt = (0.083 mole Cl 2 )(0.0821 Latm/moleK)(300K) P 1.2 atm = 1.7 L Cl 2 25 26 What volume (L) of O 2 at 24 C and 0.950 atm are needed to react with 28.0 g NH 3? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) 1. Calculate the moles of O 2 needed. 28.0 g NH 3 x 1 mole NH 3 x 5 mole O 2 17.0 g NH 3 4 mole NH 3 = 2.06 mole O 2 2. Place the moles O 2 in the ideal gas equation. V = nrt = (2.06 moles)(0.0821 L atm/molek)(297k) P 0.950 atm = 52.9 L O 2 27 28 7
V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 = V 2 T 1 Isolate T 2 by dividing through by V 1 V 1 T 2 = V 2 T 1 T 2 = V 2 T 1 V 1 V 1 V 1 29 8