1. 1 Chapter 11 Gases 11.8 The Ideal Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. 2 The ideal gas law Provides a relationship between the four properties (P, V, n, and T) of gases that can be written equal to a constant R. PV = R nT Rearranges these properties to give the ideal gas law expression. PV = nRT Ideal Gas Law

3. 3 The universal gas constant, R Can be calculated using the molar volume of a gas at STP. Calculated at STP uses 273 K,1.00 atm, 1 mol of a gas, and a molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mol) (273K) n T = 0.0821 L atm mol K Universal Gas Constant, R

4. 4 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? Learning Check

5. 5 What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? R = PV = (760 mm Hg) (22.4 L) nT (1 mol) (273 K) = 62.4 L mm Hg mol K Solution

6. 6 Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.86 mol N 2 O at 23°C, what is the pressure (mm Hg) in the tank? Learning Check Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. 7 1. Adjust the units of the given properties to match the units of R. V = 20.0 L20.0 L T = 23°C + 273 296 K n = 2.86 mol2.86 mol P = ? ? Solution

8. 8 2. Rearrange the ideal gas law for P. P = nRT V 3. Substitute quantities and solve. P = (2.86 mol)(62.4 L mm Hg)(296 K) (20.0 L) (mol K) = 2.64 x 10 3 mm Hg Solution

9. 9 Learning Check A cylinder contains 5.0 L O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

10. 10 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)(mol K) = 0.18 mol O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mol O 2 x 32.0 g O 2 = 5.8 g O 2 1 mol O 2 Solution

11. 11 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) molK = 0.00703 mol RT (0.0821 L atm)(303K) 2. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol Molar Mass of a Gas