1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.8 The Ideal Gas Law Basic Chemistry Copyright © 2011 Pearson Education, Inc. Dinitrogen oxide is used as an anesthetic in dentistry.

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 The relationship between the four properties (P, V, n, and T) of a gas can be written equal to a constant R. PV = R nT Rearranging this relationship gives the ideal gas law. PV = nRT Ideal Gas Law

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 The universal gas constant, R, can be calculated at STP using a temperature of 273 K, a pressure of 1.00 atm, a quantity of 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. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Another value for the universal gas constant is obtained using mmHg for the pressure at STP. What is the value of R when a pressure of 760 mmHg is placed in the R value expression? Learning Check

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 What is the value of R when a pressure of 760 mmHg (at STP) is placed in the R value expression? R = PV = (760 mmHg) (22.4 L) nT (1 mol) (273 K) = 62.4 L mmHg mol K Solution

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Summary of Units for the Universal Gas Constant 6

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Using the Ideal Gas Law 7

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 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 of N 2 O at 23 °C, what is the pressure (mmHg) in the tank? Learning Check Dinitrogen oxide is used as an anesthetic in dentistry.

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 STEP 1 Organize the data given for the gas. R = 62.4 L mmHg mol K V = 20.0 L T = 23°C + 273 = 296 K n = 2.86 mol P = ? Solution

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for P. P = nRT V STEP 3 Substitute gas data and calculate the unknown quantity. P = (2.86 mol)(62.4 L mmHg)(296 K) (20.0 L) (mol K) = 2.64 x 10 3 mmHg Solution (continued)

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Learning Check A cylinder contains 5.0 L of O 2 at 20 °C and 0.85 atm. How many grams of oxygen are in the cylinder?

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 STEP 1 Organize the data given for the gas. P = 0.85 atm, V = 5.0 L, T = 293 K, R = 0.0821 L atm, n ( or g =?) mol K STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for n (moles). n = PV RT Solution

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 STEP 3 Substitute gas data and calculate the unknown quantity. = (0.85 atm)(5.0 L)(mol K) = 0.18 mol of O 2 (0.0821atm L)(293 K) Convert the moles of gas to the grams of gas using its molar mass. = 0. 18 mol O 2 x 32.00 g O 2 = 5.8 g of O 2 1 mol O 2 Solution (continued)

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 What is the molar mass of a gas if 0.250 g of the gas occupies 215 mL at 0.813 atm and 30.0 °C? STEP 1 Organize the data given for the gas. R = 0.0821 L atm/mol K P = 0.813 atm V = 0.215 L n = ? mol T = 30.0 °C + 273 = 303 K Calculating the Molar Mass of a Gas

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 STEP 2 Solve the ideal gas law for the unknown. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) molK = 0.00703 mol RT (0.0821 L atm)(303 K) STEP 3 Substitute gas data and calculate the unknown quantity. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol Calculating the Molar Mass of a Gas (continued)