1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.7 Volume and Moles (Avogadro’s Law) Balloons rise in the air because helium is less dense than air.

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Avogadro's Law: Volume and Moles Avogadro’s law states that the volume of a gas is directly related to the number of moles(n) of gas T and P are constant V 1 = V 2 n 1 n 2

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Learning Check If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L Balloons rise in the air because helium is less dense than air.

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Solution STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V 1 = 1.5 L V 2 = ? V increases n 1 = 0.75 mole n 2 = 1.2 moles n increases

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Solution (continued) STEP 2 Rearrange the gas law for the unknown. V 1 = V 2 n 1 n 2 V 2 = V 1 n 2 n 1 STEP 3 Substitute values into the gas law to solve for the unknown. V 2 = 1.5 L x 1.2 mol He = 2.4 L 0.75 mol He

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 The volumes of gases can be compared at STP (standard temperature and pressure) when they have the same temperature Standard temperature (T) = 0 °C or 273 K the same pressure Standard pressure (P) = 1 atm (760 mmHg) STP

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Molar Volume The molar volume of a gas is measured at STP (standard temperature and pressure) is 22.4 L for exactly1 mol of a gas

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 The molar volume at STP has about the same volume as three basketballs can be used to write conversion factors 22.4 L and 1 mol 1 mol 22.4 L Molar Volume as a Conversion Factor The molar volume of a gas is about the same as the volume of three basketballs.

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Using Molar Volume

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Example of Using Molar Volume What is the volume occupied by 2.75 mol of N 2 gas at STP? STEP 1 Identify given and needed. Given 2.75 mol of N 2 Need liters of N 2 STEP 2 Write a plan. moles of N 2 liters of N 2

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Example of Using Molar Volume (continued) STEP 3 Write conversion factors. 1 mol of gas = 22.4 L 1 mol gas and 22.4 L 22.4 L 1 mol gas STEP 4 Set up problem with factors to cancel units. 2.75 mol N 2 x 22.4 L = 61.6 L of N 2 1 mol N 2

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 How many grams of He gas are present in 8.00 L of He at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g Learning Check

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 STEP 1 Identify given and needed. Given 8.00 L of He Need grams of He STEP 2 Write a plan. liters of He moles of He grams of He Solution

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 STEP 3 Write conversion factors. 1 mol of gas = 22.4 L 1 mol gas and 22.4 L 22.4 L 1 mol gas 1 mol of He = 32.00 g 4.003 g He and 1 mol He 1 mol He 4.003 g He STEP 4 Set up problem with factors to cancel units. 8.00 L x 1 mol He x 4.003 g He = 1.43 g of He (3) 22.4 L 1 mol He Solution (continued)

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 At STP, the density of a gas is calculated using the mass of the gas and its volume. Density = molar mass molar volume Density of a Gas at STP

16. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 Learning Check Calculate the density in g/L of O 2 gas at STP.

17. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 17 Solution STEP 1 Identify given and needed. At STP, we know that 1 mol of O 2 (32.00 g) occupies a volume of 22.4 L. Given 1 mol of O 2 ; 22.4 L Need density (g/L) STEP 2 Write a plan. Density = molar mass molar volume

18. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 18 Solution (continued) STEP 3 Write conversion factors. 1 mol of O 2 = 22.4 L 1 mol O 2 and 22.4 L 22.4 L 1 mol O 2 1 mol of O 2 = 32.00 g 32.00 g O 2 1 mol O 2 1 mol O 2 32.00 g O 2 STEP 4 Set up problem with factors to cancel units. Density of O 2 (at STP) = 32.00 g O 2 = 1.43 g/L 22.4 L (STP)