1. The Gas Laws Ch. 14- Gases

3. Boyle’s Law P V PV = k Pressure and Volume are inversely proportional. As Volume increased, pressure decreases.

4. Boyle’s Law Calculation b A quantity of gas at 20.0 o C has a volume of 120.0 dm 3 and a pressure of 93.3 kPa. Find the pressure if the volume is reduced to 30.0 dm 3 and the temperature is unchanged. b P 1 V 1 = P 2 V 2 b P 1 V 1 = P 2 V 2 = (120.0 dm 3 )(93.3kPa) = b V 2 30 dm 373 kPa

5. Boyle’s Law

6. V T Charles’ Law Volume and temperature are directly proportional.

7. V T Charles’ Law b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

8. Charles’ Law

9. Charle’s Law Calculation b V 1 /T 1 = V 2 /T 2 b Example: A balloon is filled with 1.70 grams of Methane. The initial temperature is 25.0 o C and the initial volume is 0.33 liters. Find the volume if the temp. is changed to 45.0 o C. b Remember to add 273 to the temps to convert to Kelvin. b V 1 =.33 L b T 1 = 298 K b V 2 = ? b T 2 = 318 K V 1 T 2 = V 2 T 1

10. Combined Gas Law V1/T1 = V 2 /T2 P1V1 = P2V2

11. = kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

12. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

13. V n Avogadro’s Principle b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

14. PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

15. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT You don’t need to memorize these values!

16. C. Johannesson GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

17. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3

18. Gas Stoichiometry b Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law b Non- STP Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conv.

19. 1 mol CaCO 3 100.09g CaCO 3 Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.17 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

20. WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.17 mol T = 25°C = 298 K R = 8.315 dm 3  kPa/mol  K

21. WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = 0.597 mol O 2 Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm 3  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O 3 15.0 L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT 

22. 2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.

23. Dalton’s Law of Partial Pressures b Dalton’s Law of Partial Pressures: The total pressure in a container of mixed gases is equal to the sum of all partial pressure of the gases in the container. b So: P total = P 1 + P 2 + P 3 ….

24. Dalton’s Law Example b Find the partial pressure of nitrogen and oxygen in a sample of air at 101 kPa. Assume that air is 78% nitrogen and 21% oxygen. b P nitrogen = (.78)(101 kPa) = 78.8 kPa b P oxygen = (.21)(101kPa) = 21.2 kPa