1. Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures (Dalton’s Law)

2. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2

3. Problem 1 Solve the combined gas laws for T 2.

4. Solution for Problem 1 Solve the combined gas law for T 2. (Hint: cross-multiply first.) P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 T 2 = P 2 V 2 T 1 P 1 V 1

5. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

6. Data Table Set up Data Table P 1 = 0.800 atm V 1 = 0.180 L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

7. Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K - 273 = °C

8. Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T 2 = 604 K - 273 = 331 °C

9. Try this problem A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

10. Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec

11. Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C

12. Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?

13. Answer these questions True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

14. Solution True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

15. Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V 1 = V 2 n 1 n 2 initial final

16. STP The volumes of gases can be compared when they have the same temperature and pressure. If the temperature and pressure are: Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg) then, it is represented by STP.

17. Try this problem A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? P 1 = V 1 = T 1 = K P 2 = V 2 = ?? T 2 = K V 2 = 15 L x atm x K = 6.8 L atm K

18. Solution P 1 = 1.0 atm V 1 = 15 L T 1 = 273 K P 2 = 2.0 atm V 2 = ?? T 2 = 248 K V 2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

19. Molar Volume At STP 4.0 g He 16.0 g CH 4 44.0 g CO 2 1 mole 1 mole1mole (STP) (STP)(STP) V = 22.4 L V = 22.4 L V = 22.4 L

20. Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

21. Try these problems A.What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g2) 0.357 g3) 1.43 g

22. Solution A.What is the volume at STP of 4.00 g of CH 4 ? 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. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He

23. Daltons’ Law of Partial Pressures Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 + P 3 +.....

24. Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mmHg 20.95% O 2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO 2 0.2 mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg 2 2 2 Total Pressure760 mm Hg

25. Dalton’s Law Check A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

26. Solution A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557

27. Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm 0.5 mole O 2 + 0.3 mole He + 0.2 mole Ar 1 mole H 2