Chapter 14 Gases

1. Chapter 14Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures LecturePLUS Timberlake

2. Combined Gas Law P1V1 = P2V2 T1 T2 Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1 LecturePLUS Timberlake

3. Learning Check C1 Solve the combined gas law for T2. LecturePLUS Timberlake

4. Solution C1 Solve the combined gas law for T2. (Hint: cross-multiply first.) P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1 LecturePLUS Timberlake

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? LecturePLUS Timberlake

6. Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ?? LecturePLUS Timberlake

7. Solution Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C LecturePLUS Timberlake

8. Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C LecturePLUS Timberlake

9. Learning Check C2 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? LecturePLUS Timberlake

10. Solution G9 T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 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 LecturePLUS Timberlake

11. 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 V1 = V2 n1 n2 initial final LecturePLUS Timberlake

12. 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. PT = P1 + P2 + P3 + ..... LecturePLUS Timberlake

13. Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mmHg 20.95% O2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO2 0.2 mmHg PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2 Total Pressure 760 mm Hg LecturePLUS Timberlake

14. Learning Check C6 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 2) 9.14 3) 0.109 LecturePLUS Timberlake

15. Solution C6 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 LecturePLUS Timberlake

16. Ideal Gas Law The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = nRT R = ideal gas constant Lecture PLUS Timberlake 2000

17. Ideal Gases • Behave as described by the ideal gas equation; no real gas is actually ideal • Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less • In real gases, particles attract each other reducing the pressure • Real gases behave more like ideal gases as pressure approaches zero. Lecture PLUS Timberlake 2000

18. PV = nRT R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm mol-K Lecture PLUS Timberlake 2000

19. Learning Check G15 What is the value of R when the STP value for P is 760 mmHg? Lecture PLUS Timberlake 2000

20. Solution G15 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K Lecture PLUS Timberlake 2000

21. Learning Check G16 Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office? Lecture PLUS Timberlake 2000

22. Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L 20.0 L T = 23°C + 273 296 K n = 2.86 mol 2.86 mol P = ? ? Lecture PLUS Timberlake 2000

23. Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol) = 2.64 x 103 mm Hg Lecture PLUS Timberlake 2000

24. Learning Check G17 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? Lecture PLUS Timberlake 2000

25. Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2 1 mol O2 Lecture PLUS Timberlake 2000

26. Density of a Gas Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRT PV = nRT P = n RTV RTV RT V Lecture PLUS Timberlake 2000

27. Substitute (1.00 atm ) mol-K = 0.0446 mol O2/L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O2 x 32.0 g O2 = 1.43 g/L 1 L 1 mol O2 Therefore the density of O2 gas at STP is 1.43 grams per liter Lecture PLUS Timberlake 2000