1. Gas Laws What to do when conditions are ideal

2. Boyle’s Law What was the relationship between pressure and volume? When P Then V Algebraically this is written as P=k/V When you solved for k; PV=k Therefore P 1 V 1 =P 2 V 2

3. Gas Laws – Boyle’s Law Constant: Temperature Relationship: Pressure is inversely proportional to volume Pressure a 1/volume Written As: P 1 V 1 = P 2 V 2 Pressure is typically in atm or torr

4. Charles’ Law What was the relationship between Temperature and Volume? When T Then V Algebraically this is written as V=kT When you solved for k; V/T=k Therefore V1/T1=V2/T2

5. Gas Laws – Charles’ Law Constant: Pressure Relationship: Temperature is directly proportional to volume Temp a Volume Written As: V 1 /T 1 = V 2 /T 2

6. Charles’ Law What unit of measure is needed for these calculations? C or K? Temperature is in K (K = 273 + C)

7. Gas Laws – Gay-Lussac’s Law Constant: Volume Relationship: Pressure is directly proportional to temperature Pressure a Temperature Written As: P 1 /T 1 = P 2 /T 2

8. The IDEAL GAS Law –this is what we will use When we put all three laws together: PV a nT (n= number of moles) PV = nRT (R= ideal gas law constant) R=62.4 L torr/K mol or.08206 L atm/K mol

9. Lecture PLUS Timberlake 2000 9 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.

10. Lecture PLUS Timberlake 2000 10 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

11. LecturePLUS Timberlake 11 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Isolate V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

12. Lecture PLUS Timberlake 2000 12 Learning Check G15 What is the value of R when the STP value for P is 760 mmHg?

13. Lecture PLUS Timberlake 2000 13 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

14. Lecture PLUS Timberlake 2000 14 Learning Check G16 Dinitrogen monoxide (N 2 O), 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?

15. Lecture PLUS Timberlake 2000 15 Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L20.0 L T = 23°C + 273 296 K n = 2.86 mol2.86 mol P = ? ?

16. Lecture PLUS Timberlake 2000 16 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 10 3 mm Hg

17. Lecture PLUS Timberlake 2000 17 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?

18. Lecture PLUS Timberlake 2000 18 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 O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2

19. LecturePLUS Timberlake 19 Learning Check C1 Solve the combined gas laws for T 2.

20. LecturePLUS Timberlake 20 Solution C1 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

21. LecturePLUS Timberlake 21 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?

22. LecturePLUS Timberlake 22 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 = ?? ??

23. LecturePLUS Timberlake 23 Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K - 273 = °C

24. LecturePLUS Timberlake 24 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

25. LecturePLUS Timberlake 25 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?

26. LecturePLUS Timberlake 26 Solution G9 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

27. LecturePLUS Timberlake 27 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?

28. LecturePLUS Timberlake 28 Learning Check C3 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.

29. LecturePLUS Timberlake 29 Solution C3 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.

30. LecturePLUS Timberlake 30 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

31. LecturePLUS Timberlake 31 STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

32. LecturePLUS Timberlake 32 Learning Check C4 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

33. LecturePLUS Timberlake 33 Solution C4 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

34. LecturePLUS Timberlake 34 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

35. LecturePLUS Timberlake 35 Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

36. LecturePLUS Timberlake 36 Learning Check C5 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

37. LecturePLUS Timberlake 37 Solution C5 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

38. LecturePLUS Timberlake 38 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 +.....

39. LecturePLUS Timberlake 39 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

40. LecturePLUS Timberlake 40 Learning Check C6 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

41. LecturePLUS Timberlake 41 Solution C6 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

42. LecturePLUS Timberlake 42 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

43. LecturePLUS Timberlake 43 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents.

44. LecturePLUS Timberlake 44 Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O 2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

45. LecturePLUS Timberlake 45 Solution C7 P = nRT P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L(K mol) =7.19 atm

46. Lecture PLUS Timberlake 2000 46 Molar Mass of a gas What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = 0.00703 mol RT (0.0821 L-atm/molK) (303K) Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol

47. Lecture PLUS Timberlake 2000 47 Density of a Gas Calculate the density in g/L of O 2 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 = nRTPV = nRT P = n RTV RTV RT V

48. Lecture PLUS Timberlake 2000 48 Substitute (1.00 atm ) mol-K = 0.0446 mol O 2 /L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O 2 x 32.0 g O 2 = 1.43 g/L 1 L 1 mol O 2 Therefore the density of O 2 gas at STP is 1.43 grams per liter