1. DALTON’S LAW OF PARTIAL PRESSURE, AVOGADRO’S LAW, IDEAL GAS LAW MS. ANA D. HIRANG SY 2008 -2009

2. RECALL on GAS LAWS P 1 = P 2 T 1 T 2

3. RECALL on GAS LAWS Rate of gas 1 = MW of gas 2 Rate of gas 2 MW of gas 1

4. RECALL on GAS LAWS P 1 V 1 = P 2 V 2

5. RECALL on GAS LAWS P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2

6. RECALL on GAS LAWS V 1 = V 2 T 1 T 2

7. DALTON’S LAW OF PARTIAL PRESSURE In 1801, Dalton found out that the total pressure of a mixture of gases is equal to the sum of the partial pressure exerted by each gas. This is known as Dalton’s law of partial pressure. In 1801, Dalton found out that the total pressure of a mixture of gases is equal to the sum of the partial pressure exerted by each gas. This is known as Dalton’s law of partial pressure. Each gas in the mixture exerts a pressure that is independent of the other gases present. These pressures are called partial pressures. Each gas in the mixture exerts a pressure that is independent of the other gases present. These pressures are called partial pressures.

8. DALTON’S LAW OF PARTIAL PRESSURE Thus if we have a mixture of two gases, Thus if we have a mixture of two gases, O 2 = 0.1 atm N 2 = 0.7 atm P TOTAL = 0.8 atm P TOTAL = 0.8 atm N2N2 O2O2 N 2 + O 2 Mathematically this can be stated as: P TOTAL = P 1 + P 2 + P 3 + …

9. DALTON’S LAW OF PARTIAL PRESSURE

10. Dalton’s Law of Partial Pressure can be explained by 2 concepts from the Kinetic Molecular Theory. Dalton’s Law of Partial Pressure can be explained by 2 concepts from the Kinetic Molecular Theory. 1. The pressure of a gas is caused by the collision of molecules with the walls of the container. 2. Gas molecules act independently of each other.

11. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION Gas exchange between living organisms and the environment depends on the properties of gases, in particular, partial pressure and solubility. Gas exchange between living organisms and the environment depends on the properties of gases, in particular, partial pressure and solubility. RESPIRATION is one of the most important processes because we need to breathe OXYGEN and breathe out CO 2 in order to live. RESPIRATION is one of the most important processes because we need to breathe OXYGEN and breathe out CO 2 in order to live.

12. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION GAS Partial Pressure (kPa) INHALED AIR EXHALED AIR N 2 (g) 79.375.9 O 2 (g) 21.315.5 CO 2 (g) 0.0403.7 H 2 O (g) * 0.676.2 * The quantity of water in air varies. The value used in this table is based on a relatively low humidity. TABLE 1: PARTIAL PRESSURE CHANGES DURING RESPIRATION

13. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION Oxygen is the most important gas in the atmosphere, it makes up 21 % of the volume of dry air. Oxygen is the most important gas in the atmosphere, it makes up 21 % of the volume of dry air. Partial pressure of a gas is more useful than percentage composition because it is the pressure of oxygen that determines how much oxygen is absorbed by the lungs of the person. Partial pressure of a gas is more useful than percentage composition because it is the pressure of oxygen that determines how much oxygen is absorbed by the lungs of the person.

14. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION 21 % of the volume and pressure of the atmosphere is due to OXYGEN. The partial pressure of oxygen can be calculated by multiplying the percent in decimal form by the total pressure. 21 % of the volume and pressure of the atmosphere is due to OXYGEN. The partial pressure of oxygen can be calculated by multiplying the percent in decimal form by the total pressure. P OXYGEN = 0.21 X 760 torr = 159.6 torr P OXYGEN = 0.21 X 760 torr = 159.6 torr We function best breathing this partial pressure of oxygen. We function best breathing this partial pressure of oxygen.

15. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION DID YOU KNOW DID YOU KNOW You may have about 3.5 x 10 2 tiny ALVEOLI (air sacs) in each lung. The surface area of contact with capillaries for absorbing oxygen is about 75 m 2, about four (4) times as much area as an average classroom floor! You may have about 3.5 x 10 2 tiny ALVEOLI (air sacs) in each lung. The surface area of contact with capillaries for absorbing oxygen is about 75 m 2, about four (4) times as much area as an average classroom floor!

16. X –SECTION OF AN ALVEOLI

17. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION When we live at higher elevations, the partial pressure of oxygen is lower and our bodies adjust accordingly. When we live at higher elevations, the partial pressure of oxygen is lower and our bodies adjust accordingly. On top of the highest mountain, e.g. MT. EVEREST, the total atmospheric pressure is 270 torr, so the partial pressure of oxygen is only 56.7 torr, or about 1/3 of normal. On top of the highest mountain, e.g. MT. EVEREST, the total atmospheric pressure is 270 torr, so the partial pressure of oxygen is only 56.7 torr, or about 1/3 of normal. P OXYGEN = 0.21 X 270 torr = 56.7 torr P OXYGEN = 0.21 X 270 torr = 56.7 torr

18. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION A human being cannot survive for long at such a low pressure of oxygen. A human being cannot survive for long at such a low pressure of oxygen. At that altitude, even conditioned climbers must use an oxygen tank and mask, which give an increased partial pressure of oxygen to the lungs. At that altitude, even conditioned climbers must use an oxygen tank and mask, which give an increased partial pressure of oxygen to the lungs.

19. DALTON’S LAW OF PARTIAL PRESSURE: TOTAL PRESSURE SAMPLE EXERCISES 1. An equilibrium mixture contains H 2 at 560 torr, N 2 at 180 torr and O 2 at 250 torr pressure. What is the total pressure of the gases, in mm Hg and atm, in the system?

20. DALTON’S LAW OF PARTIAL PRESSURE: TOTAL PRESSURE SAMPLE EXERCISES 2. In a compressed air tank for scuba diving to a depth of 30 m, a mixture with an oxygen partial pressure of 28 atm and a nitrogen partial pressure of 110 atm is used. What is the total pressure in the tank?

21. DALTON’S LAW OF PARTIAL PRESSURE: TOTAL PRESSURE SAMPLE EXERCISES 3. The total pressure of a gas mixture in a cylinder is 6.40 atm. Gas A exerts a pressure three times that of gas B. Gas C exerts a pressure twice that of gas A. What will be the pressure of gas A, B, C if they occupy the cylinder alone?

22. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD Another application of partial pressures is THE COLLECTION OF GAS BY THE DISPLACEMENT OF WATER. Another application of partial pressures is THE COLLECTION OF GAS BY THE DISPLACEMENT OF WATER. Hydrogen and oxygen are often generated in the lab and collected by bubbling the gases into a container filled with water. Both of this gases have a relatively low solubility in water. Hydrogen and oxygen are often generated in the lab and collected by bubbling the gases into a container filled with water. Both of this gases have a relatively low solubility in water.

23. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD HYDROGEN GAS GENERATION

24. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION However, water evaporates relatively easily and the gas collected will be mixed with some water vapor. However, water evaporates relatively easily and the gas collected will be mixed with some water vapor. Water vapor is a gas like any other gas and the pressure exerted by a gas above its liquid is called VAPOR PRESSURE. Water vapor is a gas like any other gas and the pressure exerted by a gas above its liquid is called VAPOR PRESSURE.

25. DALTON’S LAW OF PARTIAL PRESSURE: APPLICATION The vapor pressure of water at different temperature is well known. See Table 13.2, p. 253 of your textbook. The vapor pressure of water at different temperature is well known. See Table 13.2, p. 253 of your textbook. Dalton’s law of partial pressures and a table of known vapor pressures of water can be used to determine the pressure of dry gas that has been collected. Dalton’s law of partial pressures and a table of known vapor pressures of water can be used to determine the pressure of dry gas that has been collected.

26. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE PROBLEM SAMPLE PROBLEM In a laboratory, oxygen gas was collected by water displacement at an atmospheric pressure of 726 torr and a temperature of 22°C. Calculate the partial pressure of dry oxygen.

27. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SOLUTION to Sample Problem SOLUTION to Sample Problem P TOTAL = 726 torr P WATER = 19.8 torr (22 °C ) P OXYGEN = ? P TOTAL = P OXYGEN + P WATER P TOTAL = P OXYGEN + P WATER P OXYGEN = P TOTAL – P WATER P OXYGEN = P TOTAL – P WATER = 726 torr – 19.8 torr = 726 torr – 19.8 torr P OXYGEN = 706.2 torr P OXYGEN = 706.2 torr

28. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE EXERCISES SAMPLE EXERCISES 1. A sealed container of bottled water sits on a store shelf at a temperature of 23°C. What is the partial pressure of water vapor in the air space inside the container?

29. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE EXERCISES SAMPLE EXERCISES 2. Nitrogen gas is collected at 20°C and a total ambient pressure of 735.8 torr using the method of water displacement. What is the partial pressure of dry nitrogen?

30. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE EXERCISES SAMPLE EXERCISES 3. Hydrogen gas is collected over water at a total pressure of 714.4 mm Hg. The volume of hydrogen collected is 30 mL at 25°C. What is the partial pressure of hydrogen gas?

31. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE PROBLEM SAMPLE PROBLEM A 500 mL sample of oxygen was collected over water at 23°C and 760 torr pressure. What volume will the dry oxygen occupy at 23°C and 760 torr? The vapor pressure of water at 23°C is 21.1 torr (from table 13.2, TBK)

32. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SOLUTION to Sample Problem SOLUTION to Sample Problem P TOTAL = P OXYGEN - P WATER P OXYGEN = 760 torr – 21.1 torr P OXYGEN = 738.9 torr Given: V 1 = 500 mLV 2 = ? P 1 = 738.9 torrP 2 = 760 torr P 1 = 738.9 torrP 2 = 760 torr T 1 =23  c + 273 = 300K T 2 =23  c + 273 = 300K

33. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SOLUTION to Sample Problem SOLUTION to Sample Problem V 2 = P 1 V 1 P 2 P 2 = (500 mL) (739 torr) = 486 mL dry O 2 = (500 mL) (739 torr) = 486 mL dry O 2 (760 torr) (760 torr)

34. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD SAMPLE EXERCISE SAMPLE EXERCISE Oxygen gas occupies 500 mL at 20°C and 760 mm Hg. What volume will it occupy if is collected over water at 30°C and 750 mm Hg?

35. AVOGADRO’S LAW Joseph Louis Gay-Lussac of France studied the volume relationship of reacting gases. Joseph Louis Gay-Lussac of France studied the volume relationship of reacting gases. In 1809, he published his results. He summarized in a statement known as GAY- LUSSAC’S LAW OF COMBINING VOLUMES OF GASES: When measured at the same temperature and pressure, the ratios of the volumes of reacting gases are the small whole numbers. In 1809, he published his results. He summarized in a statement known as GAY- LUSSAC’S LAW OF COMBINING VOLUMES OF GASES: When measured at the same temperature and pressure, the ratios of the volumes of reacting gases are the small whole numbers.

36. AVOGADRO’S LAW H 2 + Cl 2  2 HCl 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol

37. AVOGADRO’S LAW AVOGADRO’S LAW states that equal volumes of different gases at the same temperature and pressure contain the same number of molecules. AVOGADRO’S LAW states that equal volumes of different gases at the same temperature and pressure contain the same number of molecules. If the amount of gas in a container is increased, the volume is increased. If the amount of gas in a container is increased, the volume is increased. If the amount of gas in a container is decreased, the volume is decreased. If the amount of gas in a container is decreased, the volume is decreased.

38. AVOGADRO’S LAW As you increase the amount of gas (i.e. through inhalation) the volume of the balloon increases likewise. As you increase the amount of gas (i.e. through inhalation) the volume of the balloon increases likewise.

39. AVOGADRO’S LAW This law was a real breakthrough in understanding the nature of gases. This law was a real breakthrough in understanding the nature of gases. 1. It offered a rational explanation of Gay-Lussac’s law of combining volumes of gases and indicated the diatomic nature of elemental gases, such as hydrogen, chlorine and oxygen.

40. AVOGADRO’S LAW H 2 + Cl 2  2 HCl H 2 + Cl 2  2 HCl 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 1 mol

41. AVOGADRO’S LAW 2. It provided a method for determining the molecular weights of gases of known molecular weight. 3. It afforded a firm foundation for the development of the kinetic molecular theory.

42. AVOGADRO’S LAW The mathematical form of Avogadro’s law is: V   ; V 1 = V 2 V   ; V 1 = V 2 n n 1 n 2 n n 1 n 2 SAMPLE PROBLEM 1 A sample of gas with a volume of 9.20 L is known to contain 1.225 mol. If the amount of gas is increased to 2.85 mol, what new volume will result if the pressure and temperature remain constant?

43. AVOGADRO’S LAW SOLUTION Given: V 1 = 9.20 LV 2 = ? n 1 = 1.225 moln 2 = 2.85 mol n 1 = 1.225 moln 2 = 2.85 molSolution: V 2 = n 2 V 1 = (2.85 mol) (9.20 L) = 21.4 L n 1 (1.225 mol) n 1 (1.225 mol)

44. AVOGADRO’S LAW SAMPLE EXERCISES 1. If 0.25 mol of argon gas occupies a volume of 7.62 mL at a particular temperature and pressure, what volume would 0.43 mol of argon have under the same conditions?

45. AVOGADRO’S LAW SAMPLE EXERCISES 2. At a certain temperature and pressure, a balloon with 10.0 g of oxygen has a volume of 7.00 L. What is the volume after 5.00 g of oxygen is added to the balloon?

46. Ideal Gas Law The ideal gas law was first written in 1834 by EMIL CLAPEYRON. It is a combination of all the gas laws. PV = nRT PV = nRT Where, P = pressure, V = volume, n = amount of gas expressed in mol, n = amount of gas expressed in mol, T = temperature, T = temperature, R = gas constant, 0.0821 L  atm R = gas constant, 0.0821 L  atm mol  K mol  K

47. Ideal Gas Law The kinetic molecular theory assumes that the particles of an IDEAL GAS have negligible volume and no attraction exists between molecules. The temperature, pressure and volume of an ideal gas are related to each other by the ideal gas equation.

48. Ideal Gas Law GAS EQUATION R = PV = (1.0 atm) (22.4L) nT (1.0 mol) (273K) nT (1.0 mol) (273K) R = 0.0821 L  atm mol  K mol  K

49. Ideal Gas Law SAMPLE PROBLEM 1 What pressure will be exerted by 0.400 mol of gas in a 5.00 L container at 17.0°C? Given: n = 0.400 mol V= 5.00 L T= 17.0 °C + 273 = 290 K Solution: L  atm L  atm P = nRT = (0.400 mol) (0.0821 mol  K) (290 K) = 1.9atm V (5.00 L) V (5.00 L)

50. Ideal Gas Law SAMPLE PROBLEM 2 Calculate the molecular weight of butane gas, if 4.96 g occupy 2.13 L at 20.0 °C and 1 atm pressure. Given: g = 4.96 g V= 2.13 L T= 20.0 °C + 273 = 293 K P= 1 atm Solution: L  atm L  atm MW = gRT = (4.96 g) (0.0821 mol  K) (293 K) = 56.0 g/mol PV (1 atm) (2.13 L) PV (1 atm) (2.13 L)

51. Ideal Gas Law SAMPLE PROBLEM 3 What is the density of oxygen gas at STP? Given: MW = 32 g/mol T= 273 K P= 1 atm Solution: D = MWP = (32 g/mol) (1 atm) = 1.43 g/L RT L  atm RT L  atm (0.0821 mol  K) (273 K) (0.0821 mol  K) (273 K)

52. Ideal Gas Law SAMPLE PROBLEM 4 What is the density of oxygen gas at 20°C and 750 mm Hg? Given: MW = 32 g/mol T= 20°C + 273 K= 293 K P= 750 mm Hg x 1 atm = 0.987 atm 760 mm Hg 760 mm Hg Solution: D = MWP = (32 g/mol) (0.987 atm) = 1.31 g/L RT L  atm RT L  atm (0.0821 mol  K) (293 K) (0.0821 mol  K) (293 K)

53. Ideal Gas Law SAMPLE EXERCISES 1. What volume will 1.27 mol of helium gas occupy at STP? 2. Calculate the volume of 0.360 mole H 2 at 52 °C and 1.5 atm pressure.

54. Ideal Gas Law SAMPLE EXERCISES 3. How many moles of gas are contained in a 50.0-L cylinder at a pressure of 100.0 atm and a temperature of 35.0 °C? 4. Calculate the molecular weight of a gas having a density of 2.47 g/L at 26 °C and 1 atm pressure?

55. Ideal Gas Law SAMPLE EXERCISES 5. What is the density of nitrogen gas at STP?

56. SHORT QUIZ SOLVE THE FOLLOWING PROBLEMS 1. What is the total pressure in atmospheres of a gas mixture containing argon gas at 0.28 atm, helium gas at 280 mm Hg, and nitrogen gas at 390 torr? 2. A sample of 8.00 moles of argon has a volume of 20.0 L. A small leak causes half of the molecules to escape. What is the new volume of the gas?

57. SHORT QUIZ SOLVE THE FOLLOWING PROBLEMS 3. How many moles of O 2 are present in 44.8 L of O 2 at STP?