1. Chapter 5 The Gaseous State

2. 5 | 2 Gases differ from liquids and solids: They are compressible. Pressure, volume, temperature, and amount are related.

3. 5 | 3 Pressure, P –The force exerted per unit area. –The SI unit for pressure is the pascal, Pa.

4. 5 | 4 Empirical Gas Laws –All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties. –The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century.

5. 5 | 5 Boyle’s Law –The volume of a sample of gas at constant temperature varies inversely with the applied pressure. –The mathematical relationship: –In equation form:

6. 5 | 6 When a 1.00-g sample of O 2 gas at 0°C is placed in a container at a pressure of 0.50 atm, it occupies a volume of 1.40 L. When the pressure on the O 2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

7. ? 5 | 7 A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant? V i = 38.7 mL P i = 751 mmHg T i = 21°C V f = ? P f = 359 mmHg T f = 21°C

8. 5 | 8 V i = 38.7 mL P i = 751 mmHg T i = 21°C V f = ? P f = 359 mmHg T f = 21°C = 81.0 mL 38.7 mL751 mm Hg 359 mm Hg

9. 5 | 9 Charles’s Law –The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K). –The mathematical relationship: –In equation form: V i T f = V f T i V 1 T 2 = V 2 T 1

10. 5 | 10 A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume. As the air inside warms, the balloon expands to its orginial size.

11. 5 | 11 A 1.0-g sample of O 2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L. When the absolute temperature of the sample is raised to 200 K, the volume of the O 2 is doubled to 0.52 L.

12. ? 5 | 12 You prepared carbon dioxide by adding HCl (aq) to marble chips, CaCO 3. According to your calculations, you should obtain 79.4 mL of CO 2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C? V i = 79.4 mL P i = 760 mmHg T i = 0°C = 273 K V f = ? P f = 760 mmHg T f = 27°C = 300. K

13. 300 K79.4 mL 273 K 5 | 13 V i = 79.4 mL P i = 760 mmHg T i = 0°C = 273 K V f = ? P f = 760 mmHg T f = 27°C = 300. K = 87.3 mL

14. The Empirical Gas Laws Gay-Lussac’s Law: –The pressure exerted by a gas at constant volume is directly proportional to its absolute temperature. P  T abs (constant moles and V) P 2 T 1 = P 1 T 2

15. A Problem to Consider Presentation of Lecture Outlines, 5–15 An aerosol can has a pressure of 1.4 atm at 25°C. What pressure would it attain at 1200°C, assuming the volume remained constant? 1.4 atm1473 K 298 K

16. 5 | 16 Combined Gas Law –Use this law in the event that all three parameters, P, V, and T, are changing –The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature. –The mathematical relationship: –In equation form: P 1 V 1 T 2 = P 2 V 2 T 1

17. ? 5 | 17 Divers working from a North Sea drilling platform experience pressure of 5.0 × 10 1 atm at a depth of 5.0 × 10 2 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C? V i = 5.0 L P i = 5.0 × 10 1 atm T i = 4°C = 277 K V f = ? P f = 1.0 atm T f = 11°C = 284 K

18. 5 | 18 V i = 5.0 L P i = 5.0 × 10 1 atm T i = 4°C = 277 K V f = ? P f = 1.0 atm T f = 11°C = 284 K = 2.6 x 10 2 L 284 K5.0 x 10 1 atm5.0 L 277 K1.0 atm

19. 5 | 19 Avogadro’s Law –Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules. –So, the volume of a sample of gas is directly proportional to the number of moles of gas, n.

20. 5 | 20 Standard Temperature and Pressure (STP) –The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure. –The molar volume, V m, of a gas at STP is 22.4 L/mol. The volume of the yellow box is 22.4 L. To its left is a basketball.

21. The Ideal Gas Law From the empirical gas laws, we see that volume varies in proportion to pressure, absolute temperature, and moles.

22. The Ideal Gas Law This implies that there must exist a proportionality constant governing these relationships. –Combining the three proportionalities, we can obtain the following relationship. where “R” is the proportionality constant referred to as the ideal gas constant.

23. The Ideal Gas Law The numerical value of R can be derived using Avogadro’s law, which states that one mole of any gas at STP will occupy 22.4 liters. 22.4 L1.00 atm 1 mol273 K

24. 5 | 24 Ideal Gas Law The ideal gas law is given by the equation PV=nRT The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T / P.

25. A Problem to Consider An experiment calls for 3.50 moles of chlorine, Cl 2. What volume would this be if the gas volume is measured at 34°C and 2.45 atm?

26. ? 17.1 atm50.0 Lmol K28.02 g.08206 L atm296 K1 mol 5 | 26 A 50.0-L cylinder of nitrogen, N 2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder? V = 50.0 L P = 17.1 atm T = 23°C = 296 K mass = 986 g

27. Gas Density and Molar Mass Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter). To find molar mass, find the moles of gas, and then find the ratio of mass to moles. 5 | 27

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29. A Problem to Consider A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its molecular mass. 15.5 g.08206 L atm298 K K mol 1.08 atm5.75 L

30. ? 5 | 30 What is the density of methane gas (natural gas), CH 4, at 125°C and 3.50 atm? MM = 16.04 g/mol P = 3.50 atm T = 125°C = 398 K 16.05 g3.50 atmmol K mol.08206 L atm398 K

31. ? 5 | 31 A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C 4 H 9.) What is the molecular formula of octane?

32. 1.57 g.08206 L atm368 K760 mm Hg mol K.5000 L634 mm Hg1 atm 5 | 32 P = 634 mmHg m = 1.57 g V = 500.0 mL =.5000 L T = 95.0°C = 368.2 K

33. 5 | 33 Molecular formula: C 8 H 18 Molar mass = 114 g/mol Empirical formula: C 4 H 9 Empirical formula molar mass = 57.13 g/mol

34. 5 | 34 Stoichiometry and Gas Volumes –Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.

35. ? 5 | 35 When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO 3 was required to neutralize the spill. What volume of CO 2 was released by the neutralization at 735 mmHg and 20.°C?

36. 5 | 36 First, write the balanced chemical equation: CaCO 3(s) + 2HCl (aq)  CaCl 2(aq) + H 2 O (l) + CO 2(g) Moles of CO 2 produced = 11.9992007 mol Second, calculate the moles of CO 2 produced: 1.201 kg CaCO 3 1000 g CaCO 3 1 moles CaCO 3 1 mole CO 2 1 kg CaCO 3 100.09 CaCO 3 1 mol CaCO 3

37. 11.9992007 mol.08206 L atm293 K 760 mm Hg K mol 735 mm Hg1 atm 5 | 37 n = 11.9992007 mol P = 735 mmHg T = 20°C = 293 K = 298 L

38. 5 | 38 Gas Mixtures –Dalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.

39. 5 | 39 Originally (left), flask A contains He at 152 mmHg and flask B contains O 2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg

40. 5 | 40 Partial Pressure –The pressure exerted by a particular gas in a mixture. Dalton’s Law of Partial Pressures –The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture: –P Total = P A + P B + P C +...

41. ? 5 | 41 A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain: 0.0830 g N 2, 0.0194 g O 2, 0.00640 g CO 2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?

42. 5 | 42 0.0830 g N 2 1 mol N 2.08206 L atm308 K1000 mL 28.02 g N 2 K mol 100.0 mL1 L.74867303 atm N 2 0.0194 g O 2 1 mol O 2.08206 L atm308 K1000 mL 32.00 g O 2 K mol 100.0 mL1 L.153226535 atm O 2 0.00640 g CO 2 1 mol CO 2.08206 L atm308 K1000 mL 44.01 g CO 2 K mol 100.0 mL1 L.0367545266 atm CO 2

43. 5 | 43 P = 1.0005078412 atm 0.00441 g H 2 O1 mol H 2 O.08206 L atm308 K1000 mL 18.02 g H 2 O K mol 100.0 mL1 L.0618537496 atm H 2 O.74867303 atm N 2.153226535 atm O 2.0367545266 atm CO 2.0618537496 atm H 2 O P = 1.001 atm

44. 5 | 44 Collecting Gas Over Water –Gases are often collected over water. The result is a mixture of the gas and water vapor. –The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. –The partial pressure of water depends only on temperature and is known (Table 5.6). –The pressure of the gas can then be found using Dalton’s law of partial pressures.

45. 5 | 45 The reaction of Zn (s) with HCl (aq) produces hydrogen gas according to the following reaction: Zn (s) + 2HCl (aq)  ZnCl 2(aq) + H 2(g) The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.

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