Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Similar presentations


Presentation on theme: "1 Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 1 Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Relating temperature, pressure, volume and amount of gases Heat an aerosol can Blow up a balloon and place it in a freezer Adding 5 mols of nitrogen to a tank that already contains 10 mols of nitrogen Exploding tires on a hot summer day 2

3 3 Elements that exist as gases at 25 0 C and 1 atmosphere

4 4 NO 2 gas

5 5 Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. Physical Characteristics of Gases solid liquid gas

6 Atmospheric Pressure 6 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

7 7 Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Pressure = Force Area (force = mass x acceleration) Barometer

8 8 The pressure outside of a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore the air inside of the cabin must be pressurized to protect the passengers. What is the pressure in the cabin if the barometer reading is 749 mmHg? P = 1 atm 760 mmHg = atm 1 atm = 760 mm Hg 749 mmHg x

9 Problem 5.13 Convert 562 mmHg to atm. 9

10 10 The atmospheric pressure in New York City on a given day is 744 mmHg. What was the pressure in kPa? P = kPa 760 mmHg = 99.2 kPa 1 atm = 760 mm Hg = kPa 744 mmHg x

11 Problem 5.14 The atmospheric pressure at the summit of Mt. McKinley is 606 mmHg on a certain day. What is the pressure in atm? In kPa? 11

12 12 Manometers Used to Measure Gas Pressures closed-tubeopen-tube

13 Forces Affecting Gases. 13

14 The Gas Laws Boyles Law –Robert Boyle ( ) –Relates pressure and volume at constant temperature –For a given mass of a gas at constant temperature, the volume of the gas varied inversely with pressure.

15 15 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

16 16 P 1/V P x V = k P 1 x V 1 = P 2 x V 2 Boyles Law Constant temperature Constant amount of gas

17 17 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg P x V = constant

18 The Gas Laws Charles's Law –Jacques Charles ( ) –Relates temperature and volume at constant pressure –For a given mass of a gas at constant pressure, the volume of the gas varied directly with temperature. V = b x T V 1 = V 2 T 1 T 2

19 19 As T increasesV increases Variation in Gas Volume with Temperature at Constant Pressure

20 20 Variation of Gas Volume with Temperature at Constant Pressure V T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles Law Temperature must be in Kelvin

21 21 A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K

22 Problem 5.24 Under constant pressure conditions a sample of hydrogen gas initially at 88ºC and 9.6 L is cooled until its final volume is 3.4 L. What is the final temperature (in K)? 22 V = b x T V 1 = V 2 T 1 T 2

23 The Gas Laws Gay-Lussac's Law –Joseph Gay-Lussac ( ) Gay-Lussac studied the effect of temperature on the pressure of a gas at constant volume –For a given mass of a gas at constant volume, the pressure of the gas varied directly with temperature. P 1 = P 2 T 1 T 2

24 24 A sample of nitrogen gas has a pressure of 2.50 kPa at 25 0 C. At what temperature will the gas have a pressure of 1.54 kPa if the volume remains constant? P 1 = 2.50 kPa T 1 = K P 2 = 1.54 kPa T 2 = ? T 2 = P 2 x T 1 P1P kPa x K 2.50 kPa = = 184 K P 1 /T 1 = P 2 /T 2 T 1 = 25 ( 0 C) (K) = K

25 Problem 5.20 At 46ºC a sample of ammonia gas exerts a pressure of 5.3 atm. What is the temperature when the pressure of the gas is reduced to one- tenth of the original pressure at the same volume? 25

26 Problem 5.22 A sample of air occupies 3.8 L when the pressure is 1.2 atm. (a) What volume (in L) does it occupy at 5010mmHg if the temperature remains constant? (b) At the original pressure and volume the temperature was 101ºC. What temperature is necessary to increase the pressure to 5010mmHg, but keep the volume constant? 26

27 Explain the following. State the gas law. Helium balloon pops at a party on a summer day Predict how the size of exhaled bubbles change as a scuba diver approaches the surface of the water Aerosol cans have a warning not to put near a flame. 27

28 Describe the relationships of… 1. Boyles Law 2. Charless Law 3. Gay-Lussacs Law 28

29 29 Avogadros Law V number of moles (n) V = a x n V 1 / n 1 = V 2 / n 2 Constant temperature Constant pressure

30 30 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO

31 31 Summary of Gas Laws Boyles Law

32 32 Charles Law Gay-Lussacs Law

33 33 Avogadros Law

34 34 Ideal Gas Equation Charles law: V T (at constant n and P) Avogadros law: V n (at constant P and T) Boyles law: P (at constant n and T) 1 V V nT P V = constant x = R nT P P R is the gas constant PV = nRT R = L atm / (mol K)

35 35 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = L atm / (mol K) Calculate the volume of 1 mole of an ideal gas at STP When the moles are constant, PV = nR T Therefore P 1 V 1 = P 2 V 2 T 1 T 2

36 36 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.7 L

37 Problem 5.40 Calculate the volume (in L) of 88.4 g of CO 2 at STP. 37

38 38 Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

39 Extra Practice: Problem 5.36 The temperature of 2.5 L of a gas initially at STP is raised to 250ºC at constant volume. Calculate the final pressure of the gas in atm. 39

40 Extra Practice: Problem 5.32 Given that 6.9 moles of carbon monoxide are present in a container of volume 30.4 L, what is the pressure of the gas (in atm) if the temperature is 62ºC? 40

41 41 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L n = m M

42 42 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.5 g/mol

43 Problem 5.44 At 741 torr and 44ºC 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass of the gas (in g/mol)? 43

44 44 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm = = 4.76 L

45 Problem 5.60 Ethanol (C 2 H 5 OH) burns in air: C 2 H 5 OH(l) + O 2 (g) CO 2 (g) + H 2 O(l) Determine the volume of air in liters at 35.0ºC and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent O 2 by volume. 45

46 46 Daltons Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 Studying gaseous mixtures shows that each gas behaves independantly of the others

47 47 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT

48 48 A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = x 1.37 atm= atm

49 Problem 5.63 A mixture of gases contains 0.31 mol CH 4, 0.25 mol C 2 H 6, and 0.29 mol C 3 H 8. The total pressure is 1.50 atm. Calculate the partial pressure of each gas. = _____ = ______ = ______ 49

50 50 2KClO 3 (s) 2KCl (s) + 3O 2 (g) P T = P O + P H O 22 Collecting a Gas over Water

51 51 Vapor of Water and Temperature

52 Problem 5.66 A mixture of helium and neon gasses is collected over water at 28.0ºC and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon (in mmHg)? (vp H20 = 28.3 mmHg at 28.0ºC) 52

53 Problem 5.68 A sample of zinc metal reacts completely with an excess of hydrochloric acid: Zn(s) + 2HCl(aq) The hydrogen gas produced is collected over water at 25.0ºC. The volume of the as is 7.80L, and the pressure is atm. Calculate the amount of zinc metal (in grams) consumed in the reaction. (vp H20 = 28.3 mmHg at 28.0ºC) 53


Download ppt "1 Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."

Similar presentations


Ads by Google