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Gas Behavior Chapter 12 The Beginning The first gas studied was air. The first gas studied was air. The studies were very important to understanding.

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Presentation on theme: "Gas Behavior Chapter 12 The Beginning The first gas studied was air. The first gas studied was air. The studies were very important to understanding."— Presentation transcript:

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2 Gas Behavior Chapter 12

3 The Beginning The first gas studied was air. The first gas studied was air. The studies were very important to understanding gas behavior because: The studies were very important to understanding gas behavior because: –Air is a mixture of gases. –It still behaved as one gas. –So…all gases, under similar conditions, behave similarly

4 Kinetic-Molecular Theory of Gases Assumes gas particles are in constant motion Assumes gas particles are in constant motion Used to explain gas behavior Used to explain gas behavior Based on five assumptions Based on five assumptions

5 Kinetic-Molecular Theory of Gases Assumption #1 Assumption #1 –Gases consist of large numbers of tiny particles whose sizes are negligible in comparison to their distance from each other. –Translation: Gas molecules are very small and far apart, giving the gas a low density and the property of compressibility

6 Kinetic-Molecular Theory of Gases Kinetic-Molecular Theory of Gases Assumption #2 Assumption #2 –Collisions between gas particles and collisions between gas particles and the walls of the container are elastic. –Translation: No kinetic energy is lost in the collisions; it is transferred

7 Kinetic-Molecular Theory of Gases Assumption #3 Assumption #3 –Particles of gas are continually, rapidly, and randomly moving, thereby possessing kinetic energy. –Translation: Gas particles are not attracted to each other because their kinetic energy is too strong; gas particles never stop; they are always moving

8 Kinetic-Molecular Theory of Gases Assumption #4 Assumption #4 –Gas particles have no forces of attraction or repulsion between them. –Translation: Gas particles will not stick together when they collide; they bounce off each other

9 Kinetic-Molecular Theory of Gases Assumption #5 Assumption #5 –The temperature of the gas affects the average kinetic energy of gas particles. –Translation: Temperature is a measure of average kinetic energy; higher temp.=higher KE

10 Kinetic-Molecular Theory of Gases The KMT only applies to ideal gases theoretical gases The KMT only applies to ideal gases theoretical gases Other gases are called real gases Other gases are called real gases –Real gases behave almost ideally when the pressure is not too high and/or the temperature is not too low

11 Kinetic Energy (KE) The energy of motion The energy of motion KE = ½ mv 2 KE = ½ mv 2 m = mass m = mass v = speed v = speed

12 Try This A 68 kg track runner is running at 10 m/s and a 136 kg football player is running at 5 m/s. Which has more kinetic energy? A 68 kg track runner is running at 10 m/s and a 136 kg football player is running at 5 m/s. Which has more kinetic energy? KE = ½ (68 kg)(10 m/s) 2 = 3400 J KE = ½ (68 kg)(10 m/s) 2 = 3400 J KE = ½ (136)(5 m/s) 2 = 1700 J KE = ½ (136)(5 m/s) 2 = 1700 J Track runner Track runner

13 Volume Volume is measured in: Volume is measured in: –Liters (L) –Milliliters (mL) –Cubic centimeters (cm 3 or cc)

14 Pressure Pressure is measured in: Pressure is measured in: –Atmospheres (atm) –Millimeters of mercury (mm Hg) –Kilopascals (kPa) –Torricelli (torr) –Pounds per square inch (psi)

15 Temperature Temperature is measured in: Temperature is measured in: –Celsius (°C) –Kelvin (K) –Fahrenheit (°F)

16 Standard Temperature and Pressure 0°C or 273 K 0°C or 273 K 1 atm or 760 mm Hg or 760 torr or kPa 1 atm or 760 mm Hg or 760 torr or kPa

17 The Gas Laws Developed to help create relationships between volume, pressure, temperature, and amount of a gas. Developed to help create relationships between volume, pressure, temperature, and amount of a gas. First person to collect and analyze a gas Robert Boyle First person to collect and analyze a gas Robert Boyle

18 The Gas Laws Boyles Law: volume of a fixed mass of gas varies inversely with the pressure at constant temperature Boyles Law: volume of a fixed mass of gas varies inversely with the pressure at constant temperature Translation: When pressure (P) increases, volume (V) decreases. When P decreases, V increases Translation: When pressure (P) increases, volume (V) decreases. When P decreases, V increases Mathematically: P 1 V 1 =P 2 V 2 Mathematically: P 1 V 1 =P 2 V 2

19 Important! Units must be the same thing. For instance if P 1 is in mm of Hg then P 2 must also be measured in mm of Hg. Units must be the same thing. For instance if P 1 is in mm of Hg then P 2 must also be measured in mm of Hg. If V 1 is measured in mL then V 2 must also be measured in mL. If V 1 is measured in mL then V 2 must also be measured in mL.

20 The Gas Laws Boyles Law Problem: Boyles Law Problem: –A gas at a pressure of 608 mm Hg is held in a container with a volume of 545 L. If the volume of the container is increased to 1065 L, and temperature is held constant, what is the new pressure of the gas?

21 The Gas Laws Boyles Law Solution: Boyles Law Solution: –Have: P 1 =608 mm Hg V 1 =545 L V 2 =1065 L –Want: P 2 –Use Boyles Law Equation P 1 V 1 =P 2 V 2

22 The Gas Laws Boyles Law Solution Boyles Law Solution P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 (608 mm Hg)(545 L) = (P 2 )(1065 L) (608 mm Hg)(545 L) = (P 2 )(1065 L) 311 mm Hg 311 mm Hg

23 Try this! A high- altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant) A high- altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant)

24 Answer P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 (103 kPa)(30.0 L) = (25.0 kPa)(V 2 ) (103 kPa)(30.0 L) = (25.0 kPa)(V 2 ) 124 L 124 L

25 Have you ever had the experience of buying a helium filled balloon and then taking it outside on a cold day? If you have you noticed that the balloon shrunk and looked like there was not enough helium put in it. However if you ever put a helium balloon in your car on a HOT day you will return to find that the balloon has exploded. Why do these things happen?

26 The Gas Laws Charles Lawrelationship between gas temperature and volume Charles Lawrelationship between gas temperature and volume –the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature –Translation: when V increases, T increases, and when V decreases, T decreases –Mathematically:

27 Important! The Kinetic Theory of Gases states that the kinetic energy of a gas is proportional to its temperature and using a Celsius scale would cause the kinetic energy of a gas to be negative…which is impossible! The Kinetic Theory of Gases states that the kinetic energy of a gas is proportional to its temperature and using a Celsius scale would cause the kinetic energy of a gas to be negative…which is impossible! Change the temperature to Kelvin!!!! Change the temperature to Kelvin!!!!

28 Remember…all temperatures must be in Kelvin K = ºC K = ºC + 273

29 The Gas Laws Charles Law Problem Charles Law Problem –A sample of neon gas occupies a volume of 752 mL at 25ºC. What volume will the gas occupy at 50ºC if the pressure remains constant? –Remember…the gas laws will be applied; all temperatures must be in Kelvin!

30 The Gas Laws Charles Law Solution Charles Law Solution –Have: V 1 =752 mL T 1 =25ºC + 273=298 K T 2 =50ºC + 273=323 K –Want: V 2 in mL –Use: –Solve: 752 mL = V K 323 K 298 K 323 K –815 mL

31 Try This! A balloon inflated in a room at 24°C has a volume of 4.00 L. The balloon is then heated to a temperature of 58°C. What is the new volume if the pressure remains constant? A balloon inflated in a room at 24°C has a volume of 4.00 L. The balloon is then heated to a temperature of 58°C. What is the new volume if the pressure remains constant?

32 Answer V 1 = V 2 V 1 = V 2 T 1 T 2 T 1 T 2 T 1 : 24°C = 297 K T 1 : 24°C = 297 K T 2 : 58°C = 331 K T 2 : 58°C = 331 K 4.00 L = V L = V K 331 K 297 K 331 K 4.46 L 4.46 L

33 The Gas Laws Gay-Lussacs Lawrelationship between gas pressure and temperature Gay-Lussacs Lawrelationship between gas pressure and temperature –pressure of a fixed mass of gas at a constant volume varies directly with the Kelvin temperature –Translation: When pressure increases, temp. increases; when pressure decreases, temp. decreases –Mathematically :

34 The Gas Laws Gay-Lussacs Law Problem: Gay-Lussacs Law Problem: –The gas in an aerosol can is at a pressure of 3.00 atm at 25ºC. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52ºC. What would the gas pressure in the can be at 52ºC? –Remember to convert temperatures to Kelvin!!

35 Answer P 1 = P 2 P 1 = P 2 T 1 T 2 T 1 T 2 T 1 : 25°C = 298 K T 1 : 25°C = 298 K T 2 : 52°C = 325 K T 2 : 52°C = 325 K 3.00 atm = P atm = P K 325 K 298 K 325 K 3.27 atm 3.27 atm

36 Try This! A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change?

37 Answer P 1 = P 2 P 1 = P 2 T 1 T 2 T 1 T kPa = P kPa = P K 211 K 539 K 211 K 2.58 kPa 2.58 kPa

38 The Gas Laws Combined Gas Law: combination of all three laws Combined Gas Law: combination of all three laws Mathematically: Mathematically:

39 The volume of a gas -filled balloon is 30.0 L at 40°C and 153 kPa pressure. What volume will the balloon have at standard temperature and pressure? Standard Pressure = 760 torr = 1 atm = 101.3kPa Standard Temp. = 273 K

40 Answer P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 T 1 : 40°C = 313 K T 1 : 40°C = 313 K T 2 : 0°C = 273 K T 2 : 0°C = 273 K (153 kPa)(30.0 L) = (101.3 kPa)(V 2 ) (153 kPa)(30.0 L) = (101.3 kPa)(V 2 ) 313 K 273 K 313 K 273 K 39.5 L 39.5 L

41 The Gas Laws Daltons Law of Partial Pressures: the total pressure of a mixture of gases is the sum of the pressures of each individual gas Daltons Law of Partial Pressures: the total pressure of a mixture of gases is the sum of the pressures of each individual gas Translation: the sum of the parts equals the whole Translation: the sum of the parts equals the whole Mathematically: P T = P 1 + P 2 + P 3 +… Mathematically: P T = P 1 + P 2 + P 3 +…

42 Air contains oxygen, nitrogen, carbon dioxide and trace amounts of other gases. What is the partial pressure of oxygen (P O2 ) at kPa of total pressure if the partial pressure of nitrogen,carbon dioxide and other gases are kPa, kPa and 0.94 kPa respectively. Air contains oxygen, nitrogen, carbon dioxide and trace amounts of other gases. What is the partial pressure of oxygen (P O2 ) at kPa of total pressure if the partial pressure of nitrogen,carbon dioxide and other gases are kPa, kPa and 0.94 kPa respectively.

43 Answer P T = P 1 + P 2 + P 3 +… P T = P 1 + P 2 + P 3 +… kPa = kPa kPa kPa + P O kPa = kPa kPa kPa + P O2 P O2 = 21.2 kPa P O2 = 21.2 kPa

44 Ideal Gas Law Gases behave differently under different circumstances (each gas has a different molar mass) Gases behave differently under different circumstances (each gas has a different molar mass) –Use term ideal gas to describe gas behavior under all circumstances –No such thing as ideal gas…they are real gases –In reality gases can be liquefied and sometimes solidified by cooling and by applying pressure whereas ideal gases cannot be. So real gases do not behave like ideal gases under high pressures and at low temperatures.

45 Ideal Gas Law PV = nRT PV = nRT P is pressure may be labeled kPa, atm, mm Hg, or torr P is pressure may be labeled kPa, atm, mm Hg, or torr V is volume must be labeled L V is volume must be labeled L n is moles n is moles

46 Ideal Gas Law Continued R is a constant whose value is determined by P. R is a constant whose value is determined by P. –If P is labeled kPa R = –If P is labeled atm R = –If P is labeled mm Hg or torr R = 62.4 T is temperature must be labeled K T is temperature must be labeled K

47 Try This! You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of torrs at 28°C. How many moles of nitrogen gas does the cylinder contain? You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of torrs at 28°C. How many moles of nitrogen gas does the cylinder contain? Convert Temperature to Kelvin! Convert Temperature to Kelvin!

48 Answer PV = nRT PV = nRT 28°C = 301 K 28°C = 301 K ( torr)(20.0 L) = n (62.4)(301 K) ( torr)(20.0 L) = n (62.4)(301 K) 160. moles of N moles of N 2

49 Try This! What volume is occupied by 5.03 g of hydrogen gas at 28°C and a pressure of 2.0 atm? What volume is occupied by 5.03 g of hydrogen gas at 28°C and a pressure of 2.0 atm? Hint: Convert grams to moles and °C to K! Hint: Convert grams to moles and °C to K!

50 Answer PV = nRT PV = nRT 5.03 g ÷ 2.0 g/mol = mol H g ÷ 2.0 g/mol = mol H 2 28°C = 301 K 28°C = 301 K (2 atm)(V) = (2.515 mol)(0.0821)(301 K) (2 atm)(V) = (2.515 mol)(0.0821)(301 K) 31 L 31 L

51 Ideal Gas Law Finding the molar mass Finding the molar mass M = mRT M = mRT PV PV M = molar mass and m = grams M = molar mass and m = grams What is the molar mass of a gas if 372 ml have a mass of grams at 100ºC and 108 kPa of pressure? What is the molar mass of a gas if 372 ml have a mass of grams at 100ºC and 108 kPa of pressure?

52 Answer M = mRT M = mRT PV PV What is the molar mass of a gas if 372 ml have a mass of grams at 100ºC and 108 kPa of pressure? What is the molar mass of a gas if 372 ml have a mass of grams at 100ºC and 108 kPa of pressure? 372 mL ÷ 1000 = L 372 mL ÷ 1000 = L 100°C = 373 K 100°C = 373 K M = (0.920 g)(8.314)(373 K) M = (0.920 g)(8.314)(373 K) (108 kPa)(0.372 L) (108 kPa)(0.372 L) 71.0 g/mol 71.0 g/mol

53 Try This! A container holds 2240 L of methane gas (CH 4 ) at a pressure of 1.50 kPa and a temperature of 42°C. How many grams of CH 4 does this container hold? A container holds 2240 L of methane gas (CH 4 ) at a pressure of 1.50 kPa and a temperature of 42°C. How many grams of CH 4 does this container hold?

54 Answer M = mRT M = mRT PV PV 42°C = 315 K 42°C = 315 K CH 4 = = 16.0 g/mol CH 4 = = 16.0 g/mol 16.0 g.mol = (x g)(8.314)(315 K) 16.0 g.mol = (x g)(8.314)(315 K) (1.50 kPa)(2240L) (1.50 kPa)(2240L) 20.5 g 20.5 g

55 Diffusion and Effusion Diffusion is the gradual mixing of gases due to the random, spontaneous motion of the gas particles Diffusion is the gradual mixing of gases due to the random, spontaneous motion of the gas particles Effusion is the process by which gas molecules trapped in a container randomly pass through tiny openings in the container Effusion is the process by which gas molecules trapped in a container randomly pass through tiny openings in the container What are everyday examples of diffusion or effusion? What are everyday examples of diffusion or effusion?

56 Examples Diffusion: perfume spreading, smelling cooking food, and smell something burning Diffusion: perfume spreading, smelling cooking food, and smell something burning Effusion: tire puncture and a pin hole in a balloon Effusion: tire puncture and a pin hole in a balloon

57 Diffusion and Effusion Rates of diffusion/effusion depends on the velocity of the molecules Rates of diffusion/effusion depends on the velocity of the molecules –Velocity depends on temperature and mass Would hot or cold particles move faster? Would hot or cold particles move faster? Would heavy particles move slower or faster? Would heavy particles move slower or faster?

58 Diffusion and Effusion Grahams Lawrelationship between rate of effusion (diffusion) and molar mass Grahams Lawrelationship between rate of effusion (diffusion) and molar mass –the rate of effusion of gases at the same temperature and pressure are inversely proportional to the square root of the molar mass –Mathematically:

59 Graham noticed that gases of lower molar mass effuse faster than gases of higher molar mass. Nitrogen effuses at 535 m/s. How much faster will helium gas effuse? Nitrogen effuses at 535 m/s. How much faster will helium gas effuse? 535 m/s = m/s = 4 x m/s 28 x m/s m/s ÷ 535 m/s 1415 m/s ÷ 535 m/s 2.6 times faster 2.6 times faster


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