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Ch. 14 Gas Laws

Ch. 14 Gas Laws. Properties of gases. Gases are the least dense and most mobile of the three phases of matter. Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often.

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Ch. 14 Gas Laws

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  1. Ch. 14 Gas Laws

  2. Properties of gases • Gases are the least dense and most mobile of the three phases of matter. • Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often. • Gases occupy much greater space than the same amount of liquid or solid. This is because the gas particles are spaced apart from one another and are therefore compressible. Solid or liquid particles are spaced much closer and cannot be compressed further.

  3. Properties of gases • Gases are characterized by four properties. These are: • 1. Pressure (P) • 2. Volume (V) • 3. Temperature (T) • 4. Amount

  4. Kinetic –molecular theory of gases • Scientists use the kinetic-molecular theory (KMT) to describe the behavior or gases. The KMT has 5 postulates: • 1. Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities. • 2. Gas particles have little attraction for one another. Therefore, attractive forces between gas molecules can be ignored.

  5. Kinetic –molecular theory of gases • 3. The distance between the particles is large compared to their size. Therefore the volume occupied by the gas molecules is small compared to the volume of the gas. • 4. Gas particles move in straight lines and collide with each other and the container frequently. The force of collisions of the gas particles with the walls of the container cause pressure. • 5. The average KE of gas molecules is directly proportional to the absolute temperature (K)

  6. Pressure and its measurement • Pressure is the result of collision of gas particles with the sides of the container. Pressure is defined as force per unit area. • Pressure is measured in units of atmosphere (atm) or mm Hg or torr. The SI unit of pressure is pascal (Pa) or kilopascal (kPa). • 1 atm = 760 mm Hg • 1 mm Hg = 1 torr • 1 atm = 101.3 kPa

  7. Pressure and its measurement • Atmospheric pressure can be measured with the use of a barometer. Mercury is used in a barometer due to its high density. At sea level, the mercury stands at 760 mm above its base. • The pressure of a gas is directly proportional to the number of particles (moles) present.

  8. 1 mol H2 2 mol H2 0.5 mol H2 P = 1atm P = 2 atm P = 0.5 atm

  9. Examples • Atmospheric pressure at Walnut, CA is 740. mm Hg. Calculate this pressure in torr and atm. 1 mm Hg = 1 torr therefore, 740. mm Hg = 740. torr 740. mmHg X 1 atm= 0.97 atm 760mmHg

  10. Examples • 2. The barometer at a location reads 1.12 atm. Calculate the pressure in mm Hg and torr. 1 mm Hg = 1 torr 760 mm Hg = 1 atm 1.12 atm X __________________= 851 mm Hg 851 mm Hg = 851 torr 760 mm Hg 1 atm

  11. Relationship between pressure and volume – boyle’s law • At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. • P1V1 = P2V2 • Any pressure unit /volume unit will work as long as they are identical.

  12. examples • A sample of H2 gas has a volume of 5.0 L and a pressure of 1.0 atm. What is the new pressure if the volume is decreased to 2.0 L at a constant temperature? • P1 = 1.0 atm P2 = ?? P1V1 = P2V2 • V1 = 5.0 L V2 = 2.0 L • P2 = P1V1 = (1.0 atm)(5.0 L) = 2.5 atm V2 2.0 L V2 V2

  13. Examples • 2. A sample of gas has a volume of 12-L and a pressure of 4500 mm Hg. What is the volume of the gas when the pressure is reduced to 750 mm Hg? • P1 = 4500 mm Hg P2= 750 mm Hg P1V1 = P2V2 • V1 = 12 L V2 = ??? P2 P2 V2= P1V1 = (4500 mm Hg)(12 L) = 72 L P2750 mm Hg

  14. Examples • 3. A sample of hydrogen gas occupies 4.0 L at 650 mm Hg. What volume would it occupy at 2.0 atm? 2.0 atmX760mmHg 1 atm = 1520 mm Hg Change atm to mm Hg!!! • P1 = 650 mm Hg P2= 2.0 atm P1V1= P2V2 • V1 = 4.0 L V2 = ??? 1520 mm Hg P2 P2 V2= P1V1 = (650 mm Hg)(4.0 L) = 1.7 L P21520 mm Hg

  15. Relationship between temperature and volume – Charles’s Law • At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (KELVIN!!!) • V1 = V2 • T1 T2 Note: Temperature must be in Kelvin! K=°C+273

  16. Examples • A 2.0 L sample of a gas is cooled from 298 K to 278 K at a constant pressure What is the new volume of the gas? • V1 = V2 Multiply both sides by T2 to isolate V2 • T1 T2 • T2V1= V2 • T1 • V2= V1T2 = 2.0L X 278 K = 1.9L • T1 298 K

  17. Examples • A sample of gas has a volume of 5.0L and a temperature of 20.0˚C. What is the volume of the gas when the temperature is increased to 50.0 ˚C at constant pressure? • Change ˚C to K by adding 273. • T1 = 20 ˚C + 273 = 293K, T2= 50 ˚C + 273 = 323K • V1 = 5.0 L, V2 = ? • V1 = V2 Multiply both sides by T2 to isolate V2 • T1 T2 • T2V1= V2 • T1 • V2= V1T2 = 5.0L X 323 K = 5.5L • T1 293 K

  18. Examples • If 20.0L of oxygen gas is cooled from 100 ˚C to 0 ˚C, what is the new volume ? • Change ˚C to K by adding 273. • T1 = 100 ˚C + 273 = 373K, T2= 0 ˚C + 273 = 273K • V1 = 20.0 L, V2 = ? • V1 = V2 Multiply both sides by T2 to isolate V2 • T1 T2 • T2V1= V2 • T1 • V2= V1T2 = 20.0L X 273 K = 14.5L • T1 373 K

  19. Examples • If you need T2 orT1? • V1 = V2 Cross multiply to get T2 and T1 on the top • T1T2 • T2V1= V2T1 then divide by V1 or V2 to isolate the T you want to find.

  20. Relationship between Temperature and Pressure – Gay-Lussac’s Law • At a constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature. • P1 = P2 • T1 T2 K = ˚C + 273

  21. Examples • An aerosol spray can has a pressure of 4.0 atm at 25˚C. What pressure will the can have if it is placed in a fire and reaches a temperature of 400 ˚C? • Change ˚C to K by adding 273. • T1 = 25 ˚C + 273 = 298K, T2= 400 ˚C + 273 = 673K • P1 = 4.0 atm, P2 = ? • P1= P2 Multiply both sides by T2 to isolate P2 • T1 T2 • T2P1= P2 • T1 • P2 = P1T2 = 4.0atm X 673 K = 9.0atm • T1 298 K

  22. Examples • A cylinder of gas with a volume of 15.0 L and a pressure of 965 mmHg is stored at a temperature of 55 ˚C. To what temperature must the cylinder be cooled to reach a pressure of 850.0mmHg? • Change ˚C to K by adding 273. • T1 = 55 ˚C + 273 = 328K, T2= ? • P1 = 965 mmHg, P2 = 850 mmHg • P1= P2 Cross multiply, then divide both sides by P1 • T1T2 • T1P2= T2 • P1 • T2 = T1P2 = 328 K X 850.0 mmHg = 289 K • P1 965 mmHg

  23. Examples • The pressure of a container of helium is 650.0 mmHg at 25 ˚C? If the container is cooled to 0 ˚C, what will the pressure be? • Change ˚C to K by adding 273. • T1 = 25 ˚C + 273 = 298K, T2= 0 ˚C + 273 = 273 K • P1 = 650.0 mmHg, P2 = ? • P1= P2 Multiply both sides by T2 to solve for P2 • T1T2 • T2P1= P2 • T1 • P2 = T2P1= 273 K X 650.0 mmHg = 595 mmHg • T1 298 mmHg

  24. Complete pages 10, 11, 12 and 13 in your packet!

  25. Relationship between pressure, volume, and temperature – Combined gas law • All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law. This law is useful for studying the effect of changes in two variables. • The individual gas laws studied previously are embodied in the combined gas law. • P1V1= P2V2 • T1 T2 Charles’s Law Gay-Lussac’s Law Boyle’s Law • P1V1= P2V2 • T1 T2 • P1V1= P2V2 • T1 T2 • P1 V1= P2 V2 • T1 T2

  26. Examples • A 25.0 mL sample of gas has a pressure of 4.00 atm at a temperature of 10 ˚C. What is the volume of the gas at a pressure of 1.00 atm and a temperature of 18 ˚C? • Change ˚C to K by adding 273. • T1 = 10 ˚C + 273 = 283K, T2= 18 ˚C + 273 = 291 K • V1 = 25.0 mL, V2 = ?, P1=4.0 atm, P2=1.00 atm • P1 V1 = P2 V2 To isolate V2, Multiply both T2 /P2 • T1 T2 • V1P1T2= V2 • P2T1 • V2 = V1P1T2 = 25.0 mLX4.00atmX291K= 103mL • P2T1 1.00atmX283 K

  27. Examples • A sample of ammonia has a volume of 20.0 mL at 5 ˚C and 700.0 mmHg. What is the volume of the gas at 50 ˚C and 850 torr? (Remember 1 torr = 1 mmHg) • Change ˚C to K by adding 273. • T1 = 5 ˚C + 273 = 278K, T2= 50 ˚C + 273 = 323 K • V1 = 20.0 mL, V2 = ?, P1= 700 mmHg, P2= 850 mmHg • P1 V1 = P2 V2 To isolate V2, Multiply both T2 /P2 • T1 T2 • V1P1T2= V2 • P2T1 • V2 = V1P1T2 = 20.0 mLX700mmHgX323K= 19 mL • P2T1 850mmHgX278 K

  28. Ideal Gas Law • Combining all the laws that describe the behavior of gases, one can obtain a usefulrelationship that relates the volume of a gas to the temperature, pressure and number of moles. • R=universal gas constant = 8.31LkPa/molK, n = moles • P= Pressure in kPa and Temperature is in Kelvin! • This relationship is called the Ideal Gas Law, and is commonly written as: • PV = nRT • V = nRT • P

  29. Rmember!! • 1 atm = 760 mm Hg = 760 torr • 1 mm Hg = 1 torr • 1 atm = 101.3 kPa • Pressure has to be in kPa • ADD: Volume has to be in Liters!

  30. Examples • A sample of H2 gas has a volume of 8.56 L at a temperature of 0ºC and a pressure of 1.5 atm(152 kPa). Calculate the moles of gas present. • P = 152 kPa, V = 8.56 L, T = 0ºC + 273 = 273 K, n = ?? • PV = nRT, divide both sides by RT to get n isolated • n = PV = (152kPa)(8.56 L) = 0.57 mol RT (8.31LkPa/molK)(273K)

  31. Examples • What volume does 40.0 g of N2 gas occupy at 10ºC and 99.97 kPa. • P = 99.97 kPa, V = ?, T = 10ºC + 273 = 283 K, n = ?? • 40.0g N2X(1mol N2/28.01gN2 )= 1.43 mol N2 • PV = nRT, divide both sides by P to get V isolated • V = nRT = (1.43 mol)(8.31LkPa/molK)(283 K) = 33.6 L P (99.97 kPa)

  32. Examples • A 23.8 L cylinder contains oxygen gas at 20.0ºC and 97.6 kPa. How many moles of oxygen does it contain? • P = 97.6 kPa, V = 23.8 L, T = 20ºC + 273 = 293 K, n = ?? • PV = nRT, divide both sides by RT to get n isolated • n = PV = (97.6 kPa) (23.8 L) = 0.954 mol RT (8.31LkPa/molK) (293 K)

  33. partial pressures – Dalton’s Law • Many gas samples are mixtures of gases. For example, the air we breathe is a mixture of mostly oxygen and nitrogen gases. • Since gas particles have no attractions towards one another, each gas in a mixture behaves as if it is present by itself, and is not affected by the other gases present in the mixture. • In a mixture, each gas exerts a pressure as if it was the only gas present in the container. This pressure is called the partial pressure of the gas.

  34. In a mixture, the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture. This is called Dalton’s Law of Partial Pressures. • Ptotal = P1 + P2 + P3 + … Total pressure of the gas mixture = sum of the partial pressures of the gases in the mixture. • The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture. For example, in a mixture of gases consisting of 1 mol of nitrogen and 1 mol of hydrogen gas, the partial pressure of each gas is one-half of the total pressure in the container.

  35. 4 2 6 • Twice as many particles of Ar, so twice the pressure of He. 2molAr 1 mol He PAr=4.0 atm =2.0 atm+4.0atm PHe=2.0 atm = 6.0 atm

  36. diffusion

  37. If two containers are connected, the gases mix and the volume is doubled from each gas’s original volume so the pressure on each side from each gas is halved.

  38. + = If volume is constant

  39. Examples • Two 10 L tanks, one containing propane gas at 300 torr and the other containing methane at 500 torr, are combined in a 10 L tank at the same temperature. What is the total pressure of the gas mixture? • Ptotal = P1 + P2 = 300 torr + 500 torr = 800 torr

  40. Examples • A scuba tank contains a mixture of oxygen and helium gases with a total pressure of 7.00 atm. If the partial pressure of oxygen in the tank is 1140 mmHg, what is the partial pressure of the helium in the tank? • Pressure needs to be the SAME!! • PO2 (in atm) = 1140 mmHg X (1 atm/760mmHg)=1.50 atm O2 • Ptotal = PHe + PO2 subtract PO2 from both sides • PHe= Ptotal– PO2 • PHe = 7.00 atm– 1.50 atm = 5.50 atm • PHe = 5.50 atm

  41. Examples • A mixture of gases contains 2.0mol of O2 gas and 4.0 mol of N2 gas with a total pressure of 3.0 atm. What is the partial pressure of each gas in the mixture? • 2 mol O2, 4 mol N2, 6 total moles • (2 mol O2/6 mol total )X 3.0 atm = 1 atm from O2 • (4 mol N2/6 mol total) X 3.0 atm = 2 atm from N2 • Check answer 1 atm +2 atm = 3 atm

  42. Complete pages 14, 15 and 16 Tomorrow in class (all due Friday!)

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