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Chapter 13 Gases

Chapter 13 Gases. Gases. Properties of Gases: 1) Gases have . 2) It is easy to gases. 3) Gases their containers completely. 4) Different gases can move each other quite rapidly. ( ) 5) Gases exert . 6) The pressure of a gas depends on it’s . 13-2. Gases.

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Chapter 13 Gases

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  1. Chapter 13 Gases

  2. Gases Properties of Gases: 1) Gases have. 2) It is easy to gases. 3) Gases their containers completely. 4) Different gases can move each other quite rapidly. ( ) 5) Gases exert. 6) The pressure of a gas depends on it’s. 13-2

  3. Gases The Kinetic~Molecular Theory: 1) Gases consist of very small, each of which has a. 2) The separating gas particles are relatively. 3) Gas particles are in constant, rapid, random. 4) of gas particles with each other or with the walls of the container are perfectly. 5) The average energy of gas particles depends only on the of the gas. 6) Gas particles exert no on one another. 13-3

  4. Gases 4 Important Variables Describing Gases: 1) Amount of gas ( ). = mass = m(g) molar mass M(g/mol) 2) Volume ( ) – the volume of the container in which it is placed. 3) Temperature ( ) in degrees Kelvin. T(k) = T(oC) + 273 4) Pressure ( )- the outward force spread over the area of the container. 13-4

  5. Pressure Pressure: the pressure exerted by the air in the atmosphere. N/m2 = pascal (Pa) : measures atmospheric pressure in millimeters of mercury (mm Hg). Units of Pressure: 1 atmosphere (atm) = Pa 1 atm = mm Hg 1 atm = lb./in2 1bar = 100,000 Pa = 0.9869 atm 13-5

  6. Pressure Units of Pressure: 1 atmosphere (atm) = Pa 1 atm = mm Hg 1 atm = lb./in2 1bar = 100,000 Pa = 0.9869 atm Ex. The barometer reads 758 mm Hg. What is the atmospheric pressure in Pa? 758 mm Hg x 101,325 Pa = Pa 760 mm Hg 13-6

  7. Pressure Units of Pressure: 1 atmosphere (atm) = Pa 1 atm = mm Hg 1 atm = lb./in2 1bar = 100,000 Pa = 0.9869 atm 1) The air pressure in a tire is 109 kPa. What is the pressure in atm? 109,000Pa x 1 atm = atm 101,325 Pa 13-7

  8. Boyle’s Law = standard temperature (0oC) and pressure (1.0 atm) Boyle’s Law: at constant, the pressure and volume of a sample of gas are inversely proportional to each other. ____________ 13-8

  9. Boyle’s Law P1V1 = P2V2 Ex. A gas occupies a volume of 458 ml at a pressure of 1.01 kPa and temp. 295 K. When the pressure is changed, the volume becomes 477 ml. If there is no change in temp., what is the new pressure? (1.01 KPa) (458 ml) = P2 (477 ml) P2 = kPa 13-9

  10. Boyle’s Law P1V1 = P2V2 2) A gas occupies a volume of 2.45L at a pressure of 1.03 atm and temp. 293 K. What volume will the gas occupy if the pressure changes to 0.980 atm if the temp. remains the same? (atm) (L) = (atm)V2 V2 = L 13-10

  11. Charles’s Law Charles’s Law: at constant, the volume of a fixed amount of gas is directly proportionate to it’s absolute temp. V1 = V2 T1 T2 13-11

  12. Charles’s Law V1 = V2 T1 T2 Ex. What will be the of a gas sample at 309K if it’s volume at 215K is 3.42L? Assume that the pressure is constant. 3.42L = V2 215K 309K V2 = L 13-12

  13. Charles’s Law V1 = V2 T1 T2 3) A gas sample at 83oC occupies a volume of 1400m3. At what temperature will it occupy 1200m3? Assume that the pressure is constant. 1400m3 = 1200m3 oC T2 T2 =oC 13-13

  14. Law’s Law states that equal of gases at the same temperature and pressure contain an equal number of. V = k3n Law: the of the pressures of all the components in a gas mixture is equal to the pressure of the gas mixture. PT = Pa + Pb + Pc …

  15. Law’s Ideal Gas Law: R = (1 atm)(22.4L) = 0.0821atmL/molK (1 mol)(273K) Ex. If the pressure exerted by a gas at 0oC in a volume of 0.0010L is 5.00atm, how many moles of gas are present? (5.00atm)(0.0010L) = n (0.0821atmL/molK)(273K) n = 13-15

  16. Law’s Ideal Gas Law: PV = nRT R = 0.0821atmL/molK 4) What volume would be occupied by 100. g of oxygen gas at a pressure of 1.50 atm and a temp. of 25oC? 100.g O2 x 1mol O2= mol O2 g ( )V = (mol)(atmL/molK)(298K) V = L 13-16

  17. Gas Law Summary LawStatementEquationConstant Boyle’s P inverselyP1V1 = P2V2 T, n proportional to V Charles’s V directly V1 = V2 P, n proportional to TT1 T2 Avogadro’s V directly V = k3n P, T proportional to n Dalton’s PT is the sum PT = Pa + Pb + Pc …T, V of partial pressures of components 13-17

  18. Chapter 13 The End!

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