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What are the characteristics of gases?

Chapter 13 - Gases. What are the characteristics of gases?. Ch 13 Gases Pressure Volume Temperature collapsing can Liquid Nitrogen and Balloons pumping up a tire popcorn. Gas molecules move really fast and are really far apart! For example, comparing volume : 1 mol CO 2 solid = 28 cm 3

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What are the characteristics of gases?

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  1. Chapter 13 - Gases What are the characteristics of gases?

  2. Ch 13 Gases • Pressure • Volume • Temperature • collapsing can • Liquid Nitrogen and Balloons • pumping up a tire • popcorn

  3. Gas molecules move really fast and are really far apart! • For example, comparing volume:1 mol CO2solid = 28 cm3 • 1 mol CO2gas = 25000 cm3 • Most of the volume of gas is empty space! extra room that can be used to compress the gas (e.g. tires, scuba tanks)

  4. 13.1 pressure • Force versus Pressure: Imagine balancing a bowling ball on your head. Then imagine balancing a bowling ball on a nail on your head. • Force = mass x acceleration • Pressure = Force / area • Gas pressureis like ping-pong balls hitting the walls; more hits, higher pressure; fewer hits, lower pressure

  5. The collapsing can experiment is a good intro to Atmospheric pressure • If the molecules of air hit harder and more often on the outside, the can could collapse

  6. USS Thresher

  7. Атомная подводная лодка (Kursk)

  8. In 1643 Evangelista Torricelli invented the barometer • an instrument used to measure atmospheric pressure. • It can measure the pressure exerted by the atmosphere • At sea level, a column of Hg would be 760 mm high

  9. Units of Pressure • Because barometers have a column of Hg, one unit is mm Hg • This pressure device is called a manometer

  10. Units of Pressure 1 atm = 14.7 psi = 760 mm Hg = 760.0 Torr = 101.325 kPa = 101,325 Pa

  11. Examples • The mercury has risen to a height of 729 mm. What is this pressure in kPa & psi? 729 mm 101.325 kPa = 97.2 kPa 760. mm Hg 729 mm 14.69 psi = 14.2 psi 760. mm Hg

  12. Example • The pressure reading from a barometer is 742 mm Hg. • Express this reading in • kPa • atm • torr

  13. 13.2 Pressure & VolumeBoyle’s Law • Robert Boyle in the 1600’s did an experiment w/ a tube like this • As he added Hg, the height increased (h) and the volume of the gas decreased

  14. Here’s his data; see a pattern?

  15. as the pressure increases the volume decreases • they are inversely proportionate

  16. Boyle’s Law: at constant temperature for a fixed mass, the pressure and the volume of a gas are inversely proportional. P1V1 = P2V2

  17. This plastic bottle was sealed at approximately 14,000 feet elevation, and was crushed by the increase in atmospheric pressure (at 9,000 feet and 1,000 feet) as it was brought down towards sea level.

  18. Robert Boyle

  19. Example • A gas sample occupies 100.0 cm3 when the pressure is 150.0 kPa. When the pressure is increased to 200.0 kPa, what is the new V? • Do you expect the volume will increase or decrease? P1= P2= V1= V2= 150.0 kPa 200.0 kPa 100.0 cm3 ? V2 = (150.0 kPa x 100.0 cm3) / 200.0 kPa V2 = 75.00 cm3 P1V1 = P2V2 V2 = P1V1 / P2

  20. Questions?

  21. 13.3 Volume & Temperature:Charles’ law • After Boyle, work on gases continued • Jacques Charles, a famous balloonist, also quantified a gas relationship

  22. He found that when a gas sample cools, it contracts, when heated it expands • all gases behaved very similarly

  23. When cooled, all the gas samples contracted and their linear relation took them all to the same temperature when the volume equals zero.

  24. -273˚C • The temperature at which a gas volume = 0 is absolute zero

  25. Charles’ Law: At constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature on the absolute temperature scale • Volume is directly proportional to Temperature • V1/T1 = V2/T2

  26. V1/T1 = V2/T2 (Temperatures must be in K) and K = ˚C + 273

  27. Example • A gas sample has a volume of 2.25 L at 298 K. What is the new volume when heated to 373 K? • Will the new volume be greater or less? • V1 = • T1 = • V2 = • T2 = 2.25 L 298 K ? 373 K • V1/T1 = V2/T2 • V1/T1 x T2 = V2 • (2.25L / 298K) x 373K = 2.82 L

  28. Questions?

  29. 1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg? 2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm? 3. To what pressure must a gas be compressed in order to get into a 3.00 cubic foot tank the entire weight of a gas that occupies 400.0 cu. ft. at standard pressure? 4. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm? 5. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L?

  30. 1. 8.20 L 2. 1.44 L 3. 133 atm 4. 0.520 L 5. 0.642 atm

  31. 13.4 Volume & Moles: Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain the same number of molecules. Put simply, the more of the gas there is, the greater the volume.

  32. 13.4 Volume & Moles: Avogadros Law

  33. Avogadro’s Law (equation) • V/n = k • V1/n1 = V2/n2

  34. Example • A 2.0 mol sample of hydrogen gas occupying 3.5 L has more hydrogen gas added to it until the volume reaches 7.0 L. How many mol of hydrogen are in the sample now? • V1 = • n1 = • V2 = • n2 = • V1/n1 = V2/n2 • n2 = V2 x n1 / V1 3.5 L 2.0 mol 7.0 L ? n2 = 7.0 L x 2.0 mol / 3.5 L n2 = 4.0 mol

  35. Questions?

  36. L•atm mol•K 13.5 The Ideal Gas Law • Boyle’s Law PV = k1(constant) • Charles’s Law V / T = k2 (constant) • Avogadro’s Law V / n = k3(constant) • When all three are combined we get the Ideal Gas Law PV = nRT where P is in atm, V in liters, n in mols, T in Kelvins • R is called the universal gas constant! it has a value of 0.08206 L•atm/mol•K

  37. L•atm mol•K • remember 2 basic things: 1) Rearrange PV=nRT to solve for what you are looking for 2) All variables must be in the proper units for R

  38. L•atm mol•K Example • What is the pressure of 0.412 mol He gas at 289 K with a volume of 3.25 L? • Make a list of what you are given • 0.412 mol is the… • 289 K is the… • 3.25 L is the… • Solve PV = nRT for the unknown you are looking for • P = nRT/V = 0.412 mol x 0.08206 x 289 K • 3.25L • P = 3.01 atm Number of moles (n) Temperature (T) Volume (V)

  39. L•atm mol•K Example • what is the volume of 80.0 g of oxygen gas at 25 ˚C and 104.5 kPa? • V= • n= • T= • P= • PV = nRT ? 80.0g 25oC 104.5kPa x 1 mol / 32 g = + 273 = 2.5 mol 298 K x 1 atm/101.325 kPa = 1.031 atm • V = nRT/P = 2.5 mol x 0.08206 x 298 K 1.031 atm = 59L

  40. Questions?

  41. Students know the values and meanings of standard temperature and pressure (STP). Standard temperature is 0°C standard pressure (STP) is 1 atm.

  42. 13.6 Dalton’s Law of Partial Pressure • There are mixtures of gases almost everywhere (atmosphere, scuba tanks, locker rooms) • Each gas in a mixture acts like it’s the only gas

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