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Chemistry II

Chemistry II . Unit 1 Gases. The Nature of Gases. Objectives: Use kinetic-molecular theory to explain the behavior of gases. Describe how mass affects the rates of effusion and diffusion Explain how gas pressure is measured and calculate the partial pressure of a gas.

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Chemistry II

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  1. Chemistry II Unit 1 Gases

  2. The Nature of Gases Objectives: • Use kinetic-molecular theory to explain the behavior of gases. • Describe how mass affects the rates of effusion and diffusion • Explain how gas pressure is measured and calculate the partial pressure of a gas

  3. take the shape of their container • low density • Compressible • Mixtures are • homogeneous • Fluids (flow) • Properties of Gases

  4. Diffusion and effusion • Diffusion: movement of one material through another • Effusion: a gas escapes through a tiny opening • Graham’s law of effusion • Lighter particles effuse faster than heavier particles.

  5. Gas pressure • Results from collisions of gas particles with an object. • In empty space there are no particles, there is no pressure and (vacuum.) • Atmospheric pressure (air pressure): due to atoms and molecules in air. • Barometer: used to measure atmospheric pressure.

  6. Units for measuring pressure: • Pascal (Pa) • Standard atmosphere (atm) • Millimeters of mercury (mmHg) • 1 atm = 760 mmHg = 101.3 kPa • 1kpa = 1000 pa • Standard pressure: 1 atm • Factors affecting gas pressure • Amount of gas • Volume • Temperature • Standard temperature : 0C (273K)

  7. Converting between units of pressure • A pressure gauge records a pressure of 450 kPa. What is the measurement expressed in atmospheres and millimeters of mercury? For converting to atm: 450 kpa x 1 atm = 4.4 atm 101.3kPa For converting to mmHg: 450kPa x 760 mmHg = 3.4 x 103 mmHg 101.3 kPa

  8. What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg? 51.3 kPa, 0.507 atm • The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm? 33.7 kPa is greater than 0.25 atm

  9. Reaction_to_Air_Pressure_Below_Sea_Level.asf • Dalton’s Law of Partial Pressures • The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. • Ptotal= P1 + P2 + P3 + … Pn • CW p 405 #1-3 p409 #4-7

  10. Gas Laws • Objectives • Describe the relationships among the temperature, pressure, and volume of a gas • Use the gas laws to solve problems

  11. Boyle’s Law : Pressure and Volume • States that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure. • If pressure increases, volume decreases; if pressure decreases, volume increases. • Volume could be in liters (L), mL or cm3 1L=1000 mL 1 cm3= 1 mL

  12. P1 x V1 = P2 x V2 P: pressure 1: initial condition V: volume 2: final condition YouTube - Self Inflating a Balloon YouTube - Shaving Cream Under Vacuum

  13. Using Boyle’s Law • A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)? P1 x V1 = P2 x V2 P1= 103 kPa V1= 30.0 L P2= 25.0 kPa V2=? V2 = P1V1 P2 = (103 kPa x 30.0L) 25.0 kPa = 123.6 L

  14. 2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)? P1 x V1 = P2 x V2 P1= 205 kPa V1= 4.00 L P2= ? V2=12.0 L P2 = P1V1 V2 = (205 kPa x 4.00L) 12.0L = 68.3 kPa

  15. Classwork: pg 443 #1-3

  16. Charles’s Law: Temperature and Volume • States that the temperature of an enclosed gas varies directly with the volume at constant pressure. • As temperature increases, volume increases. V1 = V2 T1 T2 V1: initial volume V2: final volume T1: initial temperature T2: final temperature Temperature has to be in Kelvin scale. K =C + 273

  17. As a gas is heated, it expands. This causes the density of the gas to decrease. YouTube - Balloon in liquid nitrogen Volume and Temperature

  18. Using Charles’s Law Ex.1 A balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ? T1= 24C T2=58C = 24C +273 = 58C +273 = 297K = 331 K V1= 4.00 L V2= ? V1T2 = V2 T1 V1 = V2 T1 T2 (4.00Lx 331K) = V2 297K 4.56 L = V2 Since temperature increases, you expect the volume to increase.

  19. Using Charles’s Law Ex.2 A sample of SO2 gas has a volume of 1.16L at a temperature of 23C. At what temperature (in C) will the gas have a volume of 1.25L? T1= 23C = 296K T2=? V1= 1.16 L V2= 1.25 T2 = V2 x T1 V1 V1 = V2 T1 T2 T2 = 1.25L x 296 K 1.16 L T2 = 319 K -273 = 46.0C Since volume increases, you expect the temperature to increase. Classwork: p 446 #4-7

  20. Combined Gas Law • Describes the relationship among the pressure, temperature and volume, when the amount of gas is constant. • P1V1 = P2V2 T1 T2 • Standard temperature and pressure (STP): 0C, 1 atm (101.3 kPa) • Useful conversions: 1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L

  21. Using the combined gas law: • The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)? T1= 313 K T2=0 C=273 K (at STP) P1= 153 kPa P2 = 101.3 kPa (at STP) V1= 30.0L V2= ? P1V1T2= V2 P2T1 P1V1= P2V2 T1 T2 (153kPa x30.0L x 273K) = V2 (101. 3 kPa x 313 K) 39.5 L = V2 Classwork: p 450 #11,12 p 984 #8,9

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