1 / 21

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.

lorene
Télécharger la présentation

Chemistry II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related