Gas Laws and Properties: Understanding the Behavior of Gases

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Gas laws govern the behavior of gases and their properties in various conditions. From the total pressure of gas mixtures to calculating partial pressures, understanding the relationship between pressure, volume, temperature, and amount of gas is crucial. Effusion and diffusion play key roles in how gases spread. The relative rates of diffusion are inversely proportional to the square root of the molecular mass, with lighter gases diffusing faster.


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  1. The sum of the individual gas pressures equals the overall pressure of the mixture of gases. + + + = ....... P P P total P 1 2 3

  2. If a container has 166 torr H2, 109 torr CO2 and 176 torr of O2, what is the total pressure of the mixture? 166 torr + 109 torr + 176 torr = 451torr

  3. Our atmosphere is made of 21% O2, 78% N2 and 1% other gases. At sea level (standard pressure), what is the partial pressure of oxygen? 760 torr x 0.21 = 159 torr O2

  4. Temperature 0oC 20oC 40oC 60oC 80oC 100oC Temperature Pressure 0.6 kPa 2.3 kPa 7.4 kPa 19.9 kPa 47.3 kPa 101.3 kPa Pressure

  5. 52.5 mL of gas is collected over water at 20oC and 100.0 kPa. What is the pressure of the dry gas?

  6. 100.0 kPa = Pgas + Pwater Pwater at 20oC = 2.3 kPa 100.0 kPa 2.3 kPa = 97.7 kPa

  7. @STP what would the volume of the gas be? Which law would we have to use? Combined Law

  8. P1 = 97.7 kPa V1 = 52.5 mL T1 = 20oC + 273K = 293K P2 = 101.3 kPa V2 = ? T2 = 273 K ( )( K 293 V2 = 47.2 mL ) ( ) 97 7 . 52 5 . 101 3 . kPa mL kPa V = 2 273 K

  9. The spontaneous spreading of particles The rate of diffusion depends on the velocities and masses of the molecules Effusion from a small hole in a container Effusion the process by which a gas escapes Lighter gases ALWAYS than heavier molecules ALWAYS diffuse/effuse faster

  10. The relative rates at which two gases, at the same temperature and pressure, will diffuse, vary inversely as the square root of the molecular mass of the gases. m v v **Always consider gas 1 the lighter gas = 1 2 m 2 1

  11. Compute the relative rates of diffusion of helium and argon. go to the periodic table for molar mass of He and Ar

  12. v m Mass He = 4 g/mol Mass Ar = 40 g/mol = 1 2 v m 2 1 40 v = = = 10 . 3 16 1 4 v 2 So, helium diffuses 3x s faster than argon. . 3 16 v = = . 3 16 1 1 v 2

  13. Compare the rates of diffusion of O2 and N2. 32 v m = = = . 1 = . 1 14 07 1 2 28 v m 2 1 Since the lighter gas always diffuses faster, N2diffuses 1.07x s faster than O2

  14. Compare the rates of diffusion of sulfur trioxide and nitrogen monoxide. 80 1 . v m = = = . 1 = . 2 67 63 1 2 30 v m 2 1 NO diffuses 1.63x sfaster than SO3

  15. Gas #1 weighs 42.0 g/mol and diffuses at a rate of 5.65 m/s through an opening in a jar. Gas #2 weighs 252.0 g/mol. At what rate would it pass through the same opening? v m . 5 65 v 252 0 . = 1 2 = v m 42 0 . 2 1 2 . 5 65 v = v2 = 2.31 m/s . 2 45 2

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