Study Notes for Chapter 9: Properties of Gases

= k $

Speed Distributions of Gas Particles

  • Not all particles travel at the same speed:

    • umu_m: The speed of the majority of particles.

    • uextavgu_{ ext{avg}}: The average speed of all gas particles.

    • uextrmsu_{ ext{rms}}: The root mean square speed, which corresponds to the average KE.

Effect of Temperature on Particle Speed Distribution

  • Higher temperatures result in higher average KE, which causes:

    • A rightward shift in the speed distribution, leading to more particles moving at higher speeds.

Effect of Mass on Speed Distribution

  • Gases of lower mass will have:

    • A rightward-shifted distribution at a constant temperature.

    • A higher uextrmsu_{ ext{rms}} value.

    • A broader range of speeds where some particles move significantly slower or faster than average.

Implications of Mass and Speed on Atmospheric Composition

  • Light gases like H2 and He escape Earth's gravitational pull due to their high speeds, making them rare in the atmosphere.

Phenomena Explained by KMT

  • Diffusion: Spontaneous mixing of gases (e.g., how gases invade each other).

  • Effusion: Gas escaping through a small opening.

  • Example: Helium balloon deflating faster than nitrogen balloon due to lighter molecular mass.

Graham’s Law of Effusion

  • Definition: The effusion rate of a gas is inversely proportional to the square root of its molar mass ( \mathrm{M}).

    • Expression: extRateofeffusionextextofgas1/extRateofeffusionextextofgas2=M<em>2/M</em>1ext{Rate of effusion} ext{ } ext{of gas 1}/ ext{Rate of effusion} ext{ } ext{of gas 2} = \sqrt{M<em>2/M</em>1}

Applications of Graham’s Law

  • Compare effusion rates of gases like He and O2 at the same conditions.

  • Real applications: In industry, separation processes can utilize effusion rates to isolate specific gas components.

Sample Problem #3 for Molar Mass Calculation

  • Experiment Setup: 93 seconds for an unknown gas vs. 117 seconds for CO2 under same conditions (25 °C, 1.0 atm).

  • Calculate molar mass of the unknown gas using effusion rates, leveraging Graham’s law to find the relationship between time and molar mass.