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Root Mean Square Velocity (u rms )

Root Mean Square Velocity (u rms ). u rms = √(3RT/M) R = 8.3145 J/K(mol) T = temp in KELVIN M = mass of one mole in KILOGRAMS (use the molar mass from the periodic table and convert to kg/mol). RMS Example.

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Root Mean Square Velocity (u rms )

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  1. Root Mean Square Velocity (urms) urms = √(3RT/M) R = 8.3145 J/K(mol) T = temp in KELVIN M = mass of one mole in KILOGRAMS (use the molar mass from the periodic table and convert to kg/mol)

  2. RMS Example Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0ºC. Answer: 461 m/s At 300ºC Answer: 668 m/s

  3. Diffusion: Movement of particles from areas of higher concentration to lower concentration Effusion: The process where molecules of a gas in a container randomly pass through a hole in the container. Graham’s Law of Effusion

  4. Graham’s Law of Effusion Rate of Effusion for gas 1 = √M2 Rate of Effusion for gas 2 √ M1 M = molar masses of the gases Example 1: Compare the rate of effusion of carbon dioxide with that of hydrogen chloride at the same temperature and pressure. Answer: CO2 effuses 0.9X as fast as HCl.

  5. Example #2 If a molecule of neon gas travels at an average of 400m/s at a given temp., estimate the average speed of a molecule of butane gas, C4H10, at the same temp. Answer: 235m/s

  6. Summary… Larger GFM = slower rate of effusion.

  7. Real Gases No gas actually follows the ideal gas law Some come close at low pressures and/or high temperatures Under certain conditions (see above), gases behave more like ideal gases

  8. Van der Waals Equation Corrects the ideal gas equation to take nonideal conditions into account (P + n2a)(V-nb) = nRT V2 P= pressure of gas (atm) V = volume of gas (L) n = moles of gas T = temperature (K) R = 0.08206 L(atm)/mol(K)

  9. Defining Variables…MORE. a = a constant, different for each gas, that takes into account the attractive forces between molecules (table 5.3 pg. 216) b = a constant, different for each gas, that takes into account the volume of each molecule (table 5.3 pg. 216)

  10. Example Calculate the pressure exerted by 0.3000 mol of He in a 0.2000 L container at -25.0ºC Using the ideal gas law Answer: 30.53 atm Using van der Waal’s equation Where a = 0.0341 and b = 0.0237 Answer: 31.60 atm You can calculate percent error: 3% |experimental – accepted|/accepted = ___ x 100 =

  11. Pollution and Atmospheric Gases Your book discusses these - you can read about them at the end of chapter 5, but we will not be going over them in class…

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