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Gas Molecules: Mixtures and movement

Gas Molecules: Mixtures and movement. Section 12.5. After reading Section 12.5, you should know:. Avogradro’s Hypothesis, Dalton’s Law and Graham’s Law How to calculate moles, mass, and volume of gases at STP How to calculate partial pressures and rates of effusion. Avogadro’s Hypothesis.

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Gas Molecules: Mixtures and movement

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  1. Gas Molecules: Mixtures and movement Section 12.5

  2. After reading Section 12.5, you should know: • Avogradro’s Hypothesis, Dalton’s Law and Graham’s Law • How to calculate moles, mass, and volume of gases at STP • How to calculate partial pressures and rates of effusion

  3. Avogadro’s Hypothesis • Equal volumes of gases at the same temperature and pressure contain equal numbers of particles • At STP, 1 mole of any gas (regardless of size) occupies 22.4 L of space

  4. Practice Problem • Determine the volume in Liters occupied by 0.202 mol of a gas at standard temperature and pressure (STP)? Known Values:Unknowns: Have 0.202 moles of gasVol= ? 1 mol of a gas = 22.4 L 0.202 mol 22.4 L = 4.52 L 1 mol

  5. Dalton’s Law • At constant volume and temperature, the total pressure exerted by a mixtures of gases is equal to the sum of the partial pressures of the component gases.

  6. Practice with Dalton’s Law • Air contains a mixture of oxygen, carbon dioxide, nitrogen and trace amounts of other gases. What is the partial pressure of oxygen (O2) at 101.3 kPa of total pressure if the partial pressures of nitrogen is 79.10 kPa, carbon dioxide is 0.040 kPa and all other gases is 0.94 kPa? Known Values: Unknowns: Ptotal = 101.3 kPa P O2 = ? P N2 = 79.10 kPa P CO2 = 0.040 kPa P other gases = 0.94 kPa

  7. Ptotal= P (CO2) + P (O2) + P (N2) + P(other gases) 101.30 kPa = 0.040 kPa + P (O2) + 79.10 kPa + 0.94 kPa Partial Pressure (O2) = 21.22 kPa

  8. Graham’s Law • Diffusion – is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout • Effusion – the process in which a gas escapes through a tiny hole in it’s container • Graham’s Law – the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass • Gases with a smaller molar mass will effuse faster than a gas with a large molar mass

  9. Equation for Graham’s Law • Rate A = √ molar mass B Rate B √ molar mass A Example: Helium has a molar mass of 4.0 grams, nitrogen (N2) has a molar mass of 28.0 grams. Therefore, helium will effuse at a faster rate because it is lighter. Rate He= √ 28.0 g = 5.3 g = 2.7 Rate N2 √ 4.0 g 2.0 g Helium will effuse at a rate of 2.7 times faster than nitrogen

  10. After reading Section 12.5, you should know: • Avogradro’s Hypothesis, Dalton’s Law and Graham’s Law • How to calculate moles, mass, and volume of gases at STP • How to calculate partial pressures and rates of effusion

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