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AP Chemistry Unit 5 – States of Matter

AP Chemistry Unit 5 – States of Matter. Lesson 2 – Ideal Gas Law Book Section: 10.4-10.5. Ideal-Gas Equation. So far we’ve seen that V  1/P (Boyle’s Law) V  T (Charles’s Law) V  n (Avogadro’s Law) Combining these, we get. Ideal-Gas Equation.

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AP Chemistry Unit 5 – States of Matter

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  1. AP ChemistryUnit 5 – States of Matter Lesson 2 – Ideal Gas Law Book Section: 10.4-10.5

  2. Ideal-Gas Equation • So far we’ve seen that • V 1/P (Boyle’s Law) • V  T (Charles’s Law) • V  n (Avogadro’s Law) • Combining these, we get

  3. Ideal-Gas Equation • To turn any proportionality into an equation, you have to multiply by a constant. • The constant of proportionality in this case is known as R, the ideal-gas constant. • R = different values depending on units of pressure: 0.0821 L-atm/mol-K (use when P = atm) 8.314 J/mol-K (use when P = Pa) 62.36 L-torr/mol-K (use when P = torr)

  4. Ideal Gas Law • The relationship • Then becomes • Or… • PV = nRT  Ideal Gas Law

  5. Ideal Gas Law and Density • If we divide both sides of the ideal gas equation by V and by RT, we get

  6. Ideal Gas Law and Density • We know that • Moles x molar mass = mass n x M = m (M = molar mass) • So, multiplying both sides by the molar mass (M) gives

  7. Ideal Gas Law and Density • Mass/volume = density • So… • One only needs to know the molar mass, pressure, and temperature to calculate the density of a gas.

  8. Molar Mass • We can manipulate the density equation to enable us to find the molar mass of a gas:

  9. 1984 MC #23 • The density of an unknown gas is 4.20 grams per liter at 3.00 atmospheres pressure and 127 ºC. What is the molecular weight of this gas? (R = 0.0821 liter-atm/mole-K) • 14.6 • 46.0 • 88.0 • 94.1 • 138

  10. 1984 MC #23 • The density of an unknown gas is 4.20 grams per liter at 3.00 atmospheres pressure and 127 ºC. What is the molecular weight of this gas? (R = 0.0821 liter-atm/mole-K) • 14.6 • 46.0 – 75% correct, easy • 88.0 • 94.1 • 138

  11. 1994 MC #24 • A sample of 0.0100 mole of oxygen gas is confined at 37 ºC and 0.216 atmosphere. What would be the pressure of this sample at 15 ºC and the same volume? • 0.0876 atm • 0.175 atm • 0.201 atm • 0.233 atm • 0.533 atm

  12. 1994 MC #24 • A sample of 0.0100 mole of oxygen gas is confined at 37 ºC and 0.216 atmosphere. What would be the pressure of this sample at 15 ºC and the same volume? • 0.0876 atm • 0.175 atm • 0.201 atm – 76% correct, easy • 0.233 atm • 0.533 atm

  13. 1994 MC #37 • A sample of 3.30 grams of an ideal gas at 150.0 ºC and 1.25 atmospheres pressure has a volume of 2.00 liters. What is the molar mass of the gas? The gas constant, R, is 0.0821 (L-atm)/(mol-K). • 0.0218 gram/mole • 16.2 grams/mole • 37.0 grams/mole • 45.8 grams/mole • 71.6 grams/mole

  14. 1994 MC #37 • A sample of 3.30 grams of an ideal gas at 150.0 ºC and 1.25 atmospheres pressure has a volume of 2.00 liters. What is the molar mass of the gas? The gas constant, R, is 0.0821 (L-atm)/(mol-K). • 0.0218 gram/mole • 16.2 grams/mole • 37.0 grams/mole • 45.8 grams/mole – 81% correct, very easy • 71.6 grams/mole

  15. Homework: 10.30, 10.34, 10.36, 10.42 • This week: • Thursday: Partial Pressures (10.6), Problem Set 3 Due • Friday: Kinetic-Molecular Theory (10.7) • Due Dates: • Volume-Temperature Behavior of Gases: 11/30 • States of Matter Exam: Monday, 12/6 • Molar Mass of Condensable Vapor: 12/8 • Problem Set 4: 12/10

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