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Chemistry 20

Chemistry 20 · Worksheets

Gases

Ten questions of mixed difficulty, covering Gases. Print it, or work through it on screen — the answer key starts on its own page.

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Gases

Chemistry 20 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. A 2.0 g sample of a gas occupies 1.24 L at SATP. Its molar mass is

    1. a) 62 g/mol
    2. b) 40 g/mol
    3. c) 24.8 g/mol
    4. d) 2.0 g/mol

  2. Absolute zero is

    1. a) −100 °C
    2. b) 0 °C
    3. c) −273.15 °C
    4. d) −373.15 °C

  3. A gas occupying 2.0 L at 100 kPa is compressed to 1.0 L at constant temperature. Its new pressure is

    1. a) 200 kPa
    2. b) 50 kPa
    3. c) 100 kPa
    4. d) 400 kPa

  4. A balloon holds 6.0 L at 300 K. Heated to 600 K at constant pressure, it holds

    1. a) 3.0 L
    2. b) 24.0 L
    3. c) 12.0 L
    4. d) 6.0 L

  5. A gas occupies 3.0 L at 200 kPa and 300 K. At 100 kPa and 400 K it occupies

    1. a) 4.5 L
    2. b) 6.0 L
    3. c) 8.0 L
    4. d) 1.5 L

  6. The amount of gas in a 5.0 L container at 150 kPa and 300 K is about

    1. a) 3.0 mol
    2. b) 0.30 mol
    3. c) 0.030 mol
    4. d) 1.2 mol

  7. Standard atmospheric pressure is

    1. a) 100.0 kPa
    2. b) 273.15 kPa
    3. c) 1.0 kPa
    4. d) 101.325 kPa

  8. Boyle's law states that, at constant temperature, the pressure of a fixed amount of gas is

    1. a) directly proportional to its volume
    2. b) inversely proportional to its volume
    3. c) equal to its volume
    4. d) independent of its volume

  9. A temperature of 25 °C is equal to

    1. a) 25 K
    2. b) 298 K
    3. c) 273 K
    4. d) 248 K

  10. The molar volume of an ideal gas at SATP is

    1. a) 101.3 L/mol
    2. b) 22.4 L/mol
    3. c) 24.8 mL/mol
    4. d) 24.8 L/mol

Answer key · Gases

Chemistry 20 · maddyhelps.com

  1. b) 40 g/mol — First the amount: n = 1.24/24.8 = 0.050 mol. Then M = m/n = 2.0/0.050 = 40 g/mol. Molar volume is the bridge from a measured volume to an amount whenever the gas is at SATP.
  2. c) −273.15 °C — It is the temperature at which particle motion would stop, and it is the zero of the Kelvin scale. Nothing colder exists, which is why gas law calculations have to use kelvins.
  3. a) 200 kPa — P₁V₁ = P₂V₂, so P₂ = (100)(2.0)/1.0 = 200 kPa. Halving the volume doubles the pressure — check that the answer moves the way the physics says before writing it down.
  4. c) 12.0 L — Charles's law: V₁/T₁ = V₂/T₂, so doubling the absolute temperature doubles the volume. This only works in kelvins — doubling 27 °C to 54 °C would not double the volume at all.
  5. c) 8.0 L — The combined gas law gives V₂ = P₁V₁T₂/(T₁P₂) = (200)(3.0)(400)/[(300)(100)] = 8.0 L. Lower pressure and higher temperature both expand the gas, so an answer larger than 3.0 L is expected.
  6. b) 0.30 mol — From PV = nRT, n = PV/(RT) = (150)(5.0)/[(8.314)(300)] ≈ 0.30 mol. The units work out because R is 8.314 kPa·L/(mol·K), so pressure in kilopascals and volume in litres are already correct.
  7. d) 101.325 kPa — One standard atmosphere is 101.325 kPa, which is also 760 mm Hg. Note that SATP uses a round 100 kPa instead, so which standard a question is using matters.
  8. b) inversely proportional to its volume — Squeeze a gas into half the space and the particles hit the walls twice as often, so the pressure doubles: P₁V₁ = P₂V₂. Inverse means the product stays constant, not the ratio.
  9. b) 298 K — Add 273: 25 + 273 = 298 K. Kelvin has no degree symbol and never goes negative. Using Celsius in a gas law is the single most common source of a wrong answer in this unit.
  10. d) 24.8 L/mol — SATP is 25 °C and 100 kPa, giving 24.8 L/mol. The familiar 22.4 L/mol is the value at STP, 0 °C and 101.325 kPa. Alberta questions usually specify which, and the two are not interchangeable.