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9702 · 15.3

Kinetic theory of gases — practice questions

Practice and worked examples for 9702 Kinetic theory of gases. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

An ideal gas of oxygen molecules (molar mass 32.0 g/mol) is at a temperature of 27.0 °C. Calculate the mean square speed (c2‾\overline{c^2}) of the oxygen molecules.

Given: Molar gas constant R=8.31 J mol−1 K−1R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} Avogadro constant NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1} Boltzmann constant k=1.38×10−23 J K−1k = 1.38 \times 10^{-23} \text{ J K}^{-1}

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  1. Convert temperature to Kelvin: T=27.0 °C+273.15=300.15 KT = 27.0 \text{ °C} + 273.15 = 300.15 \text{ K}

  2. Calculate the mass of a single oxygen molecule (mm): Molar mass M=32.0 g/mol=32.0×10−3 kg/molM = 32.0 \text{ g/mol} = 32.0 \times 10^{-3} \text{ kg/mol} m=MNA=32.0×10−3 kg mol−16.02×1023 mol−1≈5.316×10−26 kgm = \frac{M}{N_A} = \frac{32.0 \times 10^{-3} \text{ kg mol}^{-1}}{6.02 \times 10^{23} \text{ mol}^{-1}} \approx 5.316 \times 10^{-26} \text{ kg}

  3. Relate average kinetic energy to temperature and mean square speed: We know that 12mc2‾=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT.

  4. Solve for c2‾\overline{c^2}: mc2‾=3kTm\overline{c^2} = 3kT c2‾=3kTm\overline{c^2} = \frac{3kT}{m} c2‾=3×(1.38×10−23 J K−1)×(300.15 K)5.316×10−26 kg\overline{c^2} = \frac{3 \times (1.38 \times 10^{-23} \text{ J K}^{-1}) \times (300.15 \text{ K})}{5.316 \times 10^{-26} \text{ kg}} c2‾≈2.339×105 m2 s−2\overline{c^2} \approx 2.339 \times 10^5 \text{ m}^2\text{ s}^{-2}

    Therefore, the mean square speed of the oxygen molecules is approximately 2.34×105 m2 s−22.34 \times 10^5 \text{ m}^2\text{ s}^{-2} (to 3 s.f.).

Worked example 2

A sealed container of volume 2.0×10−3 m32.0 \times 10^{-3} \text{ m}^3 contains 5.0×10225.0 \times 10^{22} molecules of an ideal gas. The root-mean-square speed of the molecules is 480 m/s. The mass of one molecule is 4.7×10−264.7 \times 10^{-26} kg. Calculate the pressure of the gas.

Show solution outline
  1. Identify the given values: Volume, V=2.0×10−3 m3V = 2.0 \times 10^{-3} \text{ m}^3 Number of molecules, N=5.0×1022N = 5.0 \times 10^{22} Root-mean-square speed, crms=480 m/sc_{rms} = 480 \text{ m/s} Mass of one molecule, m=4.7×10−26 kgm = 4.7 \times 10^{-26} \text{ kg}
  2. State the relevant formula: The kinetic theory pressure equation is P=13Nmc2‾VP = \frac{1}{3}\frac{Nm\overline{c^2}}{V}.
  3. Calculate the mean square speed (c2‾\overline{c^2}): We are given the root-mean-square speed, crmsc_{rms}. The relationship is c2‾=(crms)2\overline{c^2} = (c_{rms})^2. c2‾=(480 m/s)2=230400 m2s−2\overline{c^2} = (480 \text{ m/s})^2 = 230400 \text{ m}^2\text{s}^{-2}.
  4. Substitute the values into the pressure equation: P=13×(5.0×1022)×(4.7×10−26 kg)×(230400 m2s−2)2.0×10−3 m3P = \frac{1}{3} \times \frac{(5.0 \times 10^{22}) \times (4.7 \times 10^{-26} \text{ kg}) \times (230400 \text{ m}^2\text{s}^{-2})}{2.0 \times 10^{-3} \text{ m}^3}
  5. Calculate the numerator: Numerator = (5.0×1022)×(4.7×10−26)×(230400)=0.54144(5.0 \times 10^{22}) \times (4.7 \times 10^{-26}) \times (230400) = 0.54144
  6. Calculate the pressure: P=13×0.541442.0×10−3P = \frac{1}{3} \times \frac{0.54144}{2.0 \times 10^{-3}} P=13×270.72=90240 PaP = \frac{1}{3} \times 270.72 = 90240 \text{ Pa}
  7. State the final answer with appropriate significant figures: The input values are given to 2 significant figures, so the answer should be rounded accordingly. P≈9.0×104 PaP \approx 9.0 \times 10^4 \text{ Pa} (or 90 kPa).
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