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9702 · 15.2

Equation of state — practice questions

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

Worked example 1

A cylinder contains 0.50 mol of an ideal gas at a pressure of 2.0 × 10⁵ Pa and a temperature of 27 °C. Calculate the volume of the gas.

Show solution outline
  1. Identify knowns and unknowns: n = 0.50 mol P = 2.0 × 10⁵ Pa T_C = 27 °C R = 8.31 J mol⁻¹ K⁻¹ V = ?
  2. Convert temperature to Kelvin: T_K = T_C + 273 = 27 + 273 = 300 K
  3. Select and rearrange the Ideal Gas Equation (PV = nRT) to solve for V: V=nRTPV = \frac{nRT}{P}
  4. Substitute values and calculate: V=(0.50 mol)×(8.31 J mol−1 K−1)×(300 K)2.0×105 PaV = \frac{(0.50 \text{ mol}) \times (8.31 \text{ J mol}^{-1}\text{ K}^{-1}) \times (300 \text{ K})}{2.0 \times 10^{5} \text{ Pa}} V=1246.52.0×105V = \frac{1246.5}{2.0 \times 10^{5}} V=0.0062325 m3V = 0.0062325 \text{ m}^{3}
  5. State the final answer with appropriate significant figures: V=6.2×10−3 m3V = 6.2 \times 10^{-3} \text{ m}^{3}

Worked example 2

A sealed container of volume 1.5 × 10⁻² m³ contains an ideal gas at a pressure of 3.0 × 10⁵ Pa and a temperature of 350 K. Calculate the number of gas molecules in the container. (Boltzmann constant, k = 1.38 × 10⁻²³ J K⁻¹)

Show solution outline
  1. Identify knowns and unknowns: V = 1.5 × 10⁻² m³ P = 3.0 × 10⁵ Pa T = 350 K k = 1.38 × 10⁻²³ J K⁻¹ N = ?
  2. Select the appropriate form of the Ideal Gas Equation: Since the question asks for the number of molecules (N) and provides the Boltzmann constant (k), we use the microscopic form: PV = NkT.
  3. Rearrange the equation to solve for N: N=PVkTN = \frac{PV}{kT}
  4. Substitute the given values into the equation: N=(3.0×105 Pa)×(1.5×10−2 m3)(1.38×10−23 J K−1)×(350 K)N = \frac{(3.0 \times 10^{5} \text{ Pa}) \times (1.5 \times 10^{-2} \text{ m}^{3})}{(1.38 \times 10^{-23} \text{ J K}^{-1}) \times (350 \text{ K})}
  5. Calculate the result: N=45004.83×10−21N = \frac{4500}{4.83 \times 10^{-21}} N=9.3167...×1023N = 9.3167... \times 10^{23}
  6. State the final answer with appropriate significant figures (2 s.f. from the input values): N = 9.3 × 10²³ molecules
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