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9702 · 14.2

Temperature scales — practice questions

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

Worked example 1

A gas is cooled from 27 °C to -10 °C. Calculate this temperature change in Kelvin.

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  1. First, find the temperature change in Celsius: ΔC=Initial TempFinal Temp=27°C(10°C)=37°C\Delta C = \text{Initial Temp} - \text{Final Temp} = 27 °C - (-10 °C) = 37 °C.
  2. Since a change of 1 Kelvin is equal to a change of 1 degree Celsius (ΔK=Δ°C\Delta K = \Delta °C), the temperature change in Kelvin is ΔK=37 K\Delta K = 37 \text{ K}.

Worked example 2

A sealed container holds 0.50 moles of an ideal gas at a pressure of 1.5 x 10^5 Pa. The volume of the container is 0.012 m³. Calculate the temperature of the gas in degrees Celsius (°C). (The ideal gas constant, R, is 8.31 J K⁻¹ mol⁻¹).

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  1. State the Ideal Gas Law: The relationship between pressure (p), volume (V), number of moles (n), and temperature (T) is given by the ideal gas law: pV=nRTpV = nRT. We need to find the temperature, T.
  2. Rearrange the formula for Temperature (T): T=pVnRT = \frac{pV}{nR}.
  3. Substitute the given values:
    • p=1.5×105 Pap = 1.5 \times 10^5 \text{ Pa}
    • V=0.012 m3V = 0.012 \text{ m}^3
    • n=0.50 moln = 0.50 \text{ mol}
    • R=8.31 J K1 mol1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} T=(1.5×105 Pa)×(0.012 m3)(0.50 mol)×(8.31 J K1 mol1)T = \frac{(1.5 \times 10^5 \text{ Pa}) \times (0.012 \text{ m}^3)}{(0.50 \text{ mol}) \times (8.31 \text{ J K}^{-1} \text{ mol}^{-1})}.
  4. Calculate the temperature in Kelvin: T=18004.155433.2 KT = \frac{1800}{4.155} \approx 433.2 \text{ K}. Note: The temperature calculated from the ideal gas law is always in Kelvin.
  5. Convert the temperature from Kelvin to Celsius: The question asks for the temperature in degrees Celsius. We use the conversion formula C=K273.15C = K - 273.15. C=433.2273.15=160.05°CC = 433.2 - 273.15 = 160.05 °C.
  6. Final Answer: The temperature of the gas is approximately 160 °C (to 3 significant figures).