Ideal gas molecules: Kelvin scale, p/T and pV relationships – IGCSE Physics exam questions with mark scheme

1Grade 4–56 marks

A sealed metal can contains air at a pressure of 150 kPa and a temperature of 27 °C. The can is heated to 127 °C. The volume of the can does not change.

(a) Calculate the new pressure of the air in the can. [3]

(b) Explain, in terms of particles, why the pressure of the air increases. [3]

(a) converts both temperatures to kelvin: 300 K and 400 K (allow 300.15 K and 400.15 K)(1)
(a) substitution into $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$: $\dfrac{150}{300} = \dfrac{p_2}{400}$(1)
(a) 200 kPa (allow 200 000 Pa); Celsius used throughout (answer 706 kPa) scores 0 for this question part(1)
(b) particles move faster / have more kinetic energy(1)
(b) particles collide with the walls more often / more frequently / more collisions per second; ignore 'more collisions' with no reference to time(1)
(b) each collision exerts a greater force (on the walls) so pressure (= force ÷ area) increases; accept larger change in momentum per collision(1)
Answer: (a) 200 kPa (b) faster particles collide more often and harder with the walls
Examiner tip: Forgetting to convert to kelvin is the most common way to lose marks; the gas laws only work with absolute temperature.

2Grade 6–75 marks

An air bubble is released at the bottom of a lake, 15 m below the surface. The volume of the bubble at the bottom of the lake is 2.0 cm³.

Atmospheric pressure = $1.0 \times 10^{5}$ Pa, density of water = 1000 kg/m³, $g = 10$ N/kg.

(a) Show that the total pressure at the bottom of the lake is $2.5 \times 10^{5}$ Pa. [2]

(b) Calculate the volume of the bubble when it reaches the surface. [2]

(a) pressure due to water = 15 × 1000 × 10 = 150 000 Pa(1)
(a) adds atmospheric pressure: 150 000 + 100 000 = 250 000 Pa(1)
(b) substitution: $2.5 \times 10^{5} \times 2.0 = 1.0 \times 10^{5} \times V_2$(1)
(b) 5.0 cm³ (allow 5 cm³)(1)
(c) temperature (of the air in the bubble) stays constant; accept mass / amount of gas in the bubble stays constant (no gas dissolves or escapes)(1)
Answer: (a) 15 × 1000 × 10 = 1.5 × 10⁵ Pa; 1.5 × 10⁵ + 1.0 × 10⁵ = 2.5 × 10⁵ Pa (b) 5.0 cm³ (c) the temperature of the gas stays constant
Examiner tip: In 'show that' questions write out every step, including adding atmospheric pressure; using only the water pressure (1.5 × 10⁵ Pa) in (b) gives the wrong volume.

3Grade 8–95 marks

A fixed mass of gas is kept at constant volume. A student measures its pressure at two temperatures.

Temperature / °C	Pressure / kPa
0	100.0
100	136.6

(a) The pressure changes linearly with temperature. Use the data to determine the temperature, in °C, at which the pressure of the gas would become zero. [3]

(b) Explain, in terms of the particles, what is meant by absolute zero. [2]

(a) increase in pressure per °C = 36.6 ÷ 100 = 0.366 (kPa/°C); allow any valid method, e.g. extrapolating a graph or ratio 100 ÷ 36.6 × 100(1)
(a) temperature change needed = 100 ÷ 0.366 = 273 (°C)(1)
(a) −273 °C; accept −273.2 °C; if a graph is drawn and extrapolated accept answers in the range −270 °C to −276 °C; minus sign needed for this mark. The question says 'use the data': −273 °C quoted from memory with no use of the data scores this mark only (max 1)(1)
(b) the lowest possible temperature / 0 K(1)
(b) particles have (the minimum possible or) no kinetic energy / particles stop moving; ignore 'exert no pressure'(1)
Answer: (a) −273 °C (b) the lowest possible temperature (0 K), at which the particles have no kinetic energy / stop moving
Examiner tip: The final mark needs the negative sign: the pressure falls to zero 273 °C below 0 °C.

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