KCSE 2025 Physics P1 Q16 — Gas Laws: Boyle's, Diffusion & Charles's Law
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The Question
“This is a Section B question on the behaviour of gases. (a)(i) Name the apparatus a student would need to verify Boyle's law, and (ii) state the physical quantity that must be kept constant during the experiment. (b) In a diffusion tube, two gases A and B start from opposite ends and meet where a white deposit forms; the deposit forms nearer gas B. State which gas diffuses faster, and suggest two ways to increase the rate of diffusion of the gases in the tube. (c)(i) Use the kinetic theory to explain Charles's law. (c)(ii) A gas has a volume of 20.0 cm³ at 19 °C. Calculate its new volume when it is heated to 24 °C at constant pressure.”
(a)(i) Choose apparatus to verify Boyle's law
Boyle's law relates the pressure and volume of a fixed mass of gas, so you must be able to read both. Use a pressure gauge, such as a Bourdon gauge, to measure the pressure, and a calibrated cylinder or syringe to read the volume of the trapped gas as you change it.
(a)(ii) Identify the quantity kept constant
Boyle's law only holds for a fixed mass of gas at constant temperature. Since you are deliberately varying pressure and volume, the quantity that must be held constant throughout the experiment is the temperature.
(b) Decide which gas diffuses faster
The two gases travel toward each other and react where they meet, forming a white deposit. The deposit forms nearer gas B, which means gas A has travelled the greater distance in the same time. Covering more of the tube means gas A moved faster, so gas A diffuses faster than gas B.
(b) Increase the rate of diffusion
Diffusion speeds up when the molecules move faster or meet less resistance. You can increase the rate of diffusion of the gases in the tube either by raising the temperature, which gives the molecules more kinetic energy, or by reducing the pressure, which spreads the molecules out so they collide less often.
(c)(i) Explain Charles's law using the kinetic theory
Raising the temperature makes the gas molecules move faster and carry more kinetic energy, so they would strike the walls harder and more often. To keep the pressure constant, the gas must expand: the larger volume means the number of collisions per unit area each second stays the same. Hence, at constant pressure, the volume rises as the temperature rises — which is Charles's law.
(c)(ii) Write Charles's law and convert to kelvin
At constant pressure, Charles's law states that the volume divided by the absolute temperature is constant, so V₁ over T₁ equals V₂ over T₂. Gas-law temperatures must be in kelvin, so add 273 to each Celsius reading: 19 °C becomes 292 K and 24 °C becomes 297 K.
(c)(ii) Substitute and solve for the new volume
Make V₂ the subject by multiplying the original volume by the ratio of the new temperature to the old temperature. Putting in the numbers gives the expanded volume of the gas at 24 °C.
Final Result
The new volume of the gas at 24 °C is 20.34 cm³. (Boyle's law needs a pressure gauge and a calibrated syringe with temperature held constant; gas A diffuses faster; and diffusion can be sped up by raising the temperature or lowering the pressure.)
Why this method works
Charles's law works because temperature is a measure of the average kinetic energy of the molecules. Heat the gas and the molecules speed up, so to stop the pressure from rising the gas simply takes up more room — volume and absolute temperature stay in step. The critical detail is that the law is written in terms of absolute (kelvin) temperature, not Celsius: only kelvin starts from true zero motion, so ratios like 297/292 are meaningful. Using 24/19 instead would give a wildly wrong answer because the Celsius scale has an arbitrary zero. The same kinetic picture explains diffusion: hotter, faster molecules spread out more quickly, and lighter molecules move faster than heavier ones, which is why gas A outruns gas B.
The gas was heated only slightly (5 K out of nearly 300 K), so the volume should rise by only about 1.7 % — from 20.0 cm³ to 20.34 cm³, exactly what the calculation gives.