Describe the motion of molecules in a gas, according to the kinetic theory of gases.
- (a)Describe the motion of molecules in a gas, according to the kinetic theory of gases.[2]
- (b)Describe what is observed when viewing Brownian motion that provides evidence for your answer in (a).[2]
- (c)(i)Calculate the root-mean-square (r.m.s.) speed of an atom of helium gas for a temperature of 27 °C.[3]
- (c)(ii)Using your answer in (i), calculate the r.m.s. speed of the atoms at 177°C.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] random/haphazard
- [B1] constant velocity or speed in a straight line between collisions
- [B1] or distribution of speeds/different directions
- [M1] (small) specks of light/bright specks/pollen grains/dust particles/smoke particles
- [A1] moving haphazardly/randomly/jerky/in a zigzag fashion
- [C1] pV = 1/3 Nm<c²>
- [C1] 1.05 × 10⁵ × 0.0240 = 1/3 × 4.00 × 10⁻³ × <c²>
- [C1] <c²> = 1.89 × 10⁶
- [(C1)] or 1/2 m<c²> = (3/2) kT
- [(C1)] 0.5 × (4.00 × 10⁻³/6.02 × 10²³) × <c²> = 1.5 × 1.38 × 10⁻²³ × 300
- [(C1)] <c²> = 1.87 x 10⁶
- [(C1)] or nRT = 1/3 Nm<c²>
- [(C1)] 1.00 × 8.31 × 300 = 1/3 x 4.00 × 10⁻³ × <c²>
- [(C1)] <c²> = 1.87 x 10⁶
- [A1] c_r.m.s. = 1.37 × 10³ m s⁻¹
- [C1] <c²> ∝ T
- [C1] <c²> at 177 °C = 1.89 × 10⁶ × (450 / 300)
- [A1] c_r.m.s. at 177 °C = 1.68 × 10³ m s⁻¹
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