· H2 Physics — Ten-Year Series, worked ← All 7 questions

2022 A-Level · Paper 2 · Question 4

Pressure out of collisions, and a check that molecules really are far apart

Thermal physics · ideal gases 8 marks Kinetic theory pV = NkT

Eight marks on the kinetic model of a gas. How molecules moving about produce a steady pressure on the walls; what the model assumes about their separation; and then a test of that assumption with real numbers, estimating the average spacing in a low-pressure gas and comparing it with the size of a molecule. Below is the full solution, and beside it the molecules at work and the spacing drawn to scale.

The model, live

MoleculesForce on the right-hand wall

Part (a)

3 marks

Explain how molecular movement causes the pressure exerted by a gas on the walls of its container.

Approach

Three links, one per mark: a collision changes a molecule's momentum, so a force acts on it; by Newton's third law it pushes back on the wall; and the huge number of collisions gives a steady average force per unit area.

The molecules move randomly and collide with the walls. In each collision the molecule's momentum changes (it rebounds), so the wall exerts a force on it, and by Newton's third law the molecule exerts an equal and opposite force on the wall. The very large number of collisions per second produces a steady average force on the walls, and pressure is this force per unit area.

Watch

"Molecules hit the walls, causing pressure" earns little. The marks are for the change in momentum, the force it implies, and the averaging over many collisions.

Part (b)(i)

1 mark

State the assumption of kinetic theory about the separation of the particles.

The average separation of the particles is very large compared with the size of the particles (so the total volume of the particles is negligible compared with the volume of the gas).

Part (b)(ii)

3 marks

A sealed vessel of volume 3.2×10−4 m33.2\times10^{-4}\,\text{m}^3 holds an ideal gas at 180 Pa180\,\text{Pa} and 298 K298\,\text{K}. Estimate the average distance between the particles.

Approach

Find the number of particles from pV=NkTpV = NkT, share the volume out equally so that each particle has its own small cube, and take the side of that cube as the average spacing.

N=pVkT=(180)(3.2×10−4)(1.38×10−23)(298)=1.40×1019N = \frac{pV}{kT} = \frac{(180)(3.2\times10^{-4})}{(1.38\times10^{-23})(298)} = 1.40\times10^{19}
volume per particle=3.2×10−41.40×1019=2.28×10−23 m3\text{volume per particle} = \frac{3.2\times10^{-4}}{1.40\times10^{19}} = 2.28\times10^{-23}\,\text{m}^3
average distance≈(2.28×10−23)1/3=2.8×10−8 m\text{average distance} \approx \left(2.28\times10^{-23}\right)^{1/3} = 2.8\times10^{-8}\,\text{m}

average distance =2.8×10−8 m= 2.8\times10^{-8}\,\text{m}

Note

The volume per particle is simply VN=kTp\dfrac{V}{N} = \dfrac{kT}{p}. The vessel's volume cancels out, so the spacing depends only on the pressure and temperature.

Part (b)(iii)

1 mark

A gas particle has a diameter of about 1×10−10 m1\times10^{-10}\,\text{m}. Use (b)(ii) to explain whether the assumption in (b)(i) is valid.

2.8×10−81×10−10≈280\frac{2.8\times10^{-8}}{1\times10^{-10}} \approx 280

Valid. The average separation, 2.8×10−8 m2.8\times10^{-8}\,\text{m}, is about 300 times the diameter of a particle, so the separation is very large compared with the particle size.

What this question was really testing

  1. Pressure from momentum. Change of momentum, force on the molecule, equal force on the wall, averaged over many collisions.
  2. Stating a model assumption precisely.
  3. Estimating with pV=NkTpV = NkT. Volume per particle, then a cube root.
  4. Testing a model against numbers. A ratio of a few hundred is "very large".

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