Connect gas collisions to wall force per unit area
Translate repeated microscopic impacts into an average pressure on a named wall patch.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Explain moving gas particles colliding with surfaces.
- Recall p = F/A and pressure units from section 1.8.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation relates average normal wall force, patch area and pressure?
What you will learn
- Explain collision pressure through wall forces per unit area.
- Calculate an average pressure using the actual patch and supplied total normal force.
Repeated impacts create an average wall force
During a collision a gas particle exerts a force on the wall. Many repeated impacts over the chosen wall patch give a total normal force averaged over a suitable time interval. Pressure on that patch is this average force per unit area, p = F/A. The macroscopic pressure does not require every particle to strike simultaneously or each individual impact to have an identical force.
The force is on the named wall surface. Opposite walls experience forces in opposite outward directions, yet both can have positive pressure magnitudes. A zero resultant force on the container as a whole does not imply zero pressure on every wall. Use the pressure account at the appropriate surface rather than cancelling all surface forces before calculating each local pressure.
Compare equal areas under controlled conditions
Heating a fixed gas sample at constant volume increases average particle kinetic motion. The impacts are more frequent and stronger on average, increasing the average force per equal wall area. Compressing a fixed gas sample at constant temperature increases impact frequency per equal area without requiring increased average particle speed. Both descriptions need their supplied control conditions.
For a stated average normal force of 6.0 N on a 0.0020 m² patch, pressure is 3000 Pa. Do not put the whole vessel's external area under the force assigned only to this patch. Likewise, increasing the chosen area in a uniform-pressure region increases its total force proportionally; it does not prove a new lower gas pressure merely by choosing a larger patch.
Sources for this lesson
Pause and explain
Equal uniform gas pressure acts on two patches, one with twice the area. How do their total average forces compare?
Worked example
A gas produces an average normal force of 6.0 N on a wall patch of area 0.0020 m². Calculate pressure and the force on twice the area at the same uniform pressure.
Show the worked solution
- Use the patch force and area: p = 6.0/0.0020 = 3000 Pa.
- Twice the area is 0.0040 m².
- At the same pressure F = 3000 × 0.0040 = 12 N.
Answer 3000 Pa; 12 N on twice the area at the same pressure.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate pressure for average normal 6.0 N on 0.0020 m² and relate that force to the collision model.
Give me a hint
Divide the patch force by its area and identify repeated impacts.
Compare my reasoning
- Repeated gas impacts give the average normal patch force.
- p = F/A = 6.0/0.0020.
- Pressure is 3000 N/m² = 3000 Pa.
3000 Pa from repeated impacts on the named patch.
Look for these in your work
- Calculate with the actual area.
- Connect the force to collisions.
A uniform gas pressure is 4.0 kPa. Find average normal forces on patches of 10 cm² and 30 cm².
Give me a hint
Convert squared areas before using F = pA.
Compare my reasoning
- 4.0 kPa = 4000 Pa; areas are 0.0010 and 0.0030 m².
- Forces are 4000 × 0.0010 = 4.0 N and 4000 × 0.0030 = 12 N.
- Three times the area carries three times the force at the same pressure.
4.0 N and 12 N.
Look for these in your work
- Convert the area units.
- Distinguish total force from force per area.
Opposite walls have equal and opposite gas forces, and a student concludes that gas pressure is zero. Explain the error using a named wall patch.
Give me a hint
Balance of the whole container does not remove each wall's collision force.
Compare my reasoning
- Overall forces can balance in opposite directions.
- Each named patch still receives repeated particle impacts.
- Its positive normal force per area gives pressure even when the container's total resultant is zero.
Overall force balance does not imply zero pressure on each wall.
Look for these in your work
- Keep the system and surface distinct.
- Use local force per area.
Common mistakes
- Use the force assigned to the named patch.
- Opposing wall forces can cancel overall without eliminating pressure.
- At uniform pressure a larger patch carries proportionally more total force.
What can you explain now?
Does zero overall resultant on a sealed stationary container imply zero gas pressure?
Compare with the explanation
No.
Forces on opposite walls can balance overall while each wall experiences collision pressure.
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Make it stick
Explain microscopic wall forces, calculate a patch pressure and separate balanced overall forces from local positive pressure.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.