Increasing the pressure of a gas (physics only, HT only)

⚡ Higher Tier

⚛ Physics Only

SPEC 8463 • TOPIC 3.06

Increasing the Pressure of a Gas

Boyle's law in everything but name. One equation on the formula sheet (pV = constant), two non-negotiable conditions, and one famous misconception. Master all three and HT gas calculations become free marks.

From Qualitative to Quantitative

In Lesson 3.05 we said: smaller V → particles hit walls more often per second → higher pressure. That was the qualitative story.

This lesson

Now we put a number on it. AQA gives Higher Tier students one equation that ties pressure and volume together, and questions that ask for the exact new pressure (or new volume) after compression or expansion. The equation is short. The conditions are strict. Both matter.

The Equation

p V = constant

SymbolMeaningUnits

ppressure of the gasPa (or kPa — see slide 12)

Vvolume occupied by the gasm³ (or cm³ — see slide 12)

constanta fixed value for this particular gas under stated conditionsPa m³ (or whatever combination of units p × V gives)

On the equation sheet ✓

pV = constant has been on the AQA Physics equation sheet from June 2024 onwards. You don't need to recall it — but you do need to recognise it as the one to reach for whenever a question changes the pressure or volume of a fixed mass of gas.

Two Non-Negotiable Conditions

pV = constant only works if both of these hold. If either is violated, the equation gives the wrong answer.

Condition 1 — Fixed mass

The amount of gas doesn't change between the two states. No gas added in, no gas leaks out. The same particles in both situations.

Condition 2 — Constant T

The temperature is the same in both states. Often written as "isothermal" or simply "at constant temperature".

Spot-check before solving

Read the question. Does it say "at constant temperature" or "isothermal"? Does the mass of gas change? If anything other than p and V changes, you cannot use pV = constant. (Heated gas at constant V uses different physics — see 3.05.)

The Two-State Form

Most exam questions describe a gas in state 1 (initial) and ask about state 2 (after compression or expansion). Since the value of "constant" is the same in both states:

p₁ V₁ = p₂ V₂

Use this form whenever you have before-and-after data. Three of the four quantities will be given; you rearrange to find the fourth.

Standard rearrangements

Find new pressure: p₂ = (p₁ × V₁) ÷ V₂

Find new volume: V₂ = (p₁ × V₁) ÷ p₂

Find original pressure: p₁ = (p₂ × V₂) ÷ V₁

Find original volume: V₁ = (p₂ × V₂) ÷ p₁

Inversely Proportional

pV = constant is the textbook example of an inverse proportionality: as one quantity doubles, the other halves. The graph of p against V is a hyperbola.

Reading the graph

Pick any two points on the curve. The product (pressure × volume) is identical. At point A, p₁ × V₁ might equal 8 Pa·m³. At point B, p₂ × V₂ also equals 8 Pa·m³. The constant doesn't depend on where you measure — it's a property of the gas at that temperature.

Worked Example 1 — Find New Pressure

A gas at a pressure of 100 kPa occupies a volume of 0.40 m³. The volume is reduced to 0.10 m³ at constant temperature.

Calculate the new pressure of the gas. 3 marks

E

Equation: p₁V₁ = p₂V₂, rearranged for p₂: p₂ = (p₁ × V₁) ÷ V₂

S

Substitute: p₂ = (100 × 0.40) ÷ 0.10 = 40 ÷ 0.10

U

Units: kilopascals (kPa) — same as input pressure

A

Answer: p₂ = 400 kPa (or 4.0 × 10⁵ Pa)

Sanity check

V shrank to one-quarter (0.40 → 0.10). Pressure should rise to four times (100 → 400). The inverse relationship works. If your answer doesn't match this kind of "ratio check", you've made an arithmetic slip — go back and recheck.

Worked Example 2 — Find New Volume

A fixed mass of gas at 120 kPa occupies 50 cm³ inside a sealed syringe. The gas is allowed to expand at constant temperature until its pressure falls to 80 kPa.

Calculate the new volume. 3 marks

E

Equation: p₁V₁ = p₂V₂, rearranged for V₂: V₂ = (p₁ × V₁) ÷ p₂

S

Substitute: V₂ = (120 × 50) ÷ 80 = 6000 ÷ 80

U

Units: cm³ — same as input volume

A

Answer: V₂ = 75 cm³

Notice

Pressure fell by factor 120/80 = 1.5. So volume rose by factor 1.5: 50 × 1.5 = 75 cm³. Same inverse logic — direction reversed because this is expansion, not compression.

Why pV = constant? (Particle Picture)

The equation isn't magic — it follows directly from the particle model in Lesson 3.05.

Step-by-step argument

At constant T, the mean kinetic energy of the particles is unchanged. So average particle speed is the same in both states.

Halving the volume halves the average distance a particle travels between wall collisions.

Same speed, half the distance → particles hit the walls twice as often per second.

Force per collision is unchanged (speed unchanged). With the same particles in half the volume, the rate of collisions per unit area of wall doubles.

Pressure (= force per unit area) doubles. p doubled while V halved → product pV is unchanged. That's why pV = constant.

Why "Constant T" Matters

If the temperature changes during compression, the mean KE of the particles changes too — and both the rate of collisions and the force per collision change. The simple inverse relationship breaks down.

Slow compression

If you compress slowly, heat leaks out to the surroundings. T stays roughly constant → pV = constant works.

Fast compression

If you compress fast, heat can't escape in time. Work done on the gas raises its internal energy → T rises → pV = constant fails. Pressure rises by more than the inverse relation predicts.

In an exam

Always check the wording. "At constant temperature" or "isothermal" → safe to use pV = constant. If the question implies temperature change (e.g. "the gas heats up", "no time for heat to escape") → the equation doesn't apply directly.

Work Done On a Gas — Bike Pump

This is the qualitative spec content that often appears as a 2- or 3-mark explanation question.

The physics chain

You apply a force F on the piston, moving it a distance d → work done W = F × d is transferred to the gas. The gas's internal energy increases. Since internal energy = KE + PE of particles, and PE between gas particles is negligible, the mean KE rises → the gas is at a higher temperature. You can feel this on a real bike pump.

Examiner Alert

Unit consistency on each side

p₁V₁ = p₂V₂ works in any consistent unit system. You can use kPa on both sides, or Pa on both sides — but you can't mix kPa on the left with Pa on the right. Same for volume: cm³ ↔ cm³ on both sides, or m³ ↔ m³ on both sides. If pressures are in kPa, the answer comes out in kPa.

Did you check the conditions?

Before applying pV = constant, the question must imply (or state) (1) fixed mass and (2) constant temperature. If gas is added/released, the equation fails. If the gas heats up during compression (and the question hints at this), the equation underestimates the final pressure.

Reversed-logic error

Some students write "when you compress a gas, the pressure decreases". This is the opposite of the truth. Smaller V → higher P. The inverse proportionality means as one quantity falls, the other rises — not both falling together.

Weak vs Strong Answer

"A gas in a sealed syringe is slowly pushed by a piston. The temperature stays constant. Explain, in terms of the particle model, why the pressure rises."

Weak — 1 mark

"The volume decreases so the pressure increases — they are inversely proportional, so pV stays the same."

Strong — 4 marks

"At constant temperature the mean KE of the particles is unchanged, so their average speed stays the same. In a smaller volume, particles travel less distance between wall collisions, so they hit the walls more often per second. The force per collision is unchanged, but the rate of collisions increases. The total force on the walls rises → pressure rises."

What changed

The weak answer is just the equation in words. The strong answer is the particle mechanism: same KE → same speed → less distance → more collisions per second → higher P. The mark scheme rewards mechanism, not just outcomes.

Common Errors

Students Often WriteCorrect VersionWhy It Matters

Uses pV = constant when T changesOnly apply if T is constant (isothermal) and mass is fixedEquation misapplication

"When you compress a gas, pressure decreases"Smaller V → higher P (inverse proportion)Reversed logic

Pressure in kPa on left, Pa on rightUse the same unit on both sides — same for VWrong-magnitude answer

Forgets the constant-T qualifier in the explanationAlways state "at constant T, mean KE is unchanged"Loses the M1 mark

"Particles vibrate more" when explaining compressionParticles hit walls more often per second (same speed, less distance)Wrong mechanism — vibration is solids

"Greater force per collision" cited for compressionForce per collision is unchanged at constant T; only the rate of collisions risesConfusing 3.05 (T effect) with 3.06 (V effect)

Practice Question

A weather balloon contains 0.50 m³ of helium gas at ground level, where the atmospheric pressure is 100 kPa. The balloon rises high into the atmosphere. At its peak altitude the pressure on the balloon has fallen to 25 kPa. The temperature of the helium stays approximately constant during the ascent.

Calculate the new volume of the helium gas inside the balloon. 4 marks

E

Equation: p₁V₁ = p₂V₂, rearranged for V₂: V₂ = (p₁ × V₁) ÷ p₂

S

Substitute: V₂ = (100 × 0.50) ÷ 25 = 50 ÷ 25

U

Units: m³ — same as input volume

A

Answer: V₂ = 2.0 m³

Where You See This In Real Life

pV = constant isn't a textbook abstraction. It's everywhere:

Bike pump

Compression → higher P (and warmer barrel from work done)

Syringe

Pull the plunger — V rises, P inside falls below atmospheric → liquid drawn in

Diving

Going deeper compresses any flexible air pocket (e.g. middle ear) — divers must "equalise" by letting more air in

Module 3 closing thought

This lesson ends Module 3 — Particle Model of Matter. From density (3.01) through changes of state (3.02-3.04) to gases (3.05-3.06), the same particle picture has explained every behaviour: solids, liquids, gases, melting, boiling, pressure changes. One model, the entire module.

Lesson Summary

The Equation (Equation Sheet)

pV = constant , or equivalently p₁V₁ = p₂V₂ for before-and-after problems.

Valid only at constant temperature, for a fixed mass of gas.

Inverse proportionality: halve V → double P. Graph of p vs V is a hyperbola.

The Calculation Routine

1

Check conditions : constant T? Fixed mass? If yes, proceed.

2

List p₁, V₁, p₂, V₂ — three known, one unknown.

3

Rearrange p₁V₁ = p₂V₂ to get the unknown on its own.

4

Substitute — same units on both sides.

5

Answer with units; do a ratio sanity check.

Particle-level explanation

For "explain why P rises when V falls at constant T": (1) constant T → mean KE unchanged → same particle speeds. (2) Smaller V → less distance to walls → more collisions per second. (3) Force per collision unchanged. (4) Total wall force ↑ → P ↑.