CP14 – Boyle's Law | Edexcel A Level Physics

Theory — Boyle's Law

At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.

Boyle's Law

P ∝ 1/V (at constant T) PV = constant → P₁V₁ = P₂V₂

P = pressure (Pa or kPa) · V = volume (m³ or cm³) · T = temperature (must stay constant)

Boyle's Law follows from the kinetic theory of gases. If you halve the volume, the same number of molecules occupy half the space — they hit the walls twice as often, doubling the pressure.

Linearising the Data

A plot of P vs V gives a hyperbola — hard to analyse. Linearising gives two useful graphs:

Plot P vs 1/V → straight line through origin, gradient = PV = constant Plot PV vs P → horizontal straight line, y-value = constant

Both confirm Boyle's Law. The P vs 1/V plot is more commonly used in A-level.

The Boyle's Law Apparatus

A closed tube traps a column of air above oil. The oil is connected to a pump. Pumping increases or decreases the oil pressure, which is transmitted to the trapped air. A pressure gauge reads the absolute pressure P. A millimetre scale alongside the tube measures the length L of the air column. Since the tube has uniform cross-section of area A:

V = A × L

In practice, A = π(d/2)² where d is the internal diameter of the tube. For analysis, you can use L directly as a proxy for V (since A is constant), or multiply by A to get the true volume.

Temperature must be kept constant — wait at least 30 seconds after each pressure change before reading, so the gas can return to room temperature after adiabatic compression/expansion.

Pressure Units

Pa — Pascals (SI unit): 1 Pa = 1 N m⁻²
kPa — kilopascals: 1 kPa = 1000 Pa
atm — standard atmosphere: 1 atm = 101.325 kPa
Gauge pressure — pressure above atmospheric. Absolute pressure = gauge + atmospheric (≈ 101 kPa)

Always use absolute pressure in Boyle's Law calculations, not gauge pressure.

Procedure

Using Boyle's Law apparatus with oil manometer and pressure gauge.

Equipment

Boyle's Law apparatus (sealed tube + oil reservoir + pump) · Pressure gauge (reads absolute pressure in kPa) · Millimetre scale alongside tube · Thermometer (to confirm temperature is constant)

⚠ Do not exceed the maximum pressure marked on the apparatus. Release pressure slowly.

1

Record initial conditions

Note atmospheric pressure P₀ (from a barometer or given as 101 kPa). Read the initial length L₀ of the air column at atmospheric pressure. Record room temperature — this must stay constant throughout.

💡 The pressure gauge reads absolute pressure, so the first reading should be close to 101 kPa.

2

Increase pressure using the pump

Use the pump to increase the pressure in steps. After each pump stroke, wait at least 30 seconds for the temperature to return to room temperature (compression heats the gas briefly). Then read P from the gauge and L from the scale.

💡 Wait for the pressure reading to stabilise — if it's still falling, the gas is still cooling from adiabatic compression. Read only when it has stabilised.

3

Take at least 8 readings

Collect readings from approximately 100 kPa (atmospheric) up to the maximum safe pressure (typically 400–500 kPa). Space readings evenly across this range.

💡 More readings at different pressures give a more reliable line. Include at least 2 readings below atmospheric (release pressure) and several above.

4

Calculate V = A×L

Measure the internal diameter d of the tube with a micrometer. Calculate A = πd²/4. Calculate V = A×L for each reading. If using L as proxy for V, state this clearly in your analysis.

💡 In the simulation, the cross-sectional area is fixed at A = 1.00 cm² so V (cm³) = L (cm) numerically — equivalent to using L directly.

5

Plot P vs 1/V

Calculate 1/V for each reading. Plot P (y-axis) against 1/V (x-axis). Draw a best-fit straight line through the origin. The gradient = PV = constant confirms Boyle's Law.

💡 Also plot PV vs P — this should give a horizontal straight line. Any slope indicates the temperature changed during the experiment.

6

Check temperature was constant

If the PV vs P graph is not horizontal, temperature varied during the experiment. Repeat, waiting longer between readings, or insulating the apparatus from draughts.

💡 The PV vs P plot is the most sensitive check for temperature constancy — it reveals systematic errors that P vs 1/V can hide.

      🔬 Pump: Press ▲ Increase P or ▼ Decrease P to change pressure · Wait for stabilisation · Then record each reading.
    

Pump Controls

Wait for pressure to stabilise before recording

Live Readings

101.3 kPa

Absolute Pressure P

Air column length L20.0 cm

Volume V = A×L20.0 cm³

1/V0.0500 cm⁻³

Temperature293 K

—

PV (should be constant)

●●●●●●●●●● stabilised

Readings recorded: 0/8 minimum

Results Data Table

Boyle's Law: PV = constant. Check: does PV stay approximately constant across all readings?

#	P / kPa	L / cm	V = A×L / cm³	1/V / cm⁻³	PV / kPa cm³	PV/mean PV (should≈1)
No data yet — use the Simulation tab.

Graph & Analysis

Two linearised graphs confirming Boyle's Law. Collect at least 6 readings across a wide pressure range.

P vs 1/V result

—kPa·cm³ — gradient = PV constant

Graph parameters

Gradient (= PV)—

R²—

y-intercept (→ 0)—

Interpretation

Record ≥6 readings to see analysis.

PV vs P result

—kPa·cm³ — should be constant

Graph parameters

Mean PV—

Slope (→ 0)—

R²—

Interpretation

Record ≥6 readings to see analysis.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Explain Boyle's Law in terms of the kinetic model of an ideal gas.

In the kinetic model, pressure arises from molecules colliding with the walls of the container. The pressure depends on the frequency of collisions and the force of each collision. At constant temperature, the mean speed (and therefore mean kinetic energy) of the molecules is constant — so the force per collision is unchanged. If the volume is halved, the same number of molecules occupies half the space: the molecules travel shorter distances between collisions with the walls, so the collision frequency doubles. Double the frequency with the same force per collision means double the pressure. This gives P ∝ 1/V, which is Boyle's Law.

Q2

A student uses gauge pressure instead of absolute pressure in their Boyle's Law experiment. Explain why their P vs 1/V graph would not pass through the origin, and how they could correct their results.

Boyle's Law requires absolute pressure (total pressure on the gas, including atmospheric). Gauge pressure = absolute pressure − atmospheric pressure (P₀ ≈ 101 kPa). If the student plots P_gauge vs 1/V, the relationship is: (P_abs − P₀) = k/V → P_gauge = k/V − P₀. This is of the form y = mx + c where c = −P₀ ≠ 0 — a non-zero y-intercept. The graph would not pass through the origin even if the experiment were perfect. To correct: add P₀ (atmospheric pressure) to all gauge pressure readings to get absolute pressure. The corrected P_abs vs 1/V graph should then pass through the origin.

Q3

In this experiment, the student waits 30 seconds after each pressure change before recording. Explain why this is necessary and what would happen to the PV product if they did not wait.

When gas is compressed quickly (adiabatically), work is done on the gas and its temperature rises. Conversely, when pressure is released, the gas cools. If the student reads immediately after compression, the gas is still above room temperature. From the ideal gas law PV = nRT: a higher T gives a higher PV product. As the gas cools back to room temperature over ~30 seconds, PV decreases to its correct value. If the student does not wait: at high pressures (after compression), PV readings will be too high; at low pressures (after expansion), PV readings will be too low. This would show a positive slope on the PV vs P graph — PV appears to increase with P — which would incorrectly suggest the temperature increased with pressure.

Q4

The PV vs P graph is described as a more sensitive check for temperature constancy than the P vs 1/V graph. Explain why.

In the P vs 1/V graph, a small temperature variation causes a small change in gradient (which is PV). The best-fit line still appears approximately straight through the origin because the deviations from the ideal relationship are small relative to the range of the data. It is difficult to detect a small gradient change from scatter in the data. In contrast, the PV vs P graph is specifically designed to show PV as a function of P — if temperature is truly constant, PV should be perfectly horizontal (gradient = 0). Any temperature variation — even a small one — causes a visible slope. The graph amplifies small deviations from constancy because the y-axis represents PV (which should be constant) and any trend is immediately visible as a slope rather than a subtle change in gradient.

Q5

A student records the following data: P = 150 kPa, L = 13.3 cm; P = 300 kPa, L = 6.7 cm. Cross-sectional area A = 1.00 cm². Verify that these readings are consistent with Boyle's Law and calculate the PV constant.

V₁ = A × L₁ = 1.00 × 13.3 = 13.3 cm³. V₂ = A × L₂ = 1.00 × 6.7 = 6.7 cm³. P₁V₁ = 150 × 13.3 = 1995 kPa·cm³. P₂V₂ = 300 × 6.7 = 2010 kPa·cm³. These are equal to within 0.75% — consistent with Boyle's Law given experimental uncertainty. The PV constant ≈ 2000 kPa·cm³ = 2000 × 10³ Pa × 10⁻⁶ m³ = 2.00 × 10⁻³ Pa·m³ = 2.00 × 10⁻³ J. (Note: 1 kPa·cm³ = 10³ Pa × 10⁻⁶ m³ = 10⁻³ J.)