Boyle's, Charles's and Pressure Laws

Spec mapping: OCR H556 Module 5.2 — Ideal gases (the three experimental gas laws — Boyle's law, Charles's law, the pressure law / Gay-Lussac's law — and their combination into the combined gas law $pV/T =$pV/T= constant for a fixed mass of ideal gas; the idea of an ideal gas; absolute temperature). Refer to the official OCR H556 specification document for exact wording.

Historically, the ideal gas equation $pV = nRT$pV=nRT did not spring into existence fully formed. It grew out of three simpler experimental relationships discovered over the course of the 17th, 18th and 19th centuries. Each describes the behaviour of a fixed amount of gas when one of the three variables — pressure, volume or temperature — is held constant. Together, they encapsulate the observed macroscopic behaviour of gases across a huge range of conditions.

Module 5.2 of the OCR A-Level Physics A specification (H556) requires you to know these three experimental gas laws, to use them in calculations, and to see how they combine into the single ideal gas equation that we develop in the next lesson. The three laws are empirical approximations: each is derived from a model of an ideal gas (point particles with no inter-molecular forces and only elastic collisions), and each is a particular limit of $pV = nRT$pV=nRT when one variable is held fixed.

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Setting the Scene — The Ideal-Gas Model

All three laws apply to a fixed amount of gas — the same number of molecules throughout the experiment — and work only for ideal (or near-ideal) behaviour. The ideal-gas approximation rests on four assumptions:

1. Point particles. Molecules occupy negligible volume compared with the container.
2. No intermolecular forces (except during the brief instants of collision).
3. Elastic collisions with the walls and with each other.
4. Newtonian dynamics apply between collisions.

Real gases deviate from these assumptions at high pressure (where the molecules are packed closely enough for their own size to matter) and at low temperature (where intermolecular forces become significant, and the gas approaches its liquefaction point). Under normal laboratory conditions — a few atmospheres, temperatures from about 100 K up to several hundred — the approximation is excellent.

The three variables are:

• Pressure $p$p, measured in pascals (Pa $=$= N m$^{-2}$−2). Atmospheric pressure at sea level is roughly $1.01 \times 10^{5}$1.01×105 Pa $\approx 100$≈100 kPa.
• Volume $V$V, measured in cubic metres (m$^3$3). One litre $= 10^{-3}$=10−3 m$^3$3; one cm$^3$3 $= 10^{-6}$=10−6 m$^3$3.
• Temperature $T$T, measured in kelvin (not Celsius). This is crucial. All the gas laws here fail if you use Celsius.

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Boyle's Law (constant T)

Robert Boyle, in 1662, published experimental results showing that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

Boyle's Law: For a fixed mass of ideal gas at constant temperature, $p \propto 1/V$p∝1/V, or equivalently $pV = \text{constant}$pV=constant.

If you have an initial state $(p_1, V_1)$(p1​,V1​) and a final state $(p_2, V_2)$(p2​,V2​) at the same temperature, then

$p_1 V_1 = p_2 V_2.$p1​V1​=p2​V2​.

Graphical forms

• A plot of $p$p against $V$V is a hyperbola (a curve that asymptotically approaches both axes).
• A plot of $p$p against $1/V$1/V is a straight line through the origin with slope equal to $pV = nRT$pV=nRT (the constant).
• A plot of $pV$pV against $V$V (or against $p$p) is a horizontal line.

The hyperbola at constant temperature is called an isotherm; higher temperatures correspond to isotherms further from the origin.

Physical picture

Boyle's law is easy to understand at the molecular level: if you halve the volume available to a gas, you double the frequency with which molecules strike the walls of the container, so the pressure doubles. You will derive this quantitatively from kinetic theory later in the module.

Both linear plots pass through the origin in K. Extrapolated to the Celsius axis they cross at about −273 °C.

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Worked Example 1 — Boyle's Law

A bicycle pump contains 0.20 L of air at atmospheric pressure ($1.01 \times 10^{5}$1.01×105 Pa). If the piston is pushed in until the volume is 0.050 L (the valve stays shut), what is the new pressure? Assume the temperature does not change.

$p_2 = \frac{p_1 V_1}{V_2} = \frac{(1.01 \times 10^{5})(0.20)}{0.050} = (1.01 \times 10^{5})(4) \approx 4.04 \times 10^{5} \text{ Pa} \approx 4.0 \times 10^{5} \text{ Pa}.$p2​=V2​p1​V1​​=0.050(1.01×105)(0.20)​=(1.01×105)(4)≈4.04×105 Pa≈4.0×105 Pa.

The volume was reduced by a factor of 4, so the pressure rose by the same factor, to about 4 atmospheres. (Of course, in reality a bicycle pump does heat up slightly during compression; Boyle's law strictly applies only if the compression is slow enough for heat to dissipate, keeping $T$T constant — an isothermal process. A rapid compression is adiabatic, with $T$T also rising; the result is a higher pressure than Boyle's law predicts.)

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Charles's Law (constant p)

Jacques Charles, around 1787 (the results were published much later by Gay-Lussac), discovered that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

Charles's Law: For a fixed mass of ideal gas at constant pressure, $V \propto T$V∝T, or equivalently $V/T = \text{constant}$V/T=constant.

Thus for two states of the gas at the same pressure,

$\frac{V_1}{T_1} = \frac{V_2}{T_2}.$T1​V1​​=T2​V2​​.

Absolute zero — an empirical consequence

Charles found that the volume of a gas at constant pressure decreases linearly as the temperature falls, and if you extrapolate the straight line back far enough, it crosses zero volume at around $-273\,^{\circ}$−273∘C. This is one experimental hint of absolute zero: the temperature at which the volume of an ideal gas would (if it stayed gaseous) become zero.

Of course, real gases liquefy long before this — nitrogen, for example, liquefies at $-196\,^{\circ}$−196∘C. But the extrapolation is still valid as an ideal-gas prediction and gives exactly the right value of absolute zero as the zero of the Kelvin scale.

Graphical forms

• A plot of $V$V against $T$T (in kelvin) is a straight line through the origin with slope $nR/p$nR/p.
• A plot of $V$V against $\theta$θ (in Celsius) is a straight line with a positive intercept on the volume axis; if extrapolated backwards, it crosses the temperature axis at about $-273\,^{\circ}$−273∘C.

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Worked Example 2 — Charles's Law

A helium balloon has a volume of 2.0 L at $20\,^{\circ}$20∘C ($T_1 = 293$T1​=293 K). It is left in a warm car where the temperature rises to $45\,^{\circ}$45∘C ($T_2 = 318$T2​=318 K). Assuming the pressure stays constant and the balloon does not burst, find the new volume.

$V_2 = \frac{V_1 T_2}{T_1} = \frac{(2.0)(318)}{293} \approx 2.17 \text{ L}.$V2​=T1​V1​T2​​=293(2.0)(318)​≈2.17 L.

So the balloon swells by about 8.5% — a clear illustration of why balloons left in hot cars can pop. Note how essential it was to use kelvin: the ratio $318/293 \approx 1.085$318/293≈1.085 is close to 1, but the Celsius ratio $45/20 = 2.25$45/20=2.25 is wildly wrong — Celsius is not a proportional scale because its zero is arbitrary.

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The Pressure Law (constant V)

Guillaume Amontons in 1702, and later more precisely Joseph Louis Gay-Lussac in 1809, found that at constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Pressure Law (Gay-Lussac's Law): For a fixed mass of ideal gas at constant volume, $p \propto T$p∝T, or equivalently $p/T = \text{constant}$p/T=constant.

Thus for two states at the same volume,

$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$T1​p1​​=T2​p2​​.

This law underlies the operation of the constant-volume gas thermometer, which is one of the most accurate standard thermometers known and is used to define the temperatures of fixed points on the international temperature scale.

Graphical forms

• A plot of $p$p against $T$T (in kelvin) is a straight line through the origin with slope $nR/V$nR/V.

Just as with Charles's law, the extrapolation to $p = 0$p=0 gives $T = 0$T=0 K $= -273.15\,^{\circ}$=−273.15∘C, providing another experimental path to absolute zero.

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Worked Example 3 — Pressure Law

A car tyre is inflated to $p_1 = 2.4 \times 10^{5}$p1​=2.4×105 Pa on a cool morning at $5\,^{\circ}$5∘C ($T_1 = 278$T1​=278 K). After some driving, the tyre warms up to $50\,^{\circ}$50∘C ($T_2 = 323$T2​=323 K). Assuming the tyre volume does not change significantly, what is the new pressure?

$p_2 = \frac{p_1 T_2}{T_1} = \frac{(2.4 \times 10^{5})(323)}{278} \approx 2.79 \times 10^{5} \text{ Pa}.$p2​=T1​p1​T2​​=278(2.4×105)(323)​≈2.79×105 Pa.

So the pressure rises by about 16%. Tyre pressures should always be measured when the tyres are cold, before driving, for this reason.

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Combining the Three Laws

The three laws are really three faces of the same underlying relationship. If we start from Boyle's law $pV = \text{const}$pV=const (at constant $T$T), and combine it with Charles's law $V \propto T$V∝T (at constant $p$p), we arrive at the combined gas law:

$\frac{pV}{T} = \text{constant},$TpV​=constant,

or equivalently, for two states of the same fixed mass of gas,

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}.$T1​p1​V1​​=T2​p2​V2​​.

This equation is valid for any change in the state of a fixed mass of ideal gas, no matter whether $p$p, $V$V or $T$T is held constant (or none of them). It reduces to Boyle's law when $T$T is constant, to Charles's law when $p$p is constant, and to the pressure law when $V$V is constant.

Gas-law selection decision tree

flowchart TD
    A["Fixed mass of ideal gas: what changes?"] --> B{"What is held constant?"}
    B -- "T constant" --> Boyle["Boyle: p1 V1 = p2 V2"]
    B -- "p constant" --> Charles["Charles: V1/T1 = V2/T2"]
    B -- "V constant" --> Press["Pressure law: p1/T1 = p2/T2"]
    B -- "Nothing constant" --> Combined["Combined: p1 V1 / T1 = p2 V2 / T2"]

In the next lesson we complete the story: the combined gas law, generalised from a fixed amount of gas to any amount, gives the ideal gas equation $pV = nRT$pV=nRT.

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Worked Example 4 — The Combined Gas Law

A sample of gas has pressure $p_1 = 1.2 \times 10^{5}$p1​=1.2×105 Pa, volume $V_1 = 3.0 \times 10^{-3}$V1​=3.0×10−3 m$^3$3 and temperature $T_1 = 290$T1​=290 K. It is compressed and heated to pressure $p_2 = 2.5 \times 10^{5}$p2​=2.5×105 Pa and temperature $T_2 = 350$T2​=350 K. What is the new volume?

$V_2 = \frac{p_1 V_1 T_2}{p_2 T_1} = \frac{(1.2 \times 10^{5})(3.0 \times 10^{-3})(350)}{(2.5 \times 10^{5})(290)} = \frac{1.26 \times 10^{5}}{7.25 \times 10^{7}} \approx 1.74 \times 10^{-3} \text{ m}^{3}.$V2​=p2​T1​p1​V1​T2​​=(2.5×105)(290)(1.2×105)(3.0×10−3)(350)​=7.25×1071.26×105​≈1.74×10−3 m3.

So the volume decreases from $3.0 \times 10^{-3}$3.0×10−3 m$^3$3 to about $1.74 \times 10^{-3}$1.74×10−3 m$^3$3 — a reduction of about 42%. The heating alone would have increased the volume (factor $350/290 \approx 1.21$350/290≈1.21), but the compression (factor $1.2/2.5 = 0.48$1.2/2.5=0.48) dominated.

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Experimental Verification

Each of the three laws can be verified with simple school apparatus. A typical arrangement for Boyle's law:

• A sealed glass tube containing air is connected to a reservoir of oil and a pump that can apply a known pressure.
• The pressure is varied, and the corresponding length of the air column (and hence volume) is read off.
• A graph of $p$p against $1/V$1/V is plotted; a straight line through the origin confirms Boyle's law.

For Charles's law, a gas trapped in a capillary tube is immersed in a water bath at varying temperatures; a graph of $V$V (measured from the length of the gas column) against $T$T (K) should be a straight line through the origin.

For the pressure law, gas in a sealed flask of fixed volume is connected to a pressure gauge, and the flask is placed in baths of different temperatures; a graph of $p$p against $T$T (K) is again a straight line through the origin.

Sources of error

• Temperature drift during the Boyle's law experiment means $T$T is not really constant. Minimise by working slowly and waiting for equilibrium between compressions.
• Thermal expansion of the container is usually negligible for glass but matters for plastic; correct for the container's own thermal expansion if you can.
• Reading errors in the scale measuring the length of the gas column are typically the dominant random error; use a travelling microscope or a vernier scale.
• Assumptions of ideal behaviour break down at very high pressure or very low temperature; stay within normal laboratory conditions.

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Unit Conversions — A Warning

A common source of exam errors is confused units. Remember:

Quantity	SI unit	Common alternative
Pressure	Pa	atm (1 atm $= 1.013 \times 10^{5}$=1.013×105 Pa), bar (1 bar $= 10^{5}$=105 Pa), mmHg (1 mmHg $\approx 133$≈133 Pa)
Volume	m$^3$3	L (1 L $= 10^{-3}$=10−3 m$^3$3), cm$^3$3 (1 cm$^3$3 $= 10^{-6}$=10−6 m$^3$3)
Temperature	K	$^{\circ}$∘C (subtract 273 to get back to $^{\circ}$∘C)

If you consistently use the same units on both sides of an equation like $p_1 V_1 / T_1 = p_2 V_2 / T_2$p1​V1​/T1​=p2​V2​/T2​, you do not actually have to convert everything to SI — the units cancel. But you must use kelvin for temperature, because the law is expressed in terms of the absolute temperature. Using $^{\circ}$∘C gives nonsense.