Boyle's law, Charles' law and the general gas equation: WASSCE Chemistry

A bicycle pump gets harder to push as the air inside is squeezed; a car tyre's pressure rises after a long drive in the sun; a balloon left in a cold room shrinks. Gases follow simple, predictable laws linking their pressure, volume and temperature. This lesson covers Boyle's law, Charles' law, the pressure law and the equations that combine them.

Learning objectives

By the end of this lesson you should be able to:

state the units of pressure and convert between them
state and apply Boyle's law, Charles' law and the pressure law
explain each law using the kinetic theory
use the general gas equation and convert volumes to s.t.p.
use the ideal gas equation PV = nRT

Pressure and its units

Gas pressure is caused by gas particles colliding with the walls of their container.

Unit	Standard atmospheric pressure
millimetres of mercury (mmHg)	760 mmHg
atmosphere (atm)	1 atm
pascal (Pa) = N/m²	101 325 Pa ≈ 101 kPa

Standard temperature and pressure (s.t.p.): 0 °C (273 K) and 760 mmHg (1 atm).

All gas-law calculations use temperature in kelvin: $T = \theta + 273$ .

Boyle's law

Definition

Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is constant.

P_1V_1 = P_2V_2

Left: a Boyle's law curve of pressure against volume. Right: a Charles' law straight line of volume against temperature, extended back to zero volume at −273 °C

Figure 1. Boyle's law (left) and Charles' law (right). WASSCE Tutor diagram.

Kinetic explanation: if the volume is halved, the same particles are packed into half the space, so they hit the walls twice as often, and the pressure doubles.

Worked example

200 cm³ of a gas is at 750 mmHg. What is its volume at 1500 mmHg, at the same temperature?

V_2 = \dfrac{P_1V_1}{P_2} = \dfrac{750 \times 200}{1500} = 100 \text{ cm}^3

\begin{aligned}V_2 &= \dfrac{P_1V_1}{P_2} \\ &= \dfrac{750 \times 200}{1500} \\ &= 100 \text{ cm}^3\end{aligned}

Doubling the pressure halves the volume.

Charles' law

Definition

Charles' law: the volume of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature, provided the pressure is constant.

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}

Extending the Charles' law graph backwards, the volume would reach zero at −273 °C. This temperature is absolute zero (0 K), the starting point of the kelvin scale.

Kinetic explanation: heating gives particles more kinetic energy. They hit the walls harder and more often, so the gas expands until the pressure falls back to its original value.

Worked example

A gas occupies 300 cm³ at 27 °C. Find its volume at 127 °C, at constant pressure.

$T_1 = 27 + 273 = 300$ K, $T_2 = 127 + 273 = 400$ K.

V_2 = \dfrac{V_1 \times T_2}{T_1} = \dfrac{300 \times 400}{300} = 400 \text{ cm}^3

\begin{aligned}V_2 &= \dfrac{V_1 \times T_2}{T_1} \\ &= \dfrac{300 \times 400}{300} \\ &= 400 \text{ cm}^3\end{aligned}

Note that the volume does not become 300 × 127 ÷ 27. Always use kelvin.

The pressure law

Definition

Pressure law (Gay-Lussac's law of pressures): the pressure of a fixed mass of gas is directly proportional to its absolute temperature, provided the volume is constant.

\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}

Worked example

A tyre is at 200 kPa at 27 °C. After a long drive its temperature is 57 °C. What is the new pressure, if the volume is unchanged?

P_2 = \dfrac{P_1 \times T_2}{T_1} = \dfrac{200 \times 330}{300} = 220 \text{ kPa}

\begin{aligned}P_2 &= \dfrac{P_1 \times T_2}{T_1} \\ &= \dfrac{200 \times 330}{300} \\ &= 220 \text{ kPa}\end{aligned}

The general gas equation

Combining the three laws:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}

\begin{aligned}&\dfrac{P_1V_1}{T_1} \\ &= \dfrac{P_2V_2}{T_2}\end{aligned}

It is used to correct gas volumes to s.t.p., so that volumes measured on different days can be compared.

Worked example

A gas occupies 500 cm³ at 27 °C and 700 mmHg. Find its volume at s.t.p.

$P_1 = 700$ , $V_1 = 500$ , $T_1 = 300$ K; $P_2 = 760$ , $T_2 = 273$ K.

V_2 = \dfrac{P_1V_1T_2}{T_1P_2} = \dfrac{700 \times 500 \times 273}{300 \times 760}

\begin{aligned}V_2 &= \dfrac{P_1V_1T_2}{T_1P_2} \\ &= \dfrac{700 \times 500 \times 273}{300 \times 760}\end{aligned}

V_2 \approx 419 \text{ cm}^3

The ideal gas equation

PV = nRT

where P is pressure in Pa, V is volume in m³, n is the number of moles, T is temperature in K and R is the molar gas constant, 8.314 J mol⁻¹ K⁻¹.

Worked example

What pressure is exerted by 0.5 mol of gas in a 10 dm³ container at 27 °C?

V = 10 dm³ = 0.010 m³; T = 300 K.

P = \dfrac{nRT}{V} = \dfrac{0.5 \times 8.314 \times 300}{0.010}

\begin{aligned}P &= \dfrac{nRT}{V} \\ &= \dfrac{0.5 \times 8.314 \times 300}{0.010}\end{aligned}

P \approx 1.25 \times 10^5 \text{ Pa}

Ideal and real gases. The gas laws assume an ideal gas: particles with no volume and no attractions between them. Real gases behave almost ideally at high temperature and low pressure. They deviate at high pressure (particle volume matters) and low temperature (attractions matter, and gases liquefy).

In Sierra Leone

Gas cylinders used for cooking in Sierra Leone carry warnings not to leave them in the sun. By the pressure law, the pressure inside rises as the temperature rises, and a hot cylinder can leak or burst. Tyres are best checked when cool, before a journey.

Common mistake

Never put temperatures in °C into the gas laws. Charles' law and the pressure law work only with kelvin, because 0 °C is not "zero energy". Convert first: K = °C + 273.

Key points

s.t.p.: 273 K and 760 mmHg (1 atm, 101 kPa).
Boyle's law: $P_1V_1 = P_2V_2$ at constant temperature.
Charles' law: $V_1/T_1 = V_2/T_2$ at constant pressure; absolute zero is −273 °C.
Pressure law: $P_1/T_1 = P_2/T_2$ at constant volume.
General gas equation: $P_1V_1/T_1 = P_2V_2/T_2$ . Ideal gas equation: $PV = nRT$ .
Always use kelvin.

WASSCE examination focus

WAEC tests this

Every WASSCE Chemistry paper has a gas-law calculation. Expect to state Boyle's and Charles' laws, to calculate new volumes or pressures, to correct volumes to s.t.p., and to explain the laws with the kinetic theory. Write the formula, substitute with units and convert °C to K first.

Quick check

State Boyle's law.
A gas has a volume of 60 cm³ at 2 atm. What is its volume at 3 atm, at constant temperature?
A gas occupies 250 cm³ at 77 °C. Find its volume at 7 °C, at constant pressure.
What is absolute zero in °C?

Answers:

The volume of a fixed mass of gas is inversely proportional to its pressure at constant temperature.
$V_2 = \dfrac{2 \times 60}{3} = 40$ cm³.
$T_1 = 350$ K, $T_2 = 280$ K: $V_2 = \dfrac{250 \times 280}{350} = 200$ cm³.
−273 °C.

Boyle's law, Charles' law and the general gas equation

Learning objectives

Pressure and its units

Boyle's law

Charles' law

The pressure law

The general gas equation

The ideal gas equation

Key points

WASSCE examination focus

Quick check

Keep going with gas laws

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