Gas Laws — B.3 Gas Laws — IB Physics Notes

2 Lesson 2 of 6 · B.3 Gas Laws

Gas Laws

Three simple investigations, each holding one variable fixed, reveal how pressure, volume and temperature interact for a fixed amount of gas — and combine into a single equation linking all three.

💉 Hook – block the end of a syringe and push

Seal the nozzle of a syringe with your finger and push the plunger in. At first it moves easily. The further in you push it, the harder it gets to push — and if you let go, it springs straight back out. You have trapped a fixed amount of gas and shrunk its volume, and its pressure has risen to fight back. This is exactly the apparatus used to investigate the relationship properly.

Computer, interface, data logger, pressure sensor and hand-held syringe used to compress a trapped sample of air

Fig 1 — A pressure sensor logs the pressure of the air trapped in the syringe as its volume is changed by hand.

💡 Diagnostic question (think – then read on) — the gas in the syringe is not allowed to escape and its temperature barely changes. What happens to the pressure as you halve the volume? The pressure roughly doubles. For a fixed amount of gas at constant temperature, pressure and volume are inversely proportional — halving one doubles the other.

🧪 Three classic gas law investigations

A sealed container of gas has four physical properties that are easily measured: pressure, volume, temperature and mass ($P$, $V$, $T$ and $m$). The amount of gas in moles, $n$, can be found from its mass. All gases show the same behaviour under most conditions, so air is the obvious choice for experiments.

Each of the three classic investigations below keeps the amount of gas fixed and holds one further variable constant while the other two are measured:

— How does volume change with pressure, at constant temperature? (Boyle's law)
— How does pressure change with temperature, at constant volume? (pressure law)
— How does volume change with temperature, at constant pressure? (Charles' law)

📉 Boyle's law — pressure and volume at constant temperature

Figure 2 shows apparatus for investigating how pressure affects the volume of a fixed mass of gas at constant temperature. Compressing a gas does work on it and tends to raise its temperature, so changes must be made slowly to keep the process isothermal (occurring at constant temperature).

Vertical glass tube of trapped air connected to a pressure gauge, valve, foot pump and oil reservoir

Fig 2 — A vertical glass tube of air under test, connected via oil to a pressure gauge, a valve and a foot pump.

⚠️ Safety: the gas pressure here can be high enough that the glass tube might explode, so it should be surrounded by a clear plastic protective screen.

A plot of $P$ against $V$ gives a downward curve, called an isotherm because every point is at the same temperature. Re-plotting $P$ against $1/V$ gives a straight line through the origin, showing that pressure and volume are inversely proportional.

Boyle's law · fixed amount of gas, constant temperature $$P \propto \frac{1}{V}$$

$$PV = \text{constant}, \quad P_1V_1 = P_2V_2$$

Molecular explanation: shrink the container and the molecules (moving at the same average speed) hit a given area of wall more often each second, so the pressure rises.

🌡️ Pressure law — pressure and temperature at constant volume

Figure 3 shows apparatus for this investigation: a sealed flask of air, immersed in a water bath, with a thermometer and a pressure gauge. Measurements are usually taken between 0°C and 100°C.

Round flask of trapped air in a water bath, with a thermometer and a pressure gauge connected by a short tube

Fig 3 — The flask's volume cannot change, so as the water bath heats the trapped air, only its pressure rises.

⚠️ Safety: a large beaker of very hot water could be knocked over, so keep the apparatus away from the table edge; wear gloves when pushing the bung firmly into the flask in case the glass breaks.

The raw results form a straight line of pressure against temperature in degrees Celsius (Fig 4) — but it does not pass through the origin.

Graph of pressure against temperature in degrees Celsius, a straight line rising from about 1.0 times ten to the five pascals at zero degrees to about 1.4 times ten to the five pascals at 100 degrees

Fig 4 — Pressure against temperature in °C. Extrapolated back, the line reaches $P=0$ at $-273^{\circ}\text{C}$.

Graph of pressure against temperature in kelvin, a straight line through the origin, with dashed vertical lines at 273 and 373 kelvin

Fig 5 — The same data re-plotted against T in kelvin: a straight line through the origin, so $P \propto T(\text{K})$.

📐 Tool 3 – extrapolating graphs. Extending the line in Fig 4 this far is not normally recommended — but here we are not predicting unknown gas behaviour at low temperature, only asking at what temperature the trend would put the pressure at zero.

Every gas extrapolates to the same value, $-273^{\circ}\text{C}$: the temperature at which molecular motion is assumed to (almost) stop and molecules no longer collide with the walls. This is absolute zero, the lower fixed point of the Kelvin scale — $0\ \text{K}$.

Pressure law · fixed amount of gas, constant volume $$P \propto T$$

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

Molecular explanation: cooling the gas slows the molecules, so they hit a given area of wall less often, lowering the pressure.

⚠️ Common mistake: when calculating with temperatures (not temperature changes), always use kelvin.

🎈 Charles' law — volume and temperature at constant pressure

Figure 6 shows a beaker of water heated from below, with an open-ended tube of trapped air immersed in it. A liquid plug traps the air in the tube and a scale beside the tube reads its volume; as the gas warms it pushes the plug along, keeping its pressure equal to the atmosphere outside.

Beaker of heated water with an open-ended tube of trapped air, a liquid plug, a scale and a thermometer

Fig 6 — A liquid plug traps a column of air; the scale reads how far it moves as the gas is warmed.

The concluding graph (Fig 7) is a straight line through the origin, just like Fig 5.

Graph of gas volume against absolute temperature in kelvin, a straight line through the origin, with dashed vertical lines at 273 and 373 kelvin

Fig 7 — Volume against absolute temperature: a straight line through the origin, so $V \propto T(\text{K})$.

Charles' law · fixed amount of gas, constant pressure $$V \propto T(\text{K})$$

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Molecular explanation: cooling the gas slows the molecules, so they would hit a given area less often — but shrinking the volume too keeps the collision rate, and so the pressure, constant.

⚠️ Common mistake: a gas does not automatically expand when heated. Its volume only rises if it is allowed to, at constant pressure, as in this investigation.

📈 Graph mastery – the three gas laws

Graph type	Axes (X, Y)	Shape	What to read
Boyle's law	Volume $V$ · Pressure $P$	Downward curve (isotherm)	$PV$ is the same at every point on the curve
Boyle's law, linearised	$1/V$ · Pressure $P$	Straight line through the origin	Confirms $P \propto 1/V$; gradient is the constant $PV$
Pressure law	Temperature $T/^{\circ}\text{C}$ · Pressure $P$	Straight line, does not pass through the origin	Extrapolated intercept on the $T$ axis is $-273^{\circ}\text{C}$, absolute zero
Charles' law	Temperature $T/\text{K}$ · Volume $V$	Straight line through the origin	Confirms $V \propto T$ when temperature is measured in kelvin

Graph of pressure against volume, a downward-sloping curve through red data points

Fig 8 — $P$ against $V$: pressure falls as volume rises, but not in a straight line.

Graph of pressure against one over volume, a straight line through the origin through red data points

Fig 9 — The same data re-plotted as $P$ against $1/V$: now a straight line through the origin.

✍️ Worked example 1 – Boyle's law (B3.4)

A sample of gas has a volume of 43 cm³ at normal atmospheric pressure ($1.0\times10^{5}$Pa). (a) If the pressure is increased to $3.7\times10^{5}$Pa, calculate its new volume. (b) State the assumption made in (a).

Identify: fixed amount of gas, pressure and volume both changing — use Boyle's law, $P_1V_1 = P_2V_2$.

Data: $P_1 = 1.0\times10^{5}$Pa, $V_1 = 43$cm³, $P_2 = 3.7\times10^{5}$Pa.

Equation:
$(1.0\times10^{5})\times 43 = (3.7\times10^{5})\times V_2$

Answer: $V_2 = 12\ \text{cm}^3$.
(b) The temperature of the gas did not change.

✍️ Worked example 2 – pressure law (B3.5)

Some air in a sealed container was heated from 60°C to 92°C. If its final pressure was $1.82\times10^{5}$Pa, determine its pressure at 60°C.

Identify: sealed (fixed volume) container — use the pressure law, $P_1/T_1 = P_2/T_2$, with temperatures in kelvin.

Data: $T_1 = 273+60 = 333$K, $T_2 = 273+92 = 365$K, $P_2 = 1.82\times10^{5}$Pa.

Equation:
$\dfrac{P_1}{333} = \dfrac{1.82\times10^{5}}{365}$

Answer: $P_1 = 1.7\times10^{5}\ \text{Pa}$.

✍️ Worked example 3 – Charles' law (B3.6)

In an experiment such as Fig 6, the gas temperature was increased from 5.0°C to 95.0°C. If the initial volume was 3.1 cm³, calculate the final volume.

Identify: pressure stays equal to the atmosphere throughout — use Charles' law, $V_1/T_1 = V_2/T_2$.

Data: $T_1 = 273+5.0 = 278$K, $T_2 = 273+95 = 368$K, $V_1 = 3.1$cm³.

Equation:
$\dfrac{3.1}{278} = \dfrac{V_2}{368}$

Answer: $V_2 = 4.1\ \text{cm}^3$.

🎈 Real-world: why a hot air balloon rises and falls

The balloon's envelope is open at the base, so the trapped air inside stays at roughly atmospheric pressure — the same condition as Charles' law. Heating the air with the burner raises its temperature, so (at constant pressure) it expands and some spills out of the open base, lowering the mass of air inside the envelope. The remaining hot, less dense air weighs less than the same volume of cool air outside, so the buoyant force from the surrounding air exceeds the balloon's weight and it rises. Letting the air inside cool (or venting some through a vent at the top) raises its density back up, reducing the buoyant force so the balloon descends.

Multiple colourful hot air balloons flying over a rocky landscape at Cappadocia, Turkey

Fig 10 — Hot air balloons over Cappadocia. Each balloon's lift comes from the density difference between the heated air inside and the cooler air outside.

✓ Examiner tip: an answer that just says "hot air rises" earns little. Explain it through density — heated air at the same pressure occupies more volume per unit mass (Charles' law), so it is less dense than the surrounding cool air.

🚀 HL extension – combining the three laws

The three empirical laws can be combined into one: for a fixed amount of gas, $PV \propto T$, so $PV/T$ is constant.

Combined gas law $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

This single equation reduces to Boyle's, the pressure law or Charles' law whenever one of $P$, $V$ or $T$ happens to be constant. It is also the direct forerunner of the ideal gas equation, $PV=nRT$, met in the next lesson.
Challenge: a fixed mass of gas at $1.0\times10^{5}$Pa, 2.0 m³ and 300K is compressed to 1.2 m³ while its temperature rises to 360K. Find its new pressure. Answer: $\dfrac{(1.0\times10^{5})(2.0)}{300} = \dfrac{P_2(1.2)}{360}$, so $P_2 = 2.0\times10^{5}$Pa.

⚠️ Trap corner – common mistakes

Misconception	Correct view
A gas's volume always increases when its temperature rises	Only if it is allowed to, at constant pressure (Charles' law). At constant volume, the pressure rises instead (pressure law)
Celsius temperatures can be substituted straight into $P_1/T_1=P_2/T_2$ or $V_1/T_1=V_2/T_2$	Always convert to kelvin first: $T(\text{K}) = T(^{\circ}\text{C}) + 273$
Extrapolating the $P$–$T$ line to $-273^{\circ}\text{C}$ is unreliable guesswork	It is a narrow, specific extrapolation — asking only at what temperature the existing trend would reach $P=0$, not predicting general gas behaviour there
Pressure and volume are directly proportional	Boyle's law says they are inversely proportional: $P \propto 1/V$

⚠️ Most common slip: plugging a Celsius temperature straight into a gas law ratio. Convert to kelvin as the first line of every calculation.

PV = constant P ∝ 1/V P/T = constant P ∝ T(K) V/T = constant V ∝ T(K) 0°C = 273K absolute zero = -273°C PV/T = constant

Practice check — multiple choice

1. A fixed mass of gas at constant temperature has its volume halved. Its pressure

A halves

B doubles

C stays the same

D quarters

2. A fixed mass of gas at constant volume is cooled from 127°C to 27°C. Its pressure will

A fall to 75% of its original value

B fall to 21% of its original value

C stay the same, since only the volume matters

D fall to 24% of its original value

3. On a graph of pressure (y) against temperature in °C (x) for a fixed mass of gas at constant volume, the temperature at which the extrapolated line reaches $P=0$ is

A 0°C, always

B different for every gas

C -273°C, the same for every gas

D undefined, since the line never reaches zero

4. A fixed mass of gas at constant pressure is heated from 27°C to 127°C. Its volume will

A stay the same

B increase to 133% of its original value

C increase to 470% of its original value

D decrease

5. Which combination correctly matches the law to the property held constant?

A Boyle's law – constant temperature; pressure law – constant volume; Charles' law – constant pressure

B Boyle's law – constant pressure; pressure law – constant temperature; Charles' law – constant volume

C Boyle's law – constant volume; pressure law – constant pressure; Charles' law – constant temperature

D All three hold temperature constant

Practice check — fill in the blanks

Complete the statements:

A graph on which every point is at the same temperature is called an . For a fixed mass of gas at constant temperature, pressure is proportional to volume — this is law. For a fixed mass of gas at constant volume, pressure is proportional to temperature measured in — this is the law. The temperature at which a gas's pressure would extrapolate to zero is °C, known as absolute . The three laws combine into one equation, $PV/T = $ .

Exam‑style questions

Exam question · Calculate

Q9. After the pressure on a gas was increased from $1.0\times10^{5}$Pa to $4.5\times10^{5}$Pa, its volume became 280 cm³. What was its original volume, assuming its temperature was constant? [2 marks]

✓ Boyle's law: $P_1V_1 = P_2V_2$.
✓ $(1.0\times10^{5})V_1 = (4.5\times10^{5})(280)$, so $V_1 = 1260\ \text{cm}^3$.

Exam question · Calculate

Q10. The temperature of a gas was reduced from 80°C to 20°C while keeping it at the same pressure. If the starting volume was 110 cm³, what was the final volume? [2 marks]

✓ Charles' law: $V_1/T_1 = V_2/T_2$, with $T_1 = 353$K, $T_2 = 293$K.
✓ $110/353 = V_2/293$, so $V_2 \approx 91\ \text{cm}^3$.

Exam question · Determine

Q11. A fixed volume of gas at 310K and $1.2\times10^{5}$Pa was heated in an oven. At what temperature (in °C) will the pressure have risen to $1.9\times10^{5}$Pa? [3 marks]

✓ Pressure law: $P_1/T_1 = P_2/T_2$.
✓ $(1.2\times10^{5})/310 = (1.9\times10^{5})/T_2$, so $T_2 \approx 491\ \text{K}$.
✓ In °C: $491 - 273 = 218^{\circ}\text{C}$.

Score: 0 pts

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Gas Pressure and Amount of Substance

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Ideal Gas Equation

Graph type	Axes (X, Y)	Shape	What to read
Boyle's law	Volume \(V\) · Pressure \(P\)	Downward curve (isotherm)	\(PV\) is the same at every point on the curve
Boyle's law, linearised	\(1/V\) · Pressure \(P\)	Straight line through the origin	Confirms \(P \propto 1/V\); gradient is the constant \(PV\)
Pressure law	Temperature \(T/^{\circ}\text{C}\) · Pressure \(P\)	Straight line, does not pass through the origin	Extrapolated intercept on the \(T\) axis is \(-273^{\circ}\text{C}\), absolute zero
Charles' law	Temperature \(T/\text{K}\) · Volume \(V\)	Straight line through the origin	Confirms \(V \propto T\) when temperature is measured in kelvin