Boyle's law

A gas is not a static substance sitting in its container — it is billions of molecules in constant, random motion, darting about and colliding with each other and with the container walls. Every one of those wall collisions delivers a tiny push, and the combined effect of countless pushes per second, spread over the wall's area, is what we measure as pressure. This molecular picture is the reasoning behind everything Boyle's Law predicts. Now compress that same fixed mass of gas into a smaller volume while holding its temperature steady. The molecules still move at the same average speed (temperature has not changed), but they now have less distance to travel before hitting a wall, so they strike more often in a given time. More collisions per second means higher pressure. Squeeze harder, and the pressure rises further; let the gas expand, and the molecules travel farther between collisions, so pressure falls. Boyle's Law states this precisely:

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

where: $V$ is the volume occupied by the gas, $P$ is its pressure, and the proportionality holds only when the mass of gas and its temperature are both held fixed. Because $V \propto 1/P$, the product $PV$ is a constant for that gas at that temperature — so if the state changes from an initial pressure and volume $(P_1, V_1)$ to a final pair $(P_2, V_2)$ without any change in temperature or mass:

where: $P_1, V_1$ are the pressure and volume before the change, and $P_2, V_2$ are the pressure and volume after — all four values referring to the same fixed mass of gas at the same fixed temperature.

Reading the graphs. Plotting $P$ against $V$ gives a smooth downward curve called a rectangular hyperbola: as $V$ rises, $P$ falls, but never in a straight line. Plotting $P$ against $1/V$ instead straightens the same relationship into a line through the origin, because $P$ is directly proportional to $1/V$. ICSE questions often ask you to identify which of the two graphs is which, so it is worth being able to sketch both from memory.

The result that lets you calculate unknown pressures and volumes is $P_1V_1 = P_2V_2$ for a fixed mass of gas held at constant temperature, and it deserves a short reasoned build rather than being taken purely on Boyle's authority, because ICSE examiners routinely ask you to "explain" it rather than merely "state" it.

Step 1. Fix the physical picture. A container holds $N$ molecules of gas, each colliding with the walls and rebounding; pressure is the total force per unit area these collisions exert. Temperature is a measure of the average kinetic energy of the molecules — if temperature is unchanged, the molecules' average speed is unchanged too.

Step 2. Compress the container from a larger volume $V_1$ to a smaller volume $V_2$, keeping $N$ and the temperature fixed. The same number of molecules, moving at the same average speed, are now confined to a smaller space, so each molecule covers less distance between one wall collision and the next.

Step 3. Count collisions. Less distance between collisions, at the same speed, means more collisions per second on every unit of wall area. Since pressure is built from the rate and force of these collisions, pressure must rise as volume falls, and it must fall as volume rises — the two quantities move in strictly opposite directions.

Step 4. Turn the qualitative link into a proportion. Boyle's experiments (and the reasoning above) show the relationship is not just "opposite" but precisely inverse: $V \propto 1/P$, so $PV$ stays constant for that fixed mass of gas at that fixed temperature. Writing the constant value at state 1 and state 2 and setting them equal gives $P_1V_1 = P_2V_2$ — the working form of Boyle's Law used in every numerical problem.

A gas occupies 600 cm³ at a pressure of 1 atmosphere. What will be its volume if the pressure is increased to 1.5 atmospheres, temperature remaining constant?

Step 1. List the known values: $P_1 = 1\,\text{atm}$, $V_1 = 600\,\text{cm}^3$, $P_2 = 1.5\,\text{atm}$, and $V_2$ is unknown. Temperature and mass of gas are both constant, so Boyle's Law applies directly.

Step 2. Apply Boyle's Law and substitute. $P_1V_1 = P_2V_2$ gives $1 \times 600 = 1.5 \times V_2$.

Step 3. Solve for $V_2$. $V_2 = \dfrac{600}{1.5} = 400\,\text{cm}^3$.

The volume decreases to 400 cm³. This is exactly what the molecular picture predicts: raising the pressure at constant temperature must squeeze the same gas into a smaller space.

Competitive stretch

A gas has a volume of 250 cm³ at STP (1 atmosphere). Find its volume if the pressure is reduced to half of its STP value, temperature kept constant.

Step 1. $P_1 = 1\,\text{atm}$, $V_1 = 250\,\text{cm}^3$, $P_2 = 0.5\,\text{atm}$.

Step 2. $P_1V_1 = P_2V_2 \Rightarrow 1 \times 250 = 0.5 \times V_2$.

Step 3. $V_2 = \dfrac{250}{0.5} = 500\,\text{cm}^3$. Halving the pressure exactly doubles the volume — a clean check that $P_1V_1 = 1 \times 250 = 250$ equals $P_2V_2 = 0.5 \times 500 = 250$, confirming the product stays constant.

• Forgetting "constant temperature." Boyle's Law only holds if temperature does not change during the process; if it does, the combined gas equation is needed instead.
• Treating $P$ and $V$ as directly proportional. They are inversely proportional — pressure up means volume down, and vice versa, for a fixed mass of gas.
• Mixing units across the two sides. Both $P_1$ and $P_2$ must share one unit (e.g. both in atm), and both $V_1$ and $V_2$ must share one unit (e.g. both in cm³), or $P_1V_1 = P_2V_2$ is meaningless.
• Misreading the graphs. The $P$–$V$ graph is a curve, not a straight line; only $P$ plotted against $1/V$ gives a straight line through the origin.

Ask the tutor: Want another Boyle's Law problem with different starting units, or a quick refresher on sketching the P-V and P-1/V graphs? Ask your AI tutor for a guided walkthrough.