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Boyle's Law

Calculating volume-pressure relationships

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Boyle's Law

Boyle’s Law

Boyle’s Law can be used to compare changing conditions for a gas. We use \begin{align*}P_1\end{align*} and \begin{align*}V_1\end{align*} to stand for the initial pressure and initial volume of a gas. After a change has been made, \begin{align*}P_2\end{align*} and \begin{align*}V_2\end{align*} stand for the final pressure and volume. The mathematical relationship of Boyle’s Law becomes:

\begin{align*}P_1 \times V_1=P_2 \times V_2\end{align*}

This equation can be used to calculate any one of the four quantities if the other three are known.

The pressure of a gas decreases as its volume increases

License: CC BY-NC 3.0

The pressure of a gas decreases as the volume increases, making Boyle’s law an inverse relationship. [Figure1]

Sample Problem: Boyle’s Law

A sample of oxygen gas has a volume of 425 mL when the pressure is equal to 387 kPa. The gas is allowed to expand into a 1.75 L container. Calculate the new pressure of the gas.

Step 1: List the known quantities and plan the problem.


  • \begin{align*} P_1=387 \ \text{kPa}\end{align*}
  • \begin{align*}V_1=425 \ \text{mL}\end{align*}
  • \begin{align*}V_2=1.75 \ \text{L}=1750 \ \text{mL}\end{align*}


  • \begin{align*}P_2=? \ \text{kPa}\end{align*}

Use Boyle’s Law to solve for the unknown pressure \begin{align*}(P_2)\end{align*}. It is important that the two volumes (\begin{align*}V_1\end{align*} and \begin{align*}V_2\end{align*}) are expressed in the same units, so \begin{align*}V_2\end{align*} has been converted to mL.

Step 2: Solve.

First, rearrange the equation algebraically to solve for \begin{align*}P_2\end{align*}.

\begin{align*}P_2=\frac{P_1 \times V_1}{V_2}\end{align*}

Now substitute the known quantities into the equation and solve.

\begin{align*}P_2=\frac{387 \text{ kPa} \times 425 \text{ mL}}{1750 \text{ mL}}=94.0 \text{ kPa}\end{align*}

Step 3: Think about your result.

The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about \begin{align*}\frac{1}{4}{th}\end{align*}. The pressure is in kPa and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem.


  • The volume of a gas is inversely proportional to pressure.

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  1. [1]^ License: CC BY-NC 3.0

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