P₁ • V₁ = P₂ • V₂

The formula at the top of the page is Boyle’s Law, named after the British chemist Robert Boyle (1627 – 1691). It states that the volume of a fixed mass of gas at a constant temperature is inversely proportional to the pressure of the gas.

In less formal terms, Boyle’s Law can be stated:
When pressure increases, volume decreases.

When volume increases, pressure decreases.

This calculator can solve for any one of the four variables of Boyle’s Law. You can input any type of units but you must be consistent. For example, you can’t use cubic inches for volume 1 and liters for volume 2.

Do you want to solve for:

You shouldn’t be intimidated by Boyle’s Law calculations because every Boyle’s Law word problem always gives you three of the four variables. Of those 3 variables, you have to determine which two “pair up”. (or which two of the three variables were measured at the same time) In order to get “paired” correctly, these get designated as “V1” and “P1” or “V2” and “P2”. Calling these “V1” and “P2” or “V2” and “P1” is totally incorrect.
So, let’s try our first problem.

1) Four liters of carbon dioxide have a pressure of 1.5 atmospheres. If the original pressure was .9 atmospheres, what was the original volume?

You can easily see that the four liters and 1.5 atmospheres are the two values which “pair up” (measured at the same time).
So V₂= 4 liters and P₂= 1.5 atm.
Yes, we could have called these “V1” and “P1”, but in either case, they are “paired up” properly.
We are also told the carbon dioxde’s previous pressure P₁ was .9 atmospheres and we are asked to find V₁ (the original volume).
(Yes, we could have called the previous pressure “P2”, and the original volume “V2”)
We can rearrange Boyle’s Law mathematically and obtain:
V₁= ( V₂• P₂) ÷ P₁     V₁= (4 ltr • 1.5 atm) ÷ .9 atm     V₁= 6.666… liters

As for using the calculator:
We are asked to find the previous volume, so we click on the “V1” button.
Yes, we could call the previous volume “V2”, and designate the present volume and pressure as “V1” and “P1”, but the important thing is to “pair up” the variables correctly.
When the 3 numbers are entered in the 3 boxes, make sure they are input into the correct boxes.
(In other words, do not enter 1.5 for “pressure 1” and .9 for “pressure 2”.)
Click “CALCULATE” and get your answer of 6.666… liters

2) 8 liters of a gas have a pressure of 760 torr. If the volume was originally 6 liters, what was the original pressure?

The first thing to do is to classify the data we are given.
We can see that V₂= 8 liters, P₂= 760 torr, V₁= 6 liters and we must solve for P₁
Solving Boyle’s Law for P₁ we get:
P₁= ( V₂• P₂) ÷ V₁     P₁= (8 ltr • 760 torr) ÷ 6 ltr     P₁= 1,013.333… torr

Using the calculator, we click on P₁ because this is the only variable that we don’t know.
After entering the 3 numbers into the correct boxes we click “CALCULATE” to get the answer of 1,013.3 torr

3) 1,000 cubic inches of air are under a pressure of 50 kilopascals. What is the volume if the pressure is increased to 130 kilopascals?

Classifying the data: V₁= 1,000 in³       P₁= 50 kPa       P₂= 130 kPa     and we must solve for V₂
Solving Boyle’s Law for V₂ we get:
V₂= ( V₁• P₁) ÷ P₂     V₂= (1,000 in³ • 50 kPa) ÷ 130 kPa       V₂= 384.615… in³

Using the calculator, we click on V₂ because we are calculating the present volume.
Entering the 3 numbers into the correct boxes then clicking “CALCULATE”, we get the answer of 384.62 in³

4) 3 gallons of argon were at a pressure of 14 pounds per square inch. A pressure change then reduces the volume to 2.2 gallons. What is the new pressure?

Classifying the data: V₁= 3 gallons       P₁= 14 psi       V₂= 2.2 gallons     and we must solve for P₂
Solving Boyle’s Law for P₂ we get:
P₂= ( V₁• P₁) ÷ V₂     P₂= (3 gallons • 14 psi) ÷ 2.2 gallons     P₂= 19.0909… psi

Using the calculator, we click on P₂ because we are being asked to solve for the present pressure.
Entering the 3 numbers into the correct boxes then clicking “CALCULATE”, we get the answer of 19.091 psi

The default setting is for 5 significant figures but you can change that by inputting another number in the box above.

Answers are displayed in scientific notation and for easier readability, numbers between .001 and 1,000 will be displayed in standard format (with the same number of significant figures.)
The answers should display properly but there are a few browsers that will show no output whatsoever. If so, enter a zero in the box above. This eliminates all formatting but it is better than seeing no outp