Absolute temperature is ALWAYS required with Charles’ Law.
(either Kelvin or the less common Rankine scale can be used).
For a Temperature Converter, click here.

V1          V2
——   =   ——
T1          T2

The above formula is Charles’ Law, named after the French experimenter Jacques Charles (1746-1823).

It states that the volume of a fixed mass of gas at a constant pressure is directly proportional to its absolute temperature.

Stating this in plain English,
when temperature increases, volume increases.

When temperature decreases, volume decreases.

This calculator can solve for any one of the four variables of Charles’ 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:

Similar to Boyle’s Law, every Charles’ Law word problem always gives you three of the four variables you will need. 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 “T1” or “V2” and “T2”. Pairing these as “V1” and “T2” or “V2” and “T1” is totally incorrect.
So, let’s try our first problem.

1) Five liters of a gas have a temperature of 250 Kelvin. If the previous temperature was 300 Kelvin what was its volume ? (with pressure remaining constant)

The two values which “pair up” (measured at the same time) are five liters (V₂) and 250K (T₂)
We are also told the previous temperature (T₁) was 300 degrees Kelvin and we are asked to find V₁ (the original volume).
We can rearrange Charles’ Law mathematically and obtain:
V₁= ( V₂ • T₁) ÷ T₂     V₁= (5 ltr • 300°K) ÷ 250°K     V₁= 6 liters

As for using the calculator:
We are asked to find the previous volume, so we click on the “V1” button.
When the 3 numbers are entered in the 3 boxes, we make sure they are input into the correct boxes.
(In other words, do not enter 250 for “temperature 1” and 300 for “temperature 2”.)
Click “CALCULATE” and get your answer of 6 liters.

2) 3.1 cubic meters of a gas have a temperature of 15° Celsius. What temperature is required to increase the volume to 3.5 cubic meters? (with pressure remaining constant)

As it says at the top of this page, “Absolute temperature is always required with Charles’ Law”.
The Celsius scale is not an absolute temperature scale and so we are required to change this to Kelvin which is an easy conversion only requiring the addition of 273.15 to the Celsius temperature. So, 15° Celsius equals 288.15 Kelvin.
The two values which “pair up” (measured at the same time) are 3.1 cubic meters (V₂) and 288.15K (T₂), which leaves 3.5 cubic meters as (V₁)
Solving Charles’ Law for T₁ we get:
T₁ = V₁ • T₂ ÷ V₂     T₁ = 3.5 m³ • 288.15K ÷ 3.1 m³     T₁ = 325.33 K

Using the calculator:
We want to solve for T1 so we click that button.
Looking at the previous paragraph, we see what the values are for V1, V2 and T2 and we then enter those values into the correct boxes.
Clicking on “CALCULATE” we get the answer of 325.33 liters

3) 1,300 cubic inches of a gas are at a temperature of 78° Fahrenheit. If the temperature increases to 90° F, what is the new volume?

Since we are using Charles’ Law and since we are given Fahrenheit temperatures, we must convert these to the Rankine scale. Similar to the Celsius to Kelvin conversion, converting Fahrenheit to Rankine just requires simple addition. In this case you add 459.67 to the Fahrenheit temperature to convert it to Rankine.
T₁ = 78° F + 459.67 = 537.67 Rankine
T₂ = 90° F + 459.67 = 549.67 Rankine
And the two values which “pair up” are 1,300 in³ (V₁) and 549.67 Rankine (T₂). The third variable we are given is 537.67 Rankine which gets designated as T₂ and so the remaining variable (V₂) is the one we must calculate.
Solving Charles’ Law for V₂ we get:
V₂ = V₁ • T₂ ÷ T₁     V₂ = 1,300 in³ • 549.67 R ÷ 537.67 R     = 1,329 in³

Using the calculator:
We want to solve for V2 so we click that button.
Looking at the previous paragraph, we need to convert both of the Fahrenheit temperatures to Rankine. We then enter the values of V1, T1 and T2 into the correct boxes.
Clicking on “CALCULATE” we get the answer of 1,329 in³

4) 7 liters of a gas are at a temperature of 300 K. If the volume increases to 7.5 liters, what is the new temperature of the gas?

The two values that “pair up” are 7 liters (V₁) and 300 Kelvin (T₁). We are then given the new volume of 7.5 liters (V₂) and have to determine the new temperature (T₂).
We must solve Charles’ Law for T₂
T₂ = T₁ • V₂ ÷ V₁     T₂ = 300 K • 7.5 liters ÷ 7.0 liters     T₂ = 321.43 K

Using the calculator:
We click on the T2 button.
We then enter V1, T1 and T2 into the correct boxes.
Clicking on “CALCULATE”, we get T2 = 321.43 K

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 output at all.