# Charles’s Law

Charles’s Law says the gas volume is directly related to gas temperature. As the temperature of a gas increases, so does its volume. And as the volume of a gas increases, so does its temperature.

V_{1} is the initial volume, T_{1} is the initial temperature and V_{2} is the new volume, and T_{2} is the new temperature.

Here are some examples of gas volume being directly related to gas temperature.

The volume of the above balloon will expand as the flask is heated. The hot air balloon works by that same principle. Hot air balloons are inflated by firing up the air inside of them, thereby increasing their volume.

The following graph illustrates the direct relationship between volume and temperature as described by Charles’s Law.

As temperature increases, so does volume.

Let’s look at some sample problems.

**Sample Problem**

The volume of a gas at 25 ºC is 250 ml. Find its volume at standard temperature if pressure is held constant.

Since pressure is held constant, we’re not going to worry about it. Pressure will be the same before and after the temperature change. So what is the temperature change? The problem asks us to find its new volume at standard temperature. Standard temperature is actually a value. Standard temperature is equal to 0 ºC. And, since this is a gas law problem, and ALL gas law problems must be computed with the Kelvin temperature scale, we will use 273.15 K as standard temperature.

Charles’s Law is as follows:

We are given a temperature of 25 ºC, which we’ll call T_{1}, and a volume at that temperature of 250 mL, which we’ll call V_{1}. We want to find out the new volume V_{2} when the temperature is changed to standard temperature, 273.15 K.

T_{1} = 25 ºC

T_{2} = 273.15 K

V_{1} = 250 mL

V_{2} = ?

Before we do anything, we need to convert all temperatures to the Kelvin scale. Therefore, we need to convert 20 ºC to K. We do this by adding to it 273.15.

T_{1} = 25 ºC = 298.15 K

T_{2} = 273.15 K

V_{1} = 250 mL

V_{2} = ?

To solve for V_{2} we need to isolate the variable. We’ll do that by first multiplying each side by T_{2}.

T_{2} cancels out on the right hand side and our equation becomes:

Plugging in our values for V_{1}, T_{2} and T_{1}, we get:

*The new volume is 229 mL.*

Here’s another sample problem.

**Sample Problem**

5.00 L of a gas is collected at 100. K and then allowed to expand to 20.0 L. What is the new temperature in order to maintain the same pressure?

We are given an initial volume of 5.00 L, so that will be V_{1}. The temperature at that initial volume is T_{1}, 100. K. The gas is then allowed to expand to 20.0 L, which is V_{2}.

V_{1} = 5.00 L

T_{1} = 100. K

V_{2} = 20.0 L

T_{2} = ?

Charles’s Law is:

To solve for T_{2}, we first need to get T_{2} out of the denominator. We do that by cross-multiplying.

We get:

Finally, to isolate T_{2}, divide each side by V_{1}:

And we get:

Plugging in our values for T_{1}, V_{2} and V_{1}, we get:

*The new temperature required to maintain the same temperature is 400. K.*