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Gas Laws: Charles’ Law

Core Concepts

In this article, you will learn how temperature relates to volume, and how to use Charles’ Law to solve problems regarding the change in temperature and volume of a system.

Topics Covered in Other Articles

Important Things to Consider

The gas law described in this article only applies to ideal gases, which you can read about on our article, The Ideal Gas Law.

Increasing Temperature

Consider a piston containing 1 liter of gas. Because the piston is free to change volume, the pressure remains constant.

Now, consider what happens when we increase the temperature of the gas. From Kinetic Molecular Theory, we know that the temperature of a sample of gas is directly related to its velocity.

From our article on pressure, What is Pressure, you know that the pressure of a gas is a result of force exerted when colliding with a wall. This force, as stated in the article on pressure, can be calculated with the equation F=m\cdot2v\sin(\theta).

As temperature increases, the value of v increases, and therefore the force increases. However, because the piston is free to change volume, the pressure remains constant. From Boyle’s Law, we know that pressure is inversely proportional to volume, and thus the piston would increase its volume to account for the change in temperature.

Charles’ Law

French physicist Jacques Charles also noticed this phenomenon, and discovered the relation between temperature and volume. This is what is know as Charle’s Law, and can be written as: V\propto T.

Applications

Because of this proportionality, we can use Charle’s law to see how a system would react to a change in temperature, or how a system would react to a change in volume (both with constant pressure). As temperature increases, volume increases, and vice versa.

Example Problem

Here’s how you can solve a problem regarding Charles’ Law.

What is the change in temperature when a 2.00 L piston of oxygen gas at 21 °C is compressed to 1.00 L.

     \begin{align*} V&\propto T\\ \implies \frac{V}{T}&=c\\ \implies \frac{V_1}{T_1}&=\frac{V_2}{T_2}\\ \therefore \frac{2.00\text{ L}}{(21+273)\text{ K}}&=\frac{1.00\text{ L}}{(x+273)}\\ \therefore x&=\frac{(21+273)\text{ K}}{2.00\text{ L}}\cdot1.00\text{ L}-273\\ x&=-126\text{ $\degree$C}\\ T_{f}&=T_{i}+\Delta T\\ -126\text{ $\degree$C}&=21\text{ $\degree$C}+\Delta T\\ \implies \Delta T&=\boxed{-147\text{ $\degree$C}} \end{align*}

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