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×07th Nov 2019 @ 4 min read

The equation of Charles's law is *V* = *kT*.

As we can see from the above equation, the law relates the volume of gas to its temperature. The law was discovered by Jacques Charles in the late 1700s. It states the volume of a gas is directly proportional to its temperature unless the pressure and the amount of the gas remain constant.

Let the volume of a gas be *V* and temperature, *T*. According to the law at a constant pressure for a fixed amount of gas,

The above expression is the mathematical expression of Charles's law. Removing the proportionality,

Here, *k* is a proportional constant. Finally, rearranging the above equation,

Thus, the ratio of volume to temperature is constant for a constant pressure and a fixed amount of gas

Note: The above expression is valid unless pressure and amount of gas are constant. Also, temperature is in an absolute scale (in the kelvin or rankine).

We can express the law in the celsius scale. The celsius scale uses the freezing point of water (0 °C) and the boiling point of water (100 °C) as reference points. The celsius is related to the kelvin as follows:

Let *V*_{0} be the volume of gas at 0 °C i.e. the freezing point of water. And *V* is the volume at temperature *t* (in °C). As per Charles's law,

Substituting *T* = *t* + 273.15 and *T*_{0} = *t*_{0} + 273.15,

We also know, *t*_{0} = 0 °C.

Rearranging the above expression,

This is the equation of Charles's law in the celsius scale.

Note: When temperature is expressed in the celsius scale, a small *t* is used. But when temperature is expressed in the kelvin scale, a capital *T* is used.

Let *V*_{1} and *V*_{2} be volume at temperature *T*_{1} and *T*_{2} such that pressure and amount of gas is constant. *V*_{1}, *T*_{1} is volume and temperature at condition 1, and *V*_{2}, *T*_{2} is volume and temperature at condition 2.

We can establish the relationship between the two conditions using the law.

Combining both equations,

Using the above formula, we can determine unknown volume or temperature if volume and temperature for a condition are known. Let take an example.

Consider an 130 g helium balloon. The balloon is fastened to a taffrail of a ship moving from Miami towards New York City. The volume the balloon at Miami is 1.50 L and the temperature, 27 °C. What will be the volume of the balloon at New York City if the temperature at New York City is 10 °C? Assume the pressure remains 1 atm.

First, we convert the temperatures in the kelvin from the degree celsius.

The initial volume is 1.50 L. The final volume can be calculated from Charles's law.

Thus, the volume of the balloon at New York City is 1.42 L.

If you have noticed, the volume decreased from 1.5 L to 1.42 L when the temperature of the balloon decreased from 27 °C to 10 °C. This explains Charles's law that volume is directly proportional to temperature.

We can generalise the law to any number of conditions. Consider a fixed amount of gas of pressure *P*. Let (*V*_{1}, *T*_{1}), (*V*_{2}, *T*_{2}), (*V*_{3}, *T*_{3}) … (*V _{i}*,

As per the law, the ratio of volume to temperature remains constant.

Consider the same ship in the above example. The ship after reaching New York City is scheduled to sail towards Saint John, Canada. The temperature at Saint John is 3 °C and pressure, 1 atm. Determine the volume of the balloon at Saint John.

First, we have to convert the temperature in the kelvin from the degree celsius. 3 °C is 3 + 273.15 = 276.15 K.

As per the law,

Here, *V*_{1} is the volume at Miami, *V*_{2} is the volume at New York City, which we have already calculated in the previous example, and *V*_{3} is the volume at Saint John. Rearranging the above expressions,

We can use any one of two to solve for *V*_{3}.

The volume at Saint John is 1.38 L.

As observed from the above calculations, the volume decreases even further (1.42 L to 1.38 L) with a decrease in temperature (10 °C to 3 °C). Therefore, the law holds under multiple conditions.

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Thanks for your response!

Matthew

21st Jul 2020

21st Jul 2020

Clap Clap Clap ... so easy to understand the concepts... the way you illustrate them with the help of diagrams and examples... great work, thank you !

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