Charles' Law

02nd Apr 2019 @ 10 min read

Charles' law is also known as the law of volumes. The law describes the relationship between the volume and the temperature of a gas. It is a very important law studied in chemistry and physics along with Boyle's law, Gay-Lussac's law, and Avogadro's law.

Statement

Charles' law states for a fixed amount of an ideal gas its volume is directly proportional to its temperature at constant pressure.

Equation

The equation of the law is PV = k. Here, k is a constant.

$V = k\times T$

History

This law is named after French scientist Jacques Charles. Jacques Alexandre César Charles was a scientist, inventor, and balloonist.

It was around 1787 when Charles studied the behaviour of gases with temperatures. In his experiments, he observed that the volume of different gases would increase by the equal volume when subjected to a rise in temperature. He conducted his experiment on five balloons. When the temperature of the balloons was raised, the volume of each increased by an equal amount. But his discovery remained unknown to the world until Joseph Louis Gay-Lussac in 1802 published the discovery. Gay-Lussac confirmed and extended the results of Charles and credited the discovery to him.

Explanation

As per the law, the volume of a gas is directly proportional to its temperature for a fixed amount of the gas at constant pressure. The statement can mathematically be expressed as:

$V \propto T$

By removing the proportionality,

$V = k \times T$

Here, k is a constant of proportionality, V is the volume, and T is the absolute temperature.

Rearranging the above expression,

$\frac{V}{T} = k$

Note: In the above expression temperature is in an absolute scale (in the kelvin or degree rankine).

The above expression is valid for a fixed amount of a gas and at constant pressure. Hence, as the temperature increases, the gas expands and as the temperature decreases, the gas contracts. We can also conclude from the above equation the ratio of the volume to absolute temperature is independent of temperature or volume at constant pressure.

Illustration:1 Sealed syringe

A sealed syringe is a classic example of Charles' law. When the tip of a syringe is sealed, gas inside the cylinder gets isolated. Thus, it acts like a piston.

Charles' law syringe example — The volume of a sealed syringe reduces when the temperature (thermometer) decreases under constant pressure (pressure gauge).

When the syringe is dipped into icy water, the temperature the air inside the syringe decreases. Consequently, the volume of the air also decreases as per the law. The pressure remains constant equal to atmospheric pressure.

We can see the above diagram, as the temperature in the thermometer falls, the plunger of the syringe moves inwards i.e. the volume decreases. This observation when plotted, will give a straight line as above. A sealed syringe follows Charles' law.

Illustration 2: Flexible chamber

Consider an enclosed chamber (below figure) with its boundary flexible i.e. it can expand or shrink. The figure below consists of a chamber with a piston on it, a water bath to heat the chamber, an electric heating source.

Charles' law example: flexible chamber — A flexible chamber follows Charles' law.

When the air inside the chamber is heated with a heating source via water, the temperature of the air increases i.e. the kinetic energy associated with the molecules of the air increases. This momentary increases the pressure inside the chamber. Now, the molecules of the air exert more force in the outwards direction on the resting piston, and the air inside the chamber expands. By the same analogy, when the air is cooled, the volume of the air shrinks. Therefore, it obeys the law.

For a given amount ideal gas, V₁, T₁ and V₂, T₂ are the volumes and temperatures of condition 1 and condition 2 at constant pressure. From Charles' law,

$\frac{V_1}{T_1}= \frac{V_2}{T_2}$

This formula is very useful. We can calculate the volume or temperature at any unknown condition if the volume and temperature at a condition is known.

Example 1

Statement: If the absolute temperature of a gas is doubled at constant pressure. Determine the change in the volume?

Solution: The absolute temperature of the gas is doubled. Thus, T₂ = 2T₁.

According to the law,

$\frac{V_1}{T_1}= \frac{V_2}{T_2}$

Substituting T₂ = 2T₁,

$V_2&=\frac{V_1T_2}{T_1}\\&=\frac{V_1 \times 2T_1}{T_1}\\&=2V_1$

Thus, the volume gets double when the temperature of a gas is doubled.

Example 2:

Statement: The initial and final temperature of neon is 289 K and 323 K. Find the initial volume if the final volume is 2.3 L?

Solution: T₁ = 289 K, T₂ = 323 K, and V₂ = 2.3 L.

According to the law,

$V_1&=\frac{V_2T_1}{T_2}\\&=\frac{2.3 \times 289}{323}\\&=2.0\,\text{L}$

Therefore, the final volume is 2.0 L.

To know more, continue reading on: The equation of Charles' law »

Alternative version of Charles' law

Charles from his experiments concluded that at constant pressure, the volume of a fixed amount of a gas increases or decreases by ¹⁄₂₇₃ (now ¹⁄_273.15) times the volume at 0 °C for every 1 °C rise or fall in temperature. This statement can mathematically be expressed as:

$V - V_0= \frac{1}{273.15} \times V_0 \times t$

Here, V is the final volume or the volume at T, V₀ is the volume at 0 °C, t is the temperature expressed in degree Celsius.

Rearranging the above expression,

$V =\frac{1}{273.15} \times V_0 \times t + V_0 =V_0 \left(1+\frac{t}{273.15} \right)$ $V =V_0 \left( \frac{273.15+t}{273.15} \right) =V_0 \left( \frac{T}{273.15} \right) =V_0 \left( \frac{T}{T_0} \right)$

Here, T is the temperature in the kelvin, and T₀ is the temperature in the kelvin equal to 273.15 K.

$\frac{V_T}{V_0} = \frac{T}{T_0}$

Thus, from the above equation, we arrive at Charles' law,

Charles' law and absolute zero

As per Charles' law when the absolute temperature approaches zero, the volume of the gas should approach zero. According to Gay-Lussac, this could only happen if the gas is an ideal gas.

Keep reading: Charles' law and absolute zero »

Graphical representation

The graphical representation of Charles' law is demonstrated in the below graphs.

Charles' law graph — The graph of volume vs temperature (in K) is a straight line passing through the origin.

From the above graph, the volume increases linearly with an increase in the temperature. The temperature is in an absolute scale. It can also be observed that as the temperature approaches zero, the volume also approaches zero. All the above four lines are plotted at a constant pressure i.e isobaric lines.

The above line is plotted from using the below equation. The x-intercept in the above figure is −273.15 °C and the y-intercept is V₀.

$V_T =V_0+\frac{V_0}{273} t$

Keep reading: Graphs of Charles' law »

Limitations

The limitations of Charles' law are as follows:

Charles' law is applicable to only ideal gases.
Charles' law holds good for real gases only at high temperatures and low pressures.
The relationship between the volume and temperature is not linear in nature at high pressures. At high pressures, the repulsive forces among the molecules are significantly increased which causes the expansion of the volume.

Real-world applications

Applications of Charles' law can be observed in day-to-day life. Below are some of them.

Hot air balloons

We all have seen hot air balloons flying. But has anyone wondered how it works? An air balloon consists of a bag, a basket to carry passengers, and a source of heat. The heat source is usually a fuel like propane which on burning causes the heating of the air surrounding it. When the air is heated, as per Charles' law, results in expansion of air i.e. an increase in volume. And with an increase in volume, the density of the air decreases. Hence, the buoyant force acts in upwards direction on the balloon, and it flies.

Tyres

In the summer season, sometimes tyres of vehicles get overinflated. It is because of Charles' law. When a vehicle is left untouched for a prolonged period. As the summer season approaches, there is a rise in temperature which causes the air inside tyres to expand.

Human lungs

Jogging is very difficult in colder days compare to hotter days. The reason is simply due to low temperatures our lungs shrink which decrease the human lungs' capacity.

Pool floats

When inflated pool floats are pushed into pools, they appear as under-inflated. This is because the temperature of the pool is at a lower temperature than the surrounding atmosphere. Thus, pool floats shrink by Charles' law.

Bakery

Yeast is used in preparation of many bakery products. This yeast keeps liberating carbon dioxide gas. When preprocessed cakes and bread are heated, the carbon dioxide gas expands which makes our cakes and bread fluffy.

To know in more detail check: Examples of Charles' law »

Laboratory experiment

We can verify Charles' law experimentally and determine the value of absolute zero. There are many standard experiments to prove it. However, we will briefly summarise one of them in this article.

The primary apparatus of the experiment consists of a conical flask and a beaker. The empty flask is submerged into the water-filled beaker as shown in the above diagram. When the heat is supply to the beaker by a burner, it also heats the air in the flask. As a consequence, the air inside the flask expands. This is our condition 1. The flask is later dipped in a water tank at room temperature. Now, the air in the flask contracts since the temperature is decreased. This is condition 2. By knowing the temperature and volume at both conditions, we can verify the law.

The entire experiment with procedure, observation, and calculation is discussed in the article: To verify Charles' law experimentally.

Check another experiment: To verify Charles' law by syringe experiment »

Problems

Problem 1

Statement: Consider a fixed amount of propane gas at a temperature of 30 °C and a volume 1.5 m³. The gas undergoes expansion such that the initial and final pressures remain the same. The new temperature of the gas is 75 °C. Calculate the new volume?

Solution: First, convert both temperatures in the problem statement from the degree celsius to the kelvin.

$T_1 =30+273.15 =303.15 \, \text{K}$ $T_2 =75+273.15 =348.15 \, \text{K}$

As from Charles' law at constant pressure and for a given amount a gas,

$\frac{V}{T} =k$ $\frac{V_1}{T_1} =\frac{V_2}{T_2}$ $V_2 = \frac{V_1}{T_1} \times T_2 = \frac{1.5}{303.15} \times 348.15= 1.7 \, \text{m}^3$

Therefore, the new volume after the expansion is 1.7 m³.

To ease the calculations, try out: Charles' law calculator »

Problem 2

Statement: A hot air balloon of volume 2.00 L is cooled from a temperature of 65 °C to room temperature. The final volume is 1.75 L. Estimate the room temperature?

Solution: The initial temperature (T₁) is

$T_1=65+273.15=338.15 \, \text{K}$

From Charles' law,

$\frac{V_1}{T_1}= \frac{V_2}{T_2}$ $T_2 =T_1 \times \frac{V_2}{V_1} =338.15 \times \frac{1.75}{2.00} = 295.9 \,\text{K}$ $T_2 = 295.9-273.15\approx 23 \, \degree \text{C}$

Hence, the room temperature is 23 °C.

Problem 3

Statement: A sample of nitrogen gas is placed in a mixture of ice and water at 1 atm pressure. The sample occupies 0.80 dm³ of space. The same sample is transferred into boiling acetone where it occupies the volume of 0.96 dm³. Calculate the boiling point of acetone?

Solution: The pressure is 1 atm. At this pressure, the mixture of ice and water exists only at 0 °C i.e. 273.15 K, which is the initial temperature.

From Charles' law,

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ $T_2 =T_1 \times \frac{V_2}{V_1} =273.15 \times \frac{0.96}{0.80}\approx 328 \, \text{K}$ $T_2 = 328-273.15\approx 55 \, \degree \text{C}$

Thus, the boiling point of acetone is 55 °C.

Problem 4

Statement: A balloon tied on a ship moving from the Arctic Ocean to the Atlantic Ocean arrives Miami, USA. The temperature in the Arctic Ocean is −2 °C and the corresponding volume is 1.50 L. At Miami, the temperature is 28 °C. Determine the volume occupied by the balloon at Miami?

Solution: T₁ = −2 °C, V₁ = 1.50 L, and T₂ = 28 °C.

From Charles' law,

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ $V_2 =V_1 \times \frac{T_2}{T_1} &=1.50 \times \frac{28+273.15}{-2+273.15} \\&= 1.5 \times \frac{301.15}{271.15} \\&\approx 1.67 \, \text{L}$

Therefore, the volume occupied by the balloon at Miami is 1.67 L.

Practice Problems

Problem 1

The initial and final temperature is 4 °C and 12 °C. Find the final volume if the initial volume is 230 mm³?

Problem 2

A gas shrinks from an initial volume of 9.2 dm³ to 8.5 dm³. Find the final temperature if the initial temperature is 45 °C?

Problem 3

A helium balloon expands under constant pressure. The initial and final temperature is 300 K and 430 K. Determine the initial volume if the final volume is 1.2 L.

For more problems, check: Charles' law worksheet »

Summary

According to Charles' law, the volume and the temperature of a fixed amount of gas is directly proportional to each other at constant pressure.
The equation of the law is V = kT.
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ is the formula at two different conditions.
We can determine the absolute zero through an experimental procedure.
The volume-temperature graph is a straight line with a positive slope.
The law valid for only ideal gases. It fails at high pressure/low temperature.