Gas laws are used to study the behaviour of gases. The state variables of a gas, such as pressure, volume, and temperature, show the true nature of the gas. Scientists created the gas laws near the end of the 18th century (the individual laws are named). Boyle’s Law, Charles’ Law, Gay-Law, Lussac’s Avogadro’s Law, and the Combined Gas Law are the five gas laws.

When conditions are normal, all gases appear similar. Minor changes in physical parameters such as pressure, temperature, or volume cause variations. In this article, let’s study everything about the Gas Laws and their graphical representation in detail.

What are the Gas Laws?

The gas laws are a set of rules that control how gases behave in the following conditions

Amount of space occupied by gas.
The force that a gas exerts on the container’s walls.
The gas’s absolute temperature.
The number of moles of gas

Boyle’s Law: Pressure-Volume Relationship

This law was given by Robert Boyle in \(1662\) which states that:

At constant temperature, the pressure of a fixed amount (i.e., number of moles, n) of a gas varies inversely with its volume.

Mathematically,

\({\rm{p}} \propto \frac{1}{{\rm{V}}}\)  (at constant \({\rm{T}}\) and \({\rm{n}}\))

Or,

\({\rm{p}} = {\rm{k}}\frac{1}{{\rm{V}}}\)

On rearranging, \({\rm{pV}} = {\rm{k}}\)

Here, \({\rm{k}} = \) proportionality constant, the value of which depends upon the amount of the gas, temperature of the gas and units of pressure and volume.

If temperature \({\rm{T}}\) remains constant, the process is called the isothermal process.

Thus, the law can also be given as, at a constant temperature, the product of pressure and volume of a fixed amount of a gas is constant.

If \({{\rm{V}}_1}\) is the volume of a specific amount of gas at a pressure \({{\rm{p}}_1}\) and at the same temperature, \({{\rm{V}}_2}\) is the volume of the same mass of gas at a pressure \({{\rm{p}}_2}\), then:

\({{\rm{p}}_1}{{\rm{V}}_1} = {{\rm{p}}_2}{{\rm{V}}_2}\)

Or,

\(\frac{{{{\rm{p}}_1}}}{{{{\rm{p}}_2}}} = \frac{{{{\rm{V}}_2}}}{{{{\rm{V}}_1}}}\)

Graphical Representation of Boyle’s Law

Boyle’s law can be graphically verified in the following ways:
i. Pressure-Volume graph- When the pressure of a given mass of a gas is plotted against volume at a constant temperature, the equilateral hyperbola is obtained:

ii. Pressures vs 1/V Graph- On plotting the reciprocal of volumes of a fixed mass of gas against pressure, we get straight lines passing through the origin.

iii. pV vs p Graph- When the product of pressure and volume is plotted against pressure at different temperatures, the straight line parallel to the pressure axis are obtained.

Real-life Examples of Boyle’s Law

Below are some real-life instances or applications of Boyle’s law:
i. Syringe– The amount of the fluid increases as the plunger is pulled. It causes a momentary decrease in fluid pressure, and external fluid is withdrawn as a result.

ii. Human lungs– The pressure drops for a brief moment as the lungs expand. As a result, the internal pressure will be lower than the external pressure. As a result, the ambient air enters the body.
iii. When the lungs relax, the volume of the lungs decreases, causing the pressure relative to the outside to increase for a shorter period of time. The body then exhales the air.

iv. Bicycle pump– When a bicycle’s pump handle is pulled down, the pressure inside the pump rises-  the gas inside has been compressed. As a result, pressurized gas is pumped into the vehicle’s tyre.

v. Air bubbles– As they rise in water, air bubbles expand. The surrounding pressure of the liquid lowers as the bubbles rise. The air bubbles expand according to Boyle’s law.

Solved Examples on Boyle’s Law

Example 1: A hydrogen balloon of \(0.50\;{\rm{L}}\) at a pressure of \(101.325\,{\rm{kPa}}\) expands isothermally to a final pressure of \(80.121\,{\rm{kPa}}\). Determine the final volume?

Solution: Initially, \({{\rm{p}}_1} = 101.325\,{\rm{kPa}}\) and \({{\rm{V}}_1} = 0.50\;{\rm{L}}\) and the final volume \({{\rm{p}}_2} = 80.121\,{\rm{kPa}}\).

According to the law,
\({{\rm{p}}_1}{{\rm{V}}_1} = {{\rm{p}}_2}{{\rm{V}}_2}\)

\({{\rm{V}}_2} = \frac{{{{\rm{p}}_1}{{\rm{V}}_1}}}{{{{\rm{p}}_2}}} = \frac{{101.325\,{\rm{kPa}} \times 0.50\;{\rm{L}}}}{{80.121\,{\rm{kPa}}}}\)

\( = 0.63{\rm{L}}\)

Therefore, the final volume of the balloon is \(0.63{\rm{L}}\).

Example 2: What will be the pressure required to reduce \(600\,{\rm{ml}}\) of dry gas at \(750\,{\rm{mm}}\) of Hg pressure to \(500\,{\rm{ml}}\) at the same temperature?

Solution: \({{\rm{V}}_1} = 600\,{\rm{ml}},{{\rm{V}}_2} = 500\,{\rm{ml}},{{\rm{p}}_1} = 750\,{\rm{mm}}\) of \({\rm{Hg}},\,{{\rm{p}}_2} = \)?

\({{\rm{p}}_1}{{\rm{V}}_1} = {{\rm{p}}_2}{{\rm{V}}_2}\)

\({{\rm{p}}_2} = \frac{{{{\rm{p}}_1}{{\rm{V}}_1}}}{{{{\rm{V}}_2}}} = \frac{{750\;{\rm{mm}} \times 600\,{\rm{ml}}}}{{500\,{\rm{ml}}}} = 900\;{\rm{mm}}\)

Therefore, the pressure required will be \(900\;{\rm{mm}}\) of \({\rm{Hg}}\).

Charle’s Law: Temperature-Volume Relationship

This law is named after Jacques Charles, a French scientist- and inventor and balloonist. Jacques Alexandre César Charles, around \(1787\), began researching the behaviour of gases as a function of temperature.

According to Charle’s Law:

At constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature.

All gases obey this law at very low pressures and high temperatures.

According to Charles’ experiments, the volume of a specific amount of a gas increases or decreases by \(273\) (now \(273.15\)) times the volume at \(0{}^{\rm{o}}{\rm{C}}\) for every \(1{}^{\rm{o}}{\rm{C}}\) rise or fall in temperature at constant pressure. This statement can be expressed mathematically as:

\({\rm{V}} – {{\rm{V}}_0} = \frac{1}{{273.15}} \times {{\rm{V}}_0} \times {\rm{t}}\)

Here,

1. \({\rm{V}}\) is the final volume or the volume at \({\rm{T}}\),
2. \({{\rm{V}}_0}\) is the volume at \(0{}^{\rm{o}}{\rm{C}}\),
3. \({\rm{t}}\) is the temperature change expressed in degrees Celsius.

Rearranging the above expression:

Here, \({\rm{T}}\) is the final temperature in kelvin and \({{\rm{T}}_0}\) is the temperature in kelvin which equals \(273.15{\rm{ K}}\).

Mathematically,

\({\rm{V}} \propto {\rm{T}}\)

Or,

\(\frac{{\rm{V}}}{{\rm{T}}} = \)constant

The constant pressure process is referred to as an isobaric process.

If \({{\rm{V}}_1}\) is the volume of a gas at a temperature \({{\rm{T}}_1}\) and \({{\rm{V}}_2}\) is the volume of a gas at a temperature \({{\rm{T}}_2}\) at the same pressure, then according to Charles’ law,

\(\frac{{{{\rm{V}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{V}}_2}}}{{{{\rm{T}}_2}}}\)

Graphical Representation of Charle’s Law

The plot of volume vs temperature at constant pressure is given by the following graphs:

When the volume of a fixed mass of gas is plotted with the temperature at different constant pressures, we get a straight line in each case.

On extending these lines to zero volume, all these intercept the temperature axis at \( – {273.15^{\rm{o}}}{\rm{C}}\).

Using the equation below, the above line was plotted. In the diagram above, the \({\rm{x}}\)-intercept is \( – {273.15^{\rm{o}}}{\rm{C}}\) and the y-intercept is \({{\rm{T}}_0}\).

Limitations of Charle’s Law

i. Only ideal gases are subject to Charles’ law.
ii. Only at high temperatures and low pressures does Charles’ law apply to actual gases.
iii. At high pressures, the relationship between volume and temperature is not linear. The repulsive forces between molecules are considerably increased at high pressures, causing the volume to expand.

Real-life Examples of Charle’s Law

Some of the day-to-day life applications of Charle’s Law are given below-
i. Hot air balloons- According to Charles’ law, as air is heated, it expands, resulting in an increase in volume. And as the volume of the air increases, the density of the air decreases. As a result, the buoyant force acts upwards on the balloon, causing it to fly.
ii. Tyres- During the summer, vehicle tyres can become overinflated.
iii. Bakery- Yeast is used to making a variety of bakery goods. This yeast continues to produce carbon dioxide gas. The carbon dioxide gas expands when preprocessed cakes and bread are heated, making our cakes and bread fluffy.

Solved Examples on Charle’s Law

Example 1: The initial and final temperature of neon is \(289\;{\rm{K}}\) and \(323\;{\rm{K}}\). Find the initial volume if the final volume is \(2.3\,{\rm{L}}\)?

Solution: \({{\rm{T}}_1} = 289\;{{\rm{K}}_1}\;{{\rm{T}}_2} = 323\;{\rm{K}}\) and \({{\rm{V}}_2} = 2.3\;{\rm{L}}\)

According to Charle’s Law,

\(\frac{{{{\rm{V}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{V}}_2}}}{{{{\rm{T}}_2}}}\)

\({{\rm{V}}1} = \frac{{{{\rm{V}}_2}{{\rm{T}}_1}}}{{{{\rm{T}}{\rm{2}}}}} = \frac{{2.3{\rm{L}} \times 289\;{\rm{K}}}}{{323\;{\rm{K}}}} = 2.06\;{\rm{L}}\)

Therefore, the final volume is \(2.06\;{\rm{L}}\).

Example 2: A hot air balloon of volume \(2.00\;{\rm{L}}\) is cooled from a temperature of \(65{}^{\rm{o}}{\rm{C}}\) to room temperature. The final volume is \(1.75\;{\rm{L}}\). Estimate the room temperature?

Solution: \({{\rm{T}}_1} = 65 + 273.15 = 338.15\;{\rm{K}},\,{{\rm{V}}_1} = 2\;{\rm{L}},\;\,{{\rm{V}}_2} = 1.75\;{\rm{L}}\)

According to Charle’s Law,

\(\frac{{{{\rm{V}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{V}}_2}}}{{{{\rm{T}}_2}}}\)

\({{\rm{T}}_2} = \frac{{{{\rm{V}}_2}{{\rm{T}}_1}}}{{{{\rm{V}}_1}}} = \frac{{1.75{\rm{L}} \times 338.15\,{\rm{K}}}}{{2{\rm{L}}}} = 295.9\;{\rm{K}}\)

\({{\rm{T}}_2} = 295.9 – 273.15 = {22.73^{\rm{o}}}{\rm{C}}\)

Hence, the room temperature is \({22.73^{\rm{o}}}{\rm{C}}\).

Gay-Lussac’s Law: Pressure-Temperature Relationship

Amontons’ law is another name for Gay-Lussac’s law. This law is named after Joseph Louis Gay-Lussac, a French chemist. In \(1808\) he formulated this relationship.

This law states that:

At constant volume, the pressure of a fixed amount of a gas varies directly with the absolute temperature.

Mathematically,

\({\rm{P}} \propto {\rm{T}}\)

Or,

\(\frac{{\rm{P}}}{{\rm{T}}} = \)constant

If volume remains constant, the process is called the isochoric process.

If \({{\rm{P}}_1}\) is the initial pressure of a gas at an initial temperature \({{\rm{T}}_1}{\rm{K}}\) and \({{\rm{P}}_2}\) is the final pressure of a gas at a temperature \({{\rm{T}}_2}{\rm{K}}\), then according to Gay Lussac’s law,

\(\frac{{{{\rm{P}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{P}}_2}}}{{{{\rm{T}}_2}}}\)

Graphical Representation of Gay-Lussac’s Law

Gay-Lussac’s law can be demonstrated by the following graphs:

When \({\rm{p}}/{\rm{T}}\) is plotted against temperature, a straight line parallel to \({\rm{X}}\)-axis is obtained:

Limitations of Gay-Lussac’s Law

i. This law is only applicable to ideal gases.
ii. For real gases at high temperatures and/or low pressure, Gay-law Lussac’s holds true.
iii. At high pressures, the pressure-to-temperature ratio deviates. As the pressure is increased, the ratio lowers. This drop is due to an increase in volume at high pressures, which can be explained by an increase in repulsive forces among molecules.

Real-life Examples of Gay-Lussac’s Law

Some of the day-to-day life applications are given as follows:
i. Pressure cooker- When a pressure cooker is placed on a heat source (stove), the pressure of the fluid inside the cooker rises as the temperature rises.
ii. Aerosol Can- When exposed to sun or extreme temperatures, an aerosol can explode due to increasing pressure inside the can exceeding the threshold limit.
iii. Gun bullet- When a gun’s bullet is ignited, chemical processes turn the chemical energy stored in the bullet’s shell into heat. This heat raises the temperature, which raises the pressure.  The bullet is fired from the gun due to the high pressure.
iv. Tyres- The high temperature pressurizes the air inside the tyres, causing them to explode beyond a certain point.

Solved Examples of Gay-Lussac’s Law

Example 1: A soda bottle at the room temperature of \(25{}^{\rm{o}}{\rm{C}}\) and the pressure of \(2\,{\rm{atm}}\) is heated to the temperature of \(330{}^{\rm{o}}{\rm{C}}\) at which it bursts. Calculate the pressure of the heated soda bottle.

Solution: \({{\rm{T}}_1} = 25 + 273.15 = 298.15\;{\rm{K}},\,{{\rm{T}}_2} = 330 + 273.15 = 603.15\;{\rm{K}}\)

From Gay-Lussac’s law,

\(\frac{{{{\rm{P}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{P}}_2}}}{{{{\rm{T}}_2}}}\)

\({{\rm{P}}_2} = \frac{{{{\rm{P}}_1} \times {{\rm{T}}_2}}}{{{{\rm{T}}_1}}} = \frac{{2\,{\rm{atm}} \times 603.15\,{\rm{K}}}}{{298.15\,{\rm{K}}}} = 4.05\,{\rm{atm}}\)

Thus, the soda bottle bursts at the pressure of \(4.05\,{\rm{atm}}\).

Avogadro’s Law

This law was given by Italian scientist Amedeo Avogadro in \(1811\). According to this law,

Under the same conditions of temperature and pressure, equal volumes of all gases contain an equal number of molecules, irrespective of the nature of the gas.

This means that as long as the temperature and pressure remain constant, the volume depends on the number of molecules of the gas.

Mathematically,

\({\rm{V}} \propto {\rm{n}}\)

Or,

\(\frac{{\rm{V}}}{{\rm{n}}} = \)constant

Where n is the number of moles, i.e., \({\rm{n}} = \frac{{\rm{W}}}{{\rm{M}}}\).

Graphical Representation of Avogadro’s Law

The graphs are plotted at a constant temperature and pressure and can be represented as follows:
i. The volume and mole have a linear relationship with the line of a positive slope going through the origin, as seen in the graph.

ii.  \(\frac{{\rm{V}}}{{\rm{n}}}\) has a straight line graph parallel to the \({\rm{x}}\)-axis so \(\frac{{\rm{V}}}{{\rm{n}}} = \)constant. It signifies that the volume of one-mole gas remains constant and is unaffected by changes in the mole (or volume).

Real-life Examples of Avogadro’s Law

Some of the day-to-day applications are as follows:
i. Balloons- While blowing up a balloon, we are filling more moles of air in the balloon, and it expands.

ii. Tyres– When air is pumped into deflated tyres at a gas station, the amount of gas (moles) inside the tyres is increased, increasing the volume and inflating the tyres.
iii. Human lungs- When we inhale, air enters into our lungs, expanding them, whereas when we exhale, air flows out of our lungs, shrinking them.

Solved Example on Avogadro’s Law

Example 1- Consider \(20\) moles of hydrogen gas at temperature \(0{}^{\rm{o}}{\rm{C}}\) and pressure \(1\,{\rm{atm}}\) having a volume of \(44.8\,{\rm{d}}{{\rm{m}}^3}\). Calculate the volume of \(50\) moles of nitrogen gas, at the same temperature and pressure?

Solution: \({{\rm{V}}_1} = 44.8\,{\rm{d}}{{\rm{m}}^3},\,{{\rm{n}}_1} = 20\;\,{\rm{mol}},\,{{\rm{n}}_2} = 50\;\,{\rm{mol}}\)

\(\frac{{{{\rm{V}}_1}}}{{{{\rm{n}}_1}}} = \frac{{{{\rm{V}}_2}}}{{{{\rm{n}}_2}}}\)

\({{\rm{V}}_2} = \frac{{{{\rm{V}}_1} \times {{\rm{n}}_2}}}{{{{\rm{n}}_1}}} = \frac{{44.8 \times 50}}{{20}} = 112\,{\rm{d}}{{\rm{m}}^3}\)

Therefore, the volume is \(112\,{\rm{d}}{{\rm{m}}^3}\).

Ideal Gas Law

A gas that follows Boyle’s law, Charle’s law and Avogadro’s law at all conditions of temperature and pressure are called an ideal gas. In actual practice, such gas is only hypothetical.

According to Boyle’s law, \({\rm{V}} \propto \frac{1}{{\rm{P}}}\)

According to Charle’s law, \({\rm{V}} \propto {\rm{T}}\)

According to Avogadro’s law, \({\rm{V}} \propto {\rm{n}}\)

By combining the three laws, we get  \({\rm{V}} \propto \frac{nT}{{\rm{P}}}\)

Or \({\rm{PV}} \propto {\rm{nT}}\)

\({\rm{PV}} = {\rm{nRT}}\), where \({\rm{R}} = \) universal gas constant \( = 8.314\;{\rm{J}}\;{{\rm{K}}^{ – 1}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\).

Solved Example on Ideal Gas Law

Example 1: A sample of gas occupies a volume of \(512\,{\rm{ml}}\) at \({20^{\rm{o}}}{\rm{C}}\) and \(74\;{\rm{cm}}\) of \({\rm{Hg}}\) as pressured. What volume would this gas occupy at \({\rm{STP}}\)?

Solution: \({{\rm{P}}_1} = 74\;{\rm{cm}}\) of \({\rm{Hg}},\,{{\rm{P}}_2} = 76\,{\rm{cm}}\) of \({\rm{Hg}}{{\rm{T}}_1} = 20 + 273.15 = 293.15\;{\rm{K}}\)

\({{\rm{T}}_2} = 0 + 273.15 = 273.15{\rm{K}},\,{{\rm{V}}_1} = 512\,{\rm{ml}}\)

According to the ideal gas equation \({{\rm{P}}_1}\frac{{{{\rm{V}}_1}}}{{{{\rm{T}}_1}}} = {{\rm{P}}_2}\frac{{{{\rm{V}}_2}}}{{{{\rm{T}}_2}}}\)

\(\frac{{74 \times 512}}{{293.15}} = \frac{{76 \times {{\rm{V}}_2}}}{{273.15}}\)

\({{\rm{V}}_2} = \frac{{74 \times 512 \times 273.15}}{{293.15 \times 76}} = 464.5\,{\rm{ml}}\)

Thus, the volume of the gas at \({\rm{STP}}\) is \(464.5\,{\rm{ml}}\).

Summary

In this article, we studied that there are five gas laws. The relationship between pressure, temperature, volume, and amount of gas is discovered using these fundamental gas laws.
i. Boyle’s law, which establishes a relationship between a gas’s pressure and volume.
ii. The relationship between the volume occupied by a gas and the absolute temperature is described by Charles’ Law.
iii. The relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas is known as Gay-Law. Lussac’s law.
iv. Avogadro’s law is a relationship between the volume of a gas and the amount of gaseous substance it contains.
v. The Combined Gas Law (also known as the Ideal Gas Law) is the result of merging the four laws mentioned above.

FAQs

Q.1. What are the gas laws?
Ans: The gas laws are a set of standards that control how gases behave by establishing relationships between the following:
i. The space is occupied by gas.
ii. The pressure that a gas exerts on the container’s walls.
iii. The gas’s absolute temperature.
iv. The number of moles of gas (or the amount of gaseous material).

Q.2. What are the five gas laws?
Ans: The five gas laws are as follows:
i. Boyle’s law, which establishes a relationship between a gas’s pressure and volume.
ii. The relationship between the volume occupied by a gas and the absolute temperature is described by Charles’ Law.
iii. The relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas is known as Gay-Law. Lussac’s law.
iv. Avogadro’s law is a relationship between the volume of a gas and the amount of gaseous substance it contains.
v. The Combined Gas Law (also known as the Ideal Gas Law) is the result of merging the four laws mentioned above.

Q.3. What are gas laws and why are they important?
Ans: The gas laws are a set of standards that control how gases behave by establishing relationships between the following:
i. The space is occupied by gas.
ii. The pressure that a gas exerts on the container’s walls.
iii. The gas’s absolute temperature.
iv. The number of moles of gas (or the amount of gaseous material).

Gas laws are essential because they can be used to theoretically calculate the properties of a mass of gas.

Q.4. What are ideal gas properties and characteristics?
Ans: The characteristics of an ideal gas are described by the kinetic theory of gases. The following are some of the characteristics:
i. The gas molecules are always moving in a random motion. They travel in a straight line until they come into contact with another molecule or the container’s wall.
ii. The gas molecules do not attract or repel one other.
iii. The gas particles have no volume and are point masses.
iv. The collisions are all elastic. During the collision, no energy is gained or lost.
v. The average kinetic energy of all gases at a given temperature is the same.

Q.5. What is an ideal gas?
Ans: A gas that follows Boyle’s law, Charle’s law and Avogadro’s law at all conditions of temperature and pressure are called an ideal gas. In actual practice, such gas is only hypothetical. The equation of ideal gas is:
\({\rm{PV}} = {\rm{nRT}}\), where \({\rm{R}} = \) universal gas constant \( = 8.314\;{\rm{J}}\;{{\rm{K}}^{ – 1}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\).