The Gas Laws

The relationships between a gas's pressure, volume, and temperature are governed by laws known as gas laws. Boyle's law asserts that a gas's pressure P varies inversely with its volume V at a constant temperature, or P.V. = k, where k is a constant. According to Charles' law, which was developed by J.-A.-C. Charles (1746–1823), the volume V of a gas is exactly proportionate to its absolute (Kelvin) temperature T at constant pressure, or V/T = k. An equation of state, where n is the number of gram-moles of gas and R is the universal gas constant, is a single generalisation of the behaviour of gases that is known as the ideal gas law, P.V. = nRT.

JEE Main/NEET 2027: Physics Important Formulas for Class 10

NEET 2025:  Mock Test Series | Syllabus | High Scoring Topics | PYQs

JEE Main: Study Materials | High Scoring Topics | Preparation Guide

JEE Main: Syllabus | Sample Papers | Mock Tests | PYQs

This Story also Contains

1. What is Boyle's law?
2. What is Charle's law?
3. What is Gay-Lussac’s law?
4. What is Avagadro's law?
5. What is Dalton's Law of Partial Pressure?
6. Solved Examples Based on the Gas Laws
7. Summary

In this article, we will cover the concept of the Gas Laws. This topic we study in the chapter of class 11 physics which is the Kinetic theory of gases, It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), almost four questions have been asked on this concept, and two questions were asked in NEET.

Let's read this entire article to gain an in-depth understanding of the Gas Laws.

What is Boyle's law?

It states that, for a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure.

$$$\begin{array}{rl}V& \propto \frac{1}{P}\\ \text{ or, }P.V& =\text{ constant }\\ \Rightarrow P_1{V}_1& =P_2{V}_2\end{array}$$$

We can also write the above equation as,

$$$\begin{array}{rl}PV& =P\left(\frac{m}{\rho }\right)=\text{ constant }\\ \Rightarrow \frac{P}{\rho }& =\text{ constant or }\frac{P_1}{{\rho }_1}=\frac{P_2}{{\rho }_2}\end{array}$$ $

We can represent Boyle's law through the various graphs, which are shown as

What is Charle's law?

It states that, if the pressure remains constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

From the above statement, we can conclude the following equations

$
$$\begin{array}{rl}\text{boldsymbol}V& \propto \text{boldsymbol}T\\ \frac{V}{T}& =\text{ Constant }\end{array}$$
$

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

This equation can also be written in terms of density and temperature as -

$ $$\begin{array}{rl}& \frac{V}{T}=\frac{m}{\rho T}=\text{ constant }(\text{ As volume }V=\frac{m}{\rho })\\ & \text{ or, }\rho T=\text{ constant }\Rightarrow {\rho }_1{\mathbf{T}}_1={\rho }_2{\mathbf{T}}_2\end{array}$$$

We can represent the Charle's law through the various graph, which is shown as -

Recommended Topic Video

What is Gay-Lussac’s law?

If the volume remains constant, then the pressure of a given mass of a gas is directly proportional to its absolute temperature. It is also known as pressure law.

So, We can conclude the above statement in the following equation -

$P\propto T\text{ or }\frac{P}{T}=\text{ constant }\Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2}$

The graphical representation of Gay-Lussac's law is

What is Avagadro's law?

Avogadro’s law states that an equal volume of all the gases under similar conditions of temperature and pressure contains an equal number of molecules. It implies that -

$N_1=N_2$
N = Number of molecules in a particular gas.

What is Ggaham's Law of Diffusion?

It states that when any two gases at the same pressure and temperature are allowed to diffuse into each other, then the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas.

So we can say that,

$r\propto \frac{1}{\sqrt{\rho }}\propto \frac{1}{\sqrt{M}}\propto V_{rms}$

Where, $\mathrm{r}=$ rate of diffusion of gas
$\rho =$ Density of the gas
$M=$ Molecular weight of the gas
$V_{rms}=$ Root mean square velocity

Now, from the above equation, we can write,

$\frac{r_1}{r_2}=\sqrt{\frac{{\rho }_2}{{\rho }_1}}=\sqrt{\frac{M_2}{M_1}}$

What is Dalton's Law of Partial Pressure?

It states that the total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gases exert if it alone occupies the same volume at a given temperature.

Now, let us have a mixture of 'n' gases, so from the above statement we can conclude that -

$\text{ For }n\text{ gases }P=P_1+P_2+P_3+\dots ..P_n$

Here, P = Pressure exerted by the mixture of gases

P₁, P₂. . . . . . P_n = Partial pressure of the component gases.

Solved Examples Based on the Gas Laws

Example 1: A think tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm containing mercury and two equal ends containing air at standard atmospheric pressure. if the tube is now turned to a vertical position, by what amount will the mercury be displaced?

(Given: cross-section of the tube can be assumed to be uniform)

1) 2.95 cm

2) 5.18 cm

3) 8.65 cm

4) 0.0 cm

Solution:

Let A and B be two chamber

When the tube is horizontally placed

$$$\begin{array}{rl}& P_A=76\mathrm{cm}\mathrm{Hg}\text{ and }{P}_B=76\mathrm{cm}\mathrm{Hg}\\ & V_A=40a\text{ and }{V}_B=40a\end{array}$$$

Where a is an area of chambers.

When the tube is vertically placed

let by d amount will the mercury be displaced

So new pressure in chambers are $P_A^{\prime }$ and $P_B^{\prime }$ and $V_A^{\prime }=(40+d)a$ and $V_B^{\prime }=(40-d)a$

From boyle's law:

$P_1{V}_1=P_2{V}_2$

For chamber A

$$$\begin{array}{rl}& (76\times 40a)=P_A^{\prime }\times (40+d)a\\ & P_A^{\prime }=\frac{76\times 40}{40+d}\end{array}$$$

For the chamber B

$
$$\begin{array}{rl}& (76\times 40a)=P_B^{\prime }\times (40-d)a\\ & P_B^{\prime }=\frac{76\times 40}{40-d}\end{array}$$
$

Also $P_A^{\prime }+20=P_B^{\prime }$

On Solving we get

d = 5.18 cm

Hence, the answer is option (2).

Example 2: A perfect gas changes its pressure from P to 4P at a constant temperature. If the original volume of gas is V, then its final volume (in V) will be?

1) 1

2) 2

3) 3

4) 1/4

Solution:

From boyle’s law

At Constant temperature, PV=constant

So

$ $$\begin{array}{rl}& P_1{V}_1=P_2{V}_2\\ \Rightarrow & \frac{V_1}{V_2}=\frac{P_2}{P_1}=\frac{4P}{P}=4\\ \Rightarrow & V_2=\frac{1}{4}{V}_1=\frac{1}{4}V\end{array}$$$

Hence, the answer is the option (4).

Example 3: A bottle is filled with air at atmospheric pressure and it is cooked at 35⁰ C. If the cork can come out at 3 atmospheric pressure, then up to what temperature (in ^oC) should the bottle be heated in order to remove the cork?

1) 570

2) 325

3) 376

4) 651

Solution:

From gay lussac’s law,

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

wherein

$
$$\begin{array}{rl}& P\alpha T\\ & \Rightarrow \frac{P}{T}=\text{ constant }\\ & \Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2}\end{array}$$
$

At constant volume
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$
(By Gay Lussac's law)
$
$$\begin{array}{rl}& T_2=\left(\frac{P_2}{P_1}\right)T_1\\ & T_2=\left(\frac{3P}{P}\right)\times (273+35)\\ & =3\times 308=924k={651}^0\mathrm{C}\end{array}$$
$

Hence, the answer is the option (4).

Example 4: In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T and that of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jar A and B will be:

1) $1:2$
2) $2:1$
3) $1:1$
4) $4:1$

Solution:

$$$\begin{array}{rl}& \frac{N_A}{N_B}=\frac{P_A{V}_A}{P_B{P}_B}\times \frac{T_B}{T_A}\\ & \frac{N_A}{N_B}=\frac{P\times V\times (2T)}{2P\times \frac{V}{4}\times T}=\frac{4}{1}\end{array}$$$

Hence, the answer is the option (4).

Example 5: A 5-litre flask contains 0.5 moles each of sulphur dioxide and nitrogen gas at 27°C. Calculate the total pressure (in Pa) of the gas:

1) 249000

2) 498000

3) 747000

4) 996000

Solution:

Partial pressure of ${\mathrm{SO}}_2$
$
$$\begin{array}{rl}& P_{{\mathrm{SO}}_2=\mathrm{nRT}}/\mathrm{V}=0.5\times 8.314\times 300/5\times {10}^{-3}=2.49\times {10}^5{\mathrm{Nm}}^{-2}=2.49\times \\ & {10}^5{\mathrm{Pa}}_{\mathrm{a}}\end{array}$$
$

Similarly $P_{N_2}=2.49\times {10}^5\mathrm{Pa}$
Following Dalton's Law
$
$$\begin{array}{rl}& P_{\text{Total }}=P_{S{O}_2}+P_{N_2}\\ & \Rightarrow 2.49\times {10}^5\mathrm{Pa}+2.49\times {10}^5\mathrm{Pa}=4.98\times {10}^5\mathrm{Pa}\end{array}$$
$

Hence, the answer is option (2).

Summary

There are three distinct kinds of laws: Charles' Law, Gay-Law, and Boyle's Law. Lussac's law is incorporated into the combined gas law. This formula states that a constant is equal to the product of pressure and volume times the absolute temperature of the gas. The ideal gas law is obtained by multiplying the combined gas law by Avogadro's law. It has no history of discovery. When all other variables are held constant, such as temperature, pressure, and volume, it is basically a combination of the other gas laws.