The Gas Laws

Many scientists studied and experimented about gases. As a result, different gas laws were formulated. These laws show relationship between the various state variables of a gas. These state variables include:


State Variable Symbol  Unit/s
Amount of Substance n


moles, mol




atmosphere, atm

Pascal, Pa

millimeter of mercury, mmHg





Kelvin, K





Liters, L



Avogadro’s Law

This law states that all ideal gases with equal volume at constant temperature and pressure, contain the same amount or number of molecules. It is expressed mathematically as,

\({V \over n} = k\)

Comparing substances under different conditions, the above equation can be written as

\({V_1 \over n_1} = {V_2 \over n_2}\).


A 15.0 mol of oxygen gas has a volume of 300 L. A certain amount of oxygen gas is added at constant temperature and pressure, so that the volume becomes 350 L. What is the new number of moles of oxygen gas?


n1 = 15.0 mol

V1 = 300 L

V2 = 350 L

Find: n2


To find n2, we derive its formula from the Avogadro’s law equation \({V_1 \over n_1} = {V_2 \over n_2}\).

\({V_1 \over n_1} = {V_2 \over n_2} \implies n_2 = {n_1V_2 \over V_1}\)

Substitute all the given values to get the value of n2,

\(n_2 = {(15.0\; mol)(350\;L) \over (300\;L)} = 17.5\;mol\)

There are 17.5 mol of oxygen gas at 350 L.

Boyle’s Law

Boyle’s law describes the relationship of the pressure and volume of a gas. It states that, under constant temperature and the same number of molecules, the pressure of a gas varies inversely to its volume. As pressure increases, the volume decreases and as the pressure decreases, the volume will increase. Mathematically, the relationship is written as

\(P \propto {1 \over V}\)


\(PV = k\).

For the same substance under two different conditions, the equation will become

\(P_1V_1 = P_2V_2\).

The law was named after Robert William Boyle, an Anglo-Irish chemist, inventor, natural philosopher and physicist who published the law in 1662.


The volume of a certain gas at 2.5 x 105 Pa pressure is 3 L. What is its volume if the pressure is increased to 3.5 x 105 Pa?


P1 =  2.5 x 105 Pa

V1 = 3 L

P2 = 3.5 x 105 Pa

Find: V2


We first derive the equation for V2 from \(P_1V_1 = P_2V_2\) by dividing both sides of the equation by P2. Thus, \(P_1V_1 = P_2V_2 \implies V_2={P_1V_1 \over P_2}\). Substitute the given values to the equation,

\(V_2={(2.5 x 10^5\; Pa)(3\;L) \over (3.5 x 10^5\; Pa)}=2.14\;L\).

The result shows that the volume of the gas decreases as the pressure increases.


Charles’s Law

Named after the French mathematician, physicist, and inventor Jacques-Alexandre-César Charles, this law describes the relationship between the temperature and volume of a gas at constant pressure. It states that when a sample of gas is held at constant pressure, the temperature is directly proportional to the temperature. This means that when the temperature increases, the volume of the gas will also increase. If temperature decreases, volume also decreases. This can be represented by the equation

\(V \propto T\)


\({V \over T} = k\).

When the same gas is held at two different conditions, the law is written in mathematical equation as

\({V_1 \over T_1} = {V_2 \over T_2}\).


What temperature change occurs in a sample of gas if its volume increases from 1.2 L to 2.3 L after 45 C?


V1 = 1.2 L

V2 = 2.3 L

T2 = 318 K

Find: T1


Based on the problem, the volume increases after a given temperature, thus, it is obvious to say that the missing quantity is the temperature, T1 before the increase in volume. Using the equation for Charles’s law, \({V_1 \over T_1} = {V_2 \over T_2}\), we derive the formula for T1.

\({V_1 \over T_1} = {V_2 \over T_2} \implies T_1 = {V_1T_2 \over V_2}\)

Substituting the given values, we have

\(T_1 = {(1.2\;L)(318\;K) \over 2.3\;L}=165.91\;K\)

The result shows that the volume and temperature of gas is directly proportional to each other.

Gay – Lussac’s Law

When the volume and mass of a gas is constant, its pressure varies directly to its temperature. Mathematically,

\(P \propto T\)


\({P \over T}=k\).

Comparing the same substance at two different conditions, the Gay-Lussac’s law equation is expressed as

\({P_1 \over T_1} = {P_2 \over T_2}\).


At 273 K, a tank of hydrogen gas has a pressure of 4.5 atm. In what temperature will the pressure be 4.8 atm? Will the temperature greater or lesser than the first condition?


P1 = 4.5 atm

T1 = 273 K

P2 = 4.8 atm

Find: T2


\({P_1 \over T_1} = {P_2 \over T_2} \implies T_2 = {P_2T_1 \over P_1}\)

Substitute all the given values,

\(T_2 = {(4.8\;atm)(273\;K) \over 4.5\;atm} = 291.2\;K\)

As what the law states, the pressure is directly proportional to the temperature. Since the pressure increased in the second condition, the temperature also increased.