Gay Lussac's Law - Concept And Theory

Gay Lussac’s law

This law shows the relationship between the temperature and pressure of a gas. According to Gay-Lussac’s law, the pressure of an ideal gas is directly proportional to its temperature. This law states that at constant volume of a close container, we may conclude that if the pressure inside container increases, the temperature inside the container also increases.

Formula: This law can be expressed in the equation as follows:

\(\frac{P_1}{T_1} \) = \(\frac{P_2}{T_2} \)


\(P_1\) = Initial Pressure (atm)

\(P_2\) = Final Pressure (atm)

\(T_1\) = Initial Temperature (Kelvin)

\(T_2\) = Final Temperature (Kelvin)

Gay Lussac’s law of combining gas volumes:

Key points regarding Gay Lussac’s law:

  • Volume should be constant
  • Pressure vs Temperature graph is linear for ideal gases

Gay Lussac's Law

  • Temperature and pressure are directly proportional to each other

Problems based on Gay Lussac’s law

Q. A cylinder contains a gas which has a pressure of 150kPa at a temperature of 250K. Find the temperature of the gas which has a pressure of 100kPa if the volume remains constant.
Sol: Using Gay Lussac’s law,

Formula: This law can be expressed in the equation as follows:

\(\frac{P_1}{T_1} \) = \(\frac{P_2}{T_2} \)

\(\frac{150}{250} \) = \(\frac{100}{T_2} \)

By cross multiplication we get the result as \(T_2\) = \(\frac{(100)(250)}{150}\) = \(166.6K\)

Q. If we consider a tank of gas at 1.520 torr pressure and a temperature of 200K and it is heated to 400K. What is the new pressure in atmosphere?

Sol: First we will convert pressure in torr to atmosphere (atm)

As, \(1 ~atm\) = \(760 ~torr\)

\(\frac{1520~ torr}{760~ torr}\) = \(2.00 atm\)

Now we will solve for \(P_2\) in Gay Lussac Law

\(P_2\) = \(\frac{(2atm) (400K)} {200K}\) = \)4~atm\)


1) We should always use Kelvin in all the problem solving questions and not Celsius.

2) Temperature and Pressure always relate in a direct relationship.

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