**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:

\(\frac{P_1}{T_1} \)

where,

\(P_1\)

\(P_2\)

\(T_1\)

\(T_2\)

**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

- 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:

\(\frac{P_1}{T_1} \)

\(\frac{150}{250} \)

By cross multiplication we get the result as \(T_2\)

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, \(1 ~atm\)

\(\frac{1520~ torr}{760~ torr}\)

Now we will solve for \(P_2\)

\(P_2\)

**Note**

**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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