# 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:

$\frac{P_1}{T_1}$ = $\frac{P_2}{T_2}$

where,

$P_1$ = 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

• 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\)

### 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.