# Gay Lussac Law Formula

In 1808, the French chemist Joseph Louis Gay-Lussac reported the results of new experiments together with a generalization known today as Gay-Lussac’s law of combining gases.

The volume of gas is directly proportional to the Kelvin temperature if the volume is kept constant.
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.

Where,
V1 = original volume
V2 = final volume
T1 = original temperature (K)
T2 = final temperature (K)

Gay-Lussac’s Law is applicable only to gases. The volumes of liquids or solids involved in the reactants or products are not governed by Gay Lussac’s law.

## Gay Lussac Law Formula Solved Examples

Some of the solved problems based on Gay Lussac Law Formula are given below.

Question 1: What is the volume of a quantity of a gas at 27oC if its volume was 400mL at 0oC? The pressure remains constant.
Solution:
Since the temperature increased, the volume increased. Here we must use a temperature factor greater than 1, that is 300/273

The temperature must be in degrees kelvin.
20oC + 273 = 300 K
0oC + 273 = 273 K
400mL $\times$ $\frac{300K}{273K}$ = 438mL

Question 2: Solve Gay-Lussac’s Law to get an expression for the unknown volume. Substitute the appropriate data into the formula.

V1 = 400mL
V2 = ? mL
T1 = 0oC
T2 = 27o
Solution:
V1 = 400mL
V2 = ? mL
T1 = 0oC + 273 = 273K
T2 = 27oC + 273 = 300K
Substitute all the values in the corresponding formula
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
V2 = $\frac{(400 mL)(300 K)}{273 K}$
V2 = 438 mL