# 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

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

Question 1: What is the volume of a quantity of 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 $\times$ $\frac{300K}{273K}$ = 439.5 mL

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 = 27oC

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 = (400 mL ×300 )/273

V2 = 439.56 mL