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.


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

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