Gay Lussac’s law is applicable to the reaction of gases. At constant volume, the pressure of a given amount of gas is directly proportional to the absolute temperature.
P ∝ T (at constant V)
P = KT
P / T = K
- P is the pressure exerted by the gas
- T is the absolute temperature (K)
- K is constant
Gay Lussac’s Law or Amotons Law Explanation
When gases react, they do so in volume. Suppose I have a reaction of hydrogen gas with chlorine gas to give HCl gas.
H2 (g) + Cl2 (g) → 2HCl(g)
How many volumes of hydrogen and chlorine do we have? One volume of hydrogen reacts with one volume of chlorine to give two volumes of HCl. If I have 100 ml of hydrogen gas, it will react with 100 ml of chlorine. One volume of hydrogen gives 2 volumes of HCl, so 100 ml of hydrogen gives 200 ml of HCl. This is so simple and this is the basics of Gay Lussac’s law. The gas law is applicable for reactants and products but they should be gases.
Now we solve the question with the help of an example of the complete combustion of methane.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l)
Suppose we have 100 ml of methane, how much oxygen is required, and how much CO2 is formed? We can’t talk about H2O because it is not in the gaseous state. This law is valid only for gases. In a particular reaction, one volume of methane reacts with 2 volumes of oxygen to give one volume of carbon dioxide. Now if we have 100 ml of methane, it will react with 200 ml of oxygen. What do you say about CO2? One volume of methane gives one volume of CO2, so 100 ml gives 100 ml of CO2.
Example
Let’s take another example of the formation of ammonia by nitrogen and hydrogen gas.
N2 (g) + 3H2 (g) → 2NH3 (g)
If we have 60 ml of nitrogen and 60 ml of hydrogen and they react. Find resultant?
One volume of nitrogen reacts with 3 volumes of hydrogen to give 2 volumes of ammonia. Let us start the question with nitrogen, we have 60 ml of nitrogen, as 1 volume of N2 requires 3 volumes of H2, so 60 ml of N2 requires 3×60 = 180 ml of H2. The question is how much hydrogen do you have? We have 60 ml of hydrogen (given). Is it possible? This will not happen.
Let’s start the same question from a hydrogen point of view. We have 60 ml of hydrogen, as 3 volumes of hydrogen require 1 volume of nitrogen, so 60 ml of H2 requires 60/3 = 20 ml of N2. The question is do you have 20 ml of nitrogen? Yes, we have. And how much is reacting? The reacting nitrogen is 20 ml which means some of the nitrogen will not react.
Given nitrogen = 60 ml
Reacting nitrogen = 20 ml
Unreact N2 = 60 – 20 = 40 ml
Now, how much ammonia is formed? In the above particular reaction, 1 volume forms 2 volumes of NH3, so 20 ml of nitrogen gives 40 ml of ammonia.
