Solved Exercises on Gay-Lussac`s Law

Exercise 1

The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the initial temperature of the gas if its initial pressure was 1 atm?

Given,

Initial pressure, P₁ = 1 atm

Final pressure, P₂ = 1.5 atm

Final temperature, T₂ = 250 K

As per Gay-Lussac’s Law, P₁T₂ = P₂T₁

Therefore, T₁ = (P₁T₂)/P₂ = (1*250)/(1.5) = 166.66 Kelvin.

Exercise 2

At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K.

Initial pressure, P₁ = 3 atm

Initial temperature, T₁ = 300K

Final temperature, T₂ = 900 K

Therefore, final pressure (P₂) = (P₁T₂)/T₁ = (3 atm*900K)/300K = 9 atm.