Answer is: B) it increases by a factor of four.
Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
p₁/T₁ = p₂/T₂.
For example:
p₁ = 3,6 atm.; initial pressure
T₁ = 273 K; initial temperature.
T₂ = 1092 K, final temperature
p₂ = ?.; final presure.
3,6 atm/273 K = p₂/1092K.
3,6 atm · 1092 K = 273 K · p₂.
p₂ = 3931.2 atm · K ÷ 273 K.
p₂ = 14.4 atm.
As the temperature goes up, the pressure also goes up and vice-versa.
The correct answer is option B which is it increases by a factor of four.
According to Gay-Lussac's Law:
P1/T1 = P2/T2
So lets take an example:
T1 = 1, T2 = 4
Pa = 1, P2 = ?
According to Gay-Lussac's Law:
P2 = P1/T1 * T2
= P2 = 1/1 * 4 = 4
So pressure will be 4 times the initial pressure if temperature is increased 4 times while maintaining a constant volume.
