The absolute temperature of a gas in increased four times while maintaining a constant volume. What happens to the pressure of the gas?

A) it decreases by a factor of four

B) it increases by a factor of four

C) it decreases by a factor of eight

D) it increases by a factor of eight

The correct answer is option B which is it increases by a factor of four.

Explanation:

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.

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.