What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?

Answer: 100% (double)

Explanation:

The question tells us two important things:

Mass remains constant Volume remains constant

(We can think in a gas enclosed in a closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1 State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i. e, if T2=2 T1, in order to keep both sides equal, we need that P2 = 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.