With a decrease in the volume of an ideal gas by 3.6 times, its pressure increased by 1.2 times

With a decrease in the volume of an ideal gas by 3.6 times, its pressure increased by 1.2 times by how many times the internal energy changed.

Given:
V1 = 3.6 * V2
p2 = 1.2 * p1

Internal energy of ideal gas:
U = (3/2) νRT
Mendeleev-Clapeyron equation
pV = (m / M) * R * T, where p is the gas pressure, V is the gas volume, m is the gas mass, M is the molar mass of the gas, R is the universal gas constant 8.31 J / (mol * K), T is the gas temperature.

Before the change:
U1 = (3/2) νRT1
p1V1 = (m / M) * R * T1
After the change:
U2 = (3/2) νRT2
p2V2 = (m / M) * R * T2
Divide U2 / U1 and p2V2 / p1V1:
U2 / U1 = ((3/2) νRT2) / ((3/2) νRT1) = T2 / T1
p2V2 / p1V1 = ((m / M) * R * T2) / ((m / M) * R * T1) = T2 / T1
It turns out:
U2 / U1 = p2V2 / p1V1
Let’s substitute data from given:
U2 / U1 = p2V2 / p1V1 = 1.2 * p1 * V2 / p1 * 3.6 * V2 = 0.33
Answer: the internal energy has changed 0.33 times.