How many moles of a monatomic ideal gas can be heated by 5 K by supplying 41.5 J of heat to it?

The first law of thermodynamics:
The amount of heat transferred to the gas is equal to the sum of the change in the internal energy of the gas and the work done by the gas:
Q = ∆U + A
We find the change in internal energy from the expression:
∆U = (3/2) νR∆T
Gas perfect work:
A = p * ∆V
Let’s substitute all this in the first law of thermodynamics:
Q = ∆U + A = (3/2) νR∆T + p * ∆V
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.
Our piston can move without friction, which means that the pressure is constant, we write down before and after the heat is transferred, and taking into account that ν = m / M:
pV1 = ν * R * T1
pV2 = ν * R * T2
pV2-pV1 = ν * R * T2-ν * R * T1
p * ∆V = ν * R * ∆T
Let’s substitute this into the expression of the first law of thermodynamics:
Q = (3/2) νR∆T + p * ∆V = (3/2) νR∆T + ν * R * ∆T = (5/2) νR∆T
Let us express the amount of substance from this expression, and calculate it:
ν = Q / ((5/2) R∆T) = 41.5 / ((5/2) * 8.31 * 5) = 0.4 mol.
Answer: 0.4 mol of gas.