[klaas]

i want to solve the next question

One mole of an ideal gas is compressed adaibatically in a piston/cylinder
device from 1 bar and 313 K to 4 bar. The process is irreversible and requires 20 percent more work than a reversible, adiabatic compression from the same intial state to the same final state. What is the entropy change of the gas.

Cp = 3.5R, Cv = 2.5R

Work done
From the equations of an adiabatic proces i calculated the final temperature (465 K). Then i also calculated the work done if the proces had occured reversible. This work is 3160 J. So now i know that the irreversible work equals 4107 J.

But to calculate the entropy change, i get stuck. If the proces had occured reversible the answer would become 0, becaus for an adiabatic proces Q=0.

Could anyone give me a hint.

[Yggdrasil]

Remember that for an ideal gas, you can calculate the internal energy (U) of the gas from the temperature.  Since you know the final and initial temperatures, you know the change in internal energy.  Internal energy is a state function, so ΔU is the same for the reversible and irreversible paths.  Using the first law of thermodynamics and the fact that you now know ΔU and w, you can calculate q.