Entropy change of environment during an irreversible process

[crick]

What is the change in entropy of thermodynamic environment if it changes its temperature during the process and the process is not reversible?

I'm slightly confused because, on the one side, in that case $\Delta S_{gas} \neq -\Delta S_{surroundings}$, since $\Delta S_{universe} >0$ but on the other hand $\Delta S$ does not depend on the specific process, so it should be the same as in a reversible one (and, in a reversible process, $\Delta S_{gas} =-\Delta S_{sourroundings}$).
For istance, for an ideal gas we can calculate $\Delta S$ using a reversible transformation, even if the real transformation it is not.
What am I missing here?

So how can one calculate the entropy change of the environment in such cases?

 

[Chestermiller]

If you had an irreversible process, to get the entropy change for both the system and surroundings, you will need to separate them and devise a separate reversible process for each of them to determine the entropy change for each. You will not be able to get between the initial and final states of both simultaneously with the same reversible process.