Suppose I have $n$ mol of an ideal gas in a volume $V$ at pressure $P_{1}$ and at a temperature $T$. I place a weight which exerts a pressure $P$. The gas undergoes isothermal compression until its volume reaches $V'=\frac{nRT}{P}$. I can compute its entropy change with the formula $nR \ln(\frac{V'}{V})$.

So now I want to know the entropy change of the surroundings, for the sake of simplicity let's assume that it is an ideal gas at equilibrium with the other gas, so its pressure is $P_{1}$ and its temperature is $T$, with an initial volume of $V_{e}>>V,V'$ and a final volume $V_{e}'>>V,V'$, volume conservation allows us to say that $V_{e}'=V_{e}+V-V'$. Since its temperature has not changed either (because of energy conservation) its entropy change is (or should be) $NR \ln(\frac{V_{e}'}{V_{e}})=NR \ln(\frac{V_{e}+V-V'}{V_{e}})\approx NR\frac{V-V'}{V_{e}}=\frac{P_1 (V-V')}{T}$.

However, I know that it is in fact $\frac{P (V-V')}{T}$, because modelling the surroundings as an infinite thermal reservoir all the heat exchanged is reversible and the amount of heat exchanged is $P (V-V')$

Why am I wrong?