# Entropy change of surroundings and system

The entropy change of the surroundings can be calculated by the equation $$dS_{sur}=\frac{dq}{T_{sur}}$$ regardless of the path (irreversible or reversible).

The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $$dU_{sur,rev}=dU_{sur,irrev}$$ and $$dH_{sur,rev}=dH_{sur,irrev}$$. However, $$dH=dq$$ for constant pressure, and as per definition $$dS=\frac{dq_{rev}}{T}$$, the initial equation follows.

This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $$\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$$, which some pages after the book says that is not the case and uses the fact that $$dq_{rev}\geq dq_{irrev}$$ to derive the Clausius inequality. I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.

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On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $$dq/T_{boundary}$$ than the integral of $$dq/T_{boundary}$$ for the original irreversible process.