# 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.

Similar question that has no answer: Entropy change of surroundings

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process. The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings. To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings). No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.

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.

• Could you elaborate on that separation part? I've never seen that.
– user93668
May 14, 2020 at 17:47
• That's too bad. It has annoyed me for some time that the text books out there do such a bad job of explaining how to determine the entropy change for an irreversible process. So I wrote a short primer of how to do it, which many people have found very useful: physicsforums.com/insights/grandpa-chets-entropy-recipe Please feel free to ask follow-up questions. May 14, 2020 at 18:29
• Regarding separating them: If the process was irreversible, to get the entropy change of the system and the entropy change of the surroundings, we need to devise a path that takes each of them from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state reversibly. If the original process was irreversible, there is no single path that will take them both between their initial and final states (of the irreversible process) simultaneously. So we need to separate them, and subject each of them separately to a reversible path between these states. May 15, 2020 at 14:47

The most important concept to understand is that entropy is a state function. Best explained by this image:

The change in entropy is just based on the final state and the initial state. Therefore we model all processes as a reversible process because it has a simple equation. The reversible process is the one in blue above. That allows us to use the equation for the change in entropy for reversible process:

$$\mathrm dS =\frac{\mathrm dq_\mathrm{rev}}{T}.\tag{1}$$

Hint: use the first law to get the equation for $$q.$$

So if I want to calculate change in entropy for the system I integrate from state 1 to state 2. It will give you the same answer if you want to calculate $$ds$$ for any process.

## Nuance of Surroundings:

If I want to calculate the change in entropy of the surroundings it changes, slightly, because the temperature of the surroundings is constant. That is because the surrounding is a reservoir. So we end up with

$$\int^{2}_{1}\mathrm dS_\mathrm{surr} = \int^{2}_{1}\frac{\mathrm dq_\mathrm{rev}}{T_\mathrm{surr}},\tag{2}$$

which will simplify to

$$S_2 - S_1 = \frac{q_\mathrm{rev}}{T_\mathrm{surr}}.\tag{3}$$

• "The nuance of surroundings" part is incorrect as the final result must be $\Delta S=\frac{ -q_{\text{irrevsible}}}{T}$ May 18, 2021 at 10:03
• If what you're trying to point to is that a heat transfer to the surroundings is irreversible then yes. However I wanted to explain in a way where the reader could see that this is the same q. Then they will see sign convention correctly. My point being it's the same base equation. Same heat transfer, always starting tith definition. Keeps reader focused on that then they can decide what the "system" is and sign vectors accordingly. In other words... My solution is right, your just being it from a different basis system. May 19, 2021 at 14:40