# Second law of thermodynamics: How do entropy changes in the system and the environment count in? [closed]

$$\frac{q_{rev}}{T} = \Delta S$$

Here, what does $\Delta S$ signify? Does it mean the change in entropy of the system or the surroundings?

How are the entropies of the system, the surroundings, and the universe related to each other, and which entropy is used in the Gibbs free energy equation?

consider this problem

H2O(l) -> H2O(g) at 373 k and 1 atm pressure then which of the following is correct?

a) ΔSsystem >0 ΔSsurrounding > 0

b) ΔSsystem <0 and ΔSsurrounding < 0

c) ΔSsystem>0 and ΔSsurrounding < 0

d) ΔSsystem <0 and ΔSsurrounding > 0

## closed as off-topic by orthocresol♦, ringo, M.A.R. ಠ_ಠ, Todd Minehardt, FreddyApr 26 '16 at 18:18

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• The $\Delta \mathrm{S}$ corresponds to what $\mathrm{q_{rev}}$ you are using. For example if plug in the $\mathrm{q_{rev}}$ of the system, then you get the $\Delta \mathrm{S}$ of the system. The same goes for the Gibbs Free energy equation. I don't understand what you mean by how is entropy of system,surrounding and universe related to each other. Could you please expand on what you mean. – Nanoputian Apr 26 '16 at 7:02
• @ Nanoputian sir thank you very much for replying,please see this problem – Ajay Sabarish Apr 26 '16 at 9:05
• @Nanoputian thank you very much for replying sir, please see this problem H2O(l) -> H2O(g) at 373 k and 1 atm pressure then which of the following is correct? a) ΔSsystem>0 ΔSsurrounding > 0 b) ΔSsystem<0 and ΔSsurrounding < 0 c) ΔSsystem>0 and ΔSsurrounding < 0 d) ΔSsystem <0 and ΔSsurrounding > 0 i know that ΔSsystem is positive but how to determine ΔSsurrounding? – Ajay Sabarish Apr 26 '16 at 9:18
• I may be wrong since I didn thermodynamics a while back, but I pretty sure the answer is C. Firstly I am assuming the water is heat reversibily, then $\mathrm{q_{sys} = q_{sur}}$. Therefore, $\mathrm{\Delta S_{sys} = -\Delta S_{sur}}$. Since you are convert a substance from the liquid phase to the gas phase, the entropy of the system must be increasing and hence >0. Therefore the answer is C – Nanoputian Apr 26 '16 at 10:39
• I also recommend that you edit your original question to include this problem. – Nanoputian Apr 26 '16 at 11:03

I think your first question (regarding the meaning of $\Delta S$) has been adequately covered in the comments. I hope the following will clarify your other concerns, though it isn't entirely clear what exactly you don't know.

The Second Law tells you that the total entropy of the universe will always increase, i.e $\Delta S_{universe} > 0$.

Logically, we can split the 'universe' into a system and surroundings, so:

$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$$

From the second law, this obviously implies that

$$\Delta S_{system} + \Delta S_{surroundings} > 0$$

Regarding the multiple choice question you asked in the comments, this means that we can immediately eliminate option B. If both $\Delta S_{system}$ and $\Delta S_{surroundings}$ are negative, their sum cannot be positive.

Clearly, within the system, order decreases as liquid is turned to gas, so entropy increases. So we know that $\Delta S_{system} > 0$.

In order to make the water boil, heat is transferred away from the surroundings, so $\Delta S_{surroundings} < 0$. If you knew how much water you were boiling, you could even figure out exactly how much the entropy decreases, because you know that the energy required to boil a mass of water is given by $Q=mL$, where $L$ is the latent heat of vaporisation.

• sir but at 373 k and 1 atm pressure won't the system be at equilbrium since that is just boiling point.and won't the entropy of universe be zero at equilbrium? this is my doubt. and order decreases but the system is in equilbrium so we can't say which direction reaction proceeds,so how can we say entropy of system increases? – Ajay Sabarish Apr 26 '16 at 15:40