0
$\begingroup$

In an irreversible isothermal process, what is the entropy change of the surrounding? If given $Q$ (heat) of the system is $Q_\text{sys}$.
Is the entropy change of surrounding $$\Delta S = -\frac{Q_\text{sys}}{T_\text{surr}}$$ or $$\Delta S = \frac{Q_\text{sys}}{T_\text{surr}}\quad?$$ I am confused about the sign.

$\endgroup$

2 Answers 2

1
$\begingroup$

There is only one $Q$, the magnitude of heat transfer between the system and surroundings. So $Q_{sys}=Q_{sur}=Q$.

The magnitude of the entropy change of the surroundings will depend on the initial and final equilibrium states. The sign of the entropy change will be positive if it is a compression process (heat transfer to the surroundings) and negative if it is an expansion process (heat transfer to the system).

The Temperature for the entropy transfer is the temperature at the boundary between the system and surroundings.

This means that in the isothermal irreversible process entropy of surrounding will be -q/T. In compression work done is -ve so q will be -ve(q = W) and so entropy change of surrounding will be -(-q)/T means +ve. Correct me if I am wrong

When you speak of $q$ you need to specify whether you are talking about the system or the surroundings. That is because even though the magnitude of $q$ is the same for both (the amount of heat leaving the system will equal the amount of heat entering the surroundings, and vice versa), the sign of $q$ for one will be the opposite of the sign of the other, or $q_{sys}=-q_{sur}$.

So your first equation would be correct if $q$ refers to the system. With respect to the system the sign of $q$ is negative if the process is compression (heat transfer out of the system) making the entropy change of the surroundings positive.

In your last equation the entropy change of the surroundings is correct if $q$ in the equation refers to the surroundings. Then $q$ is positive because in compression heat is transferred to the surroundings making the change in entropy positive.

You won't get tripped up on signs if you realize that the entropy of something always increases when heat is transferred to that something and decreases when heat is transferred out of that something. That "something" can be either the system or the surroundings because they are separate and distinct.

Hope this helps

$\endgroup$
2
  • $\begingroup$ This means that in the isothermal irreversible process entropy of surrounding will be -q/T. In compression work done is -ve so q will be -ve(q = W) and so entropy change of surrounding will be -(-q)/T means +ve. Correct me if I am wrong. $\endgroup$
    – Ayush
    Commented Oct 15, 2020 at 11:03
  • $\begingroup$ @rogers I have updated my answer to respond to your follow up question. $\endgroup$
    – Bob D
    Commented Oct 15, 2020 at 13:11
1
$\begingroup$

If you are considering the entropy change of the surroundings, you must treat it as a "separate system." The heat transferred from the actual system to this "separate system" is $Q_{surr}$, so the entropy change of the surroundings is $\frac{Q_{surr}}{T_{surr}}$. Since the heat transferred from the system to the surroundings is minus the heat transferred from the surroundings $Q_{surr}=-Q_{syst}$ to the system, the entropy change of the surroundings becomes $\frac{(-S_{syst})}{T_{surr}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.