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I have a question regarding entropy:

The change in intropy for a system at constant composition with no other work than volume work is:

$T_sdS=dU +pdV$, were $T_s$ is the surrounding temperature, and $p$ is the system pressure

$T_sdS= dQ -p_sdV + pdV$

if the process is done in a quasi-static manner, p_s=p (because we are pretty close to equilibrium)

T_sdS=dQ (valid for a quasi-static process)

If the process is reversible,

TdS=dQ (in this case T is the system temperature)

What I dont understand is the following: in a quasi-static process we are very close to the equilibrium position, so the system temperature is always equal to the surrounding temperature because in the equilibrium

(for the composite system) dS=0=(1/T -1/T_s)dU + (p/T - p_s/T_s)dV

because dS must vanish for every dU and dV

1/T=1/T_S

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    $\begingroup$ What is your question? $\endgroup$
    – pwf
    Oct 6, 2020 at 22:16
  • $\begingroup$ My question is, if T=T_s , then the expression for a quasi-static process is always the same that for a then the expression for a quasi-static process is always the same that for a reversible process, and that is absurd because, every reversible process is quasi-static but not every quasi-static process y reversible. $\endgroup$ Oct 7, 2020 at 2:58
  • $\begingroup$ Please provide what you consider a quasi-static process that is not reversible so that we can focus on the difference. $\endgroup$ Oct 7, 2020 at 10:44

1 Answer 1

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Your comment to @pwf stated that your question actually is

My question is, if $T=T_s$ , then the expression for a quasi-static process is always the same .......as that for a reversible process, and that is absurd because, every reversible process is quasi-static but not every quasi-static process is reversible.

An example of a quasi-static process that is irreversible is one that involves mechanical friction. In such a case you can have $T_{sys}=T_{surr}$ at the system boundary because mechanical friction generates thermal energy (generates entropy) within the system.

So the change in entropy is

$$\Delta S=\frac{Q}{T}+σ$$

Where the term $\frac{Q}{T}$ is the entropy transferred at the system boundary where $T=T_{sys}=T_{surr}$ and $σ$ is the entropy generated within the system due to mechanical friction.

Hope this helps.

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  • $\begingroup$ Thank you very much. you have been very helpfull! $\endgroup$ Oct 12, 2020 at 16:00

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