# $TdS=dU$ for a system going a quasi-static process

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

– pwf
Oct 6, 2020 at 22:16
• 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. Oct 7, 2020 at 2:58
• Please provide what you consider a quasi-static process that is not reversible so that we can focus on the difference. Oct 7, 2020 at 10:44

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.

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