Why is entropy zero in an adiabatic process?

[negation]

No heat exchange is facilitated during an adiabatic process. Change is heat is zero.
How does this relates to the entropy being zero?
∫dQ/T?
But this could really just mean that the integral is of any constant.

 

[Arcone]

Heat exchange and change in entropy are related.. dS=dQ/T. If the process is adiabatic then dQ=0 and change in entropy dS=0/T=0 or ΔS=∫dQ/T=∫0/T=0. So entropy is constant (S=constant) in reversible adiabatic process. It doesn't mean that the entropy of the system is 0, it means that the change in the entropy of the system is 0. ie. entropy of the system is same before and after the process.

 

[D H]

Heat exchange and change in entropy are related.. dS=dQ/T. If the process is adiabatic then dQ=0 and change in entropy dS=0/T=0 or ΔS=∫dQ/T=∫0/T=0. So entropy is constant (S=constant) in reversible adiabatic process. It doesn't mean that the entropy of the system is 0, it means that the change in the entropy of the system is 0. ie. entropy of the system is same before and after the process.

The correct expression is $dS \ge \frac {dQ}T$, not $dS = \frac {dQ}T$. The latter holds if and only equal if the process is reversible. Free expansion is the canonical example of a non-reversible adiabatic process. Entropy increases in free expansion even though there is no heat transfer to/from the system and no work done on/by the system.

 

[Chestermiller]

The correct expression is $dS \ge \frac {dQ}T$, not $dS = \frac {dQ}T$. The latter holds if and only equal if the process is reversible. Free expansion is the canonical example of a non-reversible adiabatic process. Entropy increases in free expansion even though there is heat transfer and work are zero.

Hi DH. I think you meant "even though there is no heat transfer..."

Chet

 

[negation]

Heat exchange and change in entropy are related.. dS=dQ/T. If the process is adiabatic then dQ=0 and change in entropy dS=0/T=0 or ΔS=∫dQ/T=∫0/T=0. So entropy is constant (S=constant) in reversible adiabatic process. It doesn't mean that the entropy of the system is 0, it means that the change in the entropy of the system is 0. ie. entropy of the system is same before and after the process.

The correct expression is $dS \ge \frac {dQ}T$, not $dS = \frac {dQ}T$. The latter holds if and only equal if the process is reversible. Free expansion is the canonical example of a non-reversible adiabatic process. Entropy increases in free expansion even though there is heat transfer and work are zero.

Thanks! The first post cleared things up. The second refined the understanding.

 

[ZapperZ]

Please note that it is the CHANGE in entropy, not the absolute value of the entropy, that is zero. This needs to be clarified in light of the title of the thread.

Zz.

 

[negation]

Please note that it is the CHANGE in entropy, not the absolute value of the entropy, that is zero. This needs to be clarified in light of the title of the thread.

Zz.

You are right. But it was a legitimate mistake. I thought the entropy was zero and so asked why could it have been other arbitrary constant. Perhaps, the mod could do some editing.

 

[D H]

Hi DH. I think you meant "even though there is no heat transfer..."

Yes, I did. Thanks. Post edited to reflect the missing "no".