Why entropy is an extensive quantity ?

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Why entropy is an extensive quantity ?
 
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Hi.
Entropy increase by volume. Gas of 2 m^3 have double entropy than gas of 1 m^3 in the same temperature and pressure.
Regards.
 
sweet springs said:
Hi.
Entropy increase by volume. Gas of 2 m^3 have double entropy than gas of 1 m^3 in the same temperature and pressure.
Regards.

But how do you know that "Entropy increase by volume. "

Let me restate my question :

How can ew know that entropy is an extensive quantity by its definition

dS=[tex]\int(dQ/T)[/tex]
 
Hi.
First law of thermodynamics, about the conservation of energy: δQ=dU - dW =dU - pdV
δQ is extensive because dU and pdV are extenxive. δQ/T and ∫δQ/T are also extensive.
Regards.
 
sweet springs said:
Hi.
First law of thermodynamics, about the conservation of energy: δQ=dU - dW =dU - pdV
δQ is extensive because dU and pdV are extenxive. δQ/T and ∫δQ/T are also extensive.
Regards.

But for different systems , their temperature T may not be the same !

First Law sates that deltaQ=dU+deltaW

We can only infer that deltaQ is additive !

Please consider my comment~

Thanks
 
Hi.

abcdefg10645 said:
But for different systems , their temperature T may not be the same !
First Law sates that deltaQ=dU+deltaW
We can only infer that deltaQ is additive !

Thanks for correcting my sign mistake of deltaW.
You are right. Additive is the essence of extensive quantity. For isolated two different systems
S1 = ∫dQ1/T1, S2 = ∫dQ2/T2 and the entropy of total system is S = S1 + S2.
Regards.
 
sweet springs said:
Hi.



Thanks for correcting my sign mistake of deltaW.
You are right. Additive is the essence of extensive quantity. For isolated two different systems
S1 = ∫dQ1/T1, S2 = ∫dQ2/T2 and the entropy of total system is S = S1 + S2.
Regards.

Ha ha~never mind ~I believe that were just some typos

But , I cannot understand why entropy is additive ?
 
Hi.
Definition ∫ δQ/T is already addition of δQ/T , isn't it ?
Why not additive you think?
Regards.
 
sweet springs said:
Hi.
Definition ∫ δQ/T is already addition of δQ/T , isn't it ?
Why not additive you think?
Regards.

Using the integral to calculate the (delta)S "OF A SYSTEM" is to calculate the difference of entropy for a particular process , I want to ask why , for different systems , it can also be added ?
 
Hi. How's that?

1 m^3 of liquid water of the room temperature and in standard pressure has entropy say S.
Let us remove half of water. Do you agree that the left water has entropy S/2 ?

We will bring removed water into another isolated room of the same temperature and pressure.
Do you agree that brought water has entropy S/2?

Water in the first room and water in the second room are different systems.
Do you agree that the integrated system of the two has entropy S?

In room 1 we increase the temperature then the entropy of water becomes S'/2
Do you agree that the integrated system of the two has entropy S/2 + S'/2 ?
Change of entropy ∫δQ/T for room 1 system is regarded as change of entropy in the integrated system, isn't it?

Please show your opinion so that I can surely get the core of your question.
Regards.
 
Last edited:
Thank u ~

I can now fully understand the reason why entropy is additive by ur explaining steps by steps !