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Topic: When can total entropy be greater than zero?  (Read 3842 times)

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Offline Burningkrome

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When can total entropy be greater than zero?
« on: October 24, 2012, 11:34:04 AM »
This is either a freshman question, or a huge can of worms.

ΔSInternal + ΔSExternal ≥ 0

When could this ever be more than zero?

As I understand it (regardless of whether using the statistical or thermodynamic definitions of entropy), this can't happen because ΔSExternal DEFINES ΔSInternal and vice-versa.

I.e.
Q = the change in thermal energy PUT INTO the system from surrounding.

ΔSInternal = Q/T
and
ΔSExternal = the heat put into the system by surrounding, or -Q/T

Q is the same energy, by definition. Thus ΔS is equal and opposite, by definition. Or not?



Offline Schrödinger

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Re: When can total entropy be greater than zero?
« Reply #1 on: October 24, 2012, 01:09:42 PM »
ΔSInternal = Q/T

What you need to understand here is that it's not just Q. ΔSInternal = Qrev/T.
Only when the process is carried out reversibly can you say that ΔSInternal = Q/T. This is when the equality with 0 holds : ΔSInternal + ΔSExternal = 0

In case the process is carried out irreversibly, then ΔSInternal + ΔSExternal > 0 (strictly positive, not equal to 0). In this case, you cannot say ΔSInternal = Q/T, because ΔSInternal ≠ Qirrev/T
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Offline curiouscat

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Re: When can total entropy be greater than zero?
« Reply #2 on: October 24, 2012, 01:22:18 PM »

In case the process is carried out irreversibly, then ΔSInternal + ΔSExternal > 0 (strictly positive, not equal to 0). In this case, you cannot say ΔSInternal = Q/T, because ΔSInternal ≠ Qirrev/T

I agree but to further the riddle: Aren't all processes reversible on a microscopic scale? If so, is the equality always valid at that scale?

Offline Burningkrome

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Re: When can total entropy be greater than zero?
« Reply #3 on: October 24, 2012, 05:41:08 PM »
So, I may be misunderstanding a number of terms. My understanding of the functional  definition of entropy is heat lost without work. Example; an exothermic reaction dissipating heat out of the system and increasing ΔSExternal.

I understand the statistical definition of entropy as (loosely) being the “probability of the system being in one of the sum of all the possible positions of the micromolecular structure times -bK”. So a spinning coin = a ΔSsystem of 2*-bK (clockwise and counterclockwise) then two spinning coins = a ΔSsystem of 4*-bK, 3 coins 6*-bK, ETC.

So, the definitions I have are…
Q = Heat ADDED to system from external.
So, Qsystem = -QExternal and Q = ΔH and –Q = -ΔH
From this ΔSsystem = Q/T and since Q = -Q then ΔSExternal = -Q/T

I’m guessing some or all of these definitions are wrong. Can you help me re-define them?

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