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Topic: Entropy  (Read 4521 times)

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« on: September 22, 2005, 08:20:55 PM »
I need to figure out why the reason for this is.

The conditions for this is adiabatic (expansion)

Normally ds> or = to qreversible/T

My questions is why is dS<qirreversible/T

I am sure it something simple but I cannot figure it out.

Offline Juan R.

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« Reply #1 on: September 24, 2005, 05:54:02 AM »
dS = diS + deS

diS is the production of entropy due to irreversible phenomena. Equilibrium textbooks only deal with equilbrium and, therefore, cannot compute this term, but the second law of nature is

diS >= 0

For a closed system

deS = dQ / T


dS >= dQ / T

that is appears in equilibrium textbooks. Take a gas if you are heating slowly, there is not dissipative phenomena in the gas and therefore

dS = dQreversible / T

The label 'reversible' used in elementary eq. textbooks is akfwul because as perfectly explained by TIP (thermodynamics of irreversible processes) the flow terms are always reversible, that is irreversible are the dissipative phenomena inside the system.

In an adiabatic expansion

dS >= 0

if expansion is slow and passing by quasiequilbrium states then

dS = 0

if not

dS > 0
« Last Edit: September 24, 2005, 05:59:44 AM by Juan R. »
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« Reply #2 on: September 26, 2005, 07:26:00 PM »

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