Increase in entropy principle

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If we consider a system to undergo an irreversible process from state 1 to state 2 and a reversible process from state 2 to state 2, then through Clausius inequality

(1to2∫dQirrev/T) + (2to1∫dQrev/T) ≤ 0
(1to2∫dQirrev/T) + s1 - s2 ≤ 0
s2-s1 ≥ (1to2∫dQirrev/T)
Δs ≥ (1to2∫dQirrev/T)

Does this mean that the entropy change for a reversible process is greater than that of an irreversible process? I'm convinced I am wrong because my notes say otherwise but isn't Δs the entropy change of a reversible process and (1to2∫dQirrev/T) the entropy change of an irreversible process?
 

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  • #2
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If we consider a system to undergo an irreversible process from state 1 to state 2 and a reversible process from state 2 to state 2, then through Clausius inequality

(1to2∫dQirrev/T) + (2to1∫dQrev/T) ≤ 0
(1to2∫dQirrev/T) + s1 - s2 ≤ 0
s2-s1 ≥ (1to2∫dQirrev/T)
Δs ≥ (1to2∫dQirrev/T)

Does this mean that the entropy change for a reversible process is greater than that of an irreversible process? I'm convinced I am wrong because my notes say otherwise but isn't Δs the entropy change of a reversible process and (1to2∫dQirrev/T) the entropy change of an irreversible process?
first there is correction in typing it should be from state 2 to state 1.

Usually the most efficient processes possible for converting energy from one form to another, are processes where the net entropy change of the system and the surroundings is zero.
These processes represent limits - the best that can be done.

one can also use entropy change as a measure of how reversible a process is-
thus the inequality given in your first equation should be changed to -greater than zero ;
i think each path has contributed to a positive change or increase in entropy ...one can try for a concrete example,
 
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stevendaryl
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Does this mean that the entropy change for a reversible process is greater than that of an irreversible process? I'm convinced I am wrong because my notes say otherwise but isn't Δs the entropy change of a reversible process and (1to2∫dQirrev/T) the entropy change of an irreversible process?
That's not quite right. Let's look at a particularly simple case: Suppose you have compressed air in a one-liter bottle. You let it expand to two liters, keeping the temperature constant. There are two ways to do this, reversibly or irreversibly.

Reversible process: You put the gas into a piston, and put the piston into a bucket of water at constant temperature. You let the piston slowly expand to twice the original volume. This causes (for an ideal gas, anyway) an entropy change of [itex]\Delta S = NR log(\frac{V_final}{V_initial}) = NR log(2)[/itex]. So the entropy of the gas inside the piston goes up. But since the piston does work as it expands, that work can be used to decrease the entropy of a second system (it can be used to isothermally compress a second piston of gas, which decreases its entropy). The total change in entropy, in a reversible process, is zero: one system's entropy increases, and another system's entropy decreases.

Irreversible process:
You can just release the one-liter volume of gas into an evacuated two-liter bottle. The gas will rapidly expand to fill the bottle. The entropy change for the gas will be the same as in the first process: [itex]\Delta S = NR log(2)[/itex]. That's because entropy is a function of state--it doesn't matter how you get to that state.

So there is no difference in the entropy change for a volume of gas, whether it's done reversibly or irreversibly. The difference is that if you make the change reversibly, then the increase in entropy for one system is balanced by a decrease in entropy for a second system. If you make the change irreversibly, then there is no compensating decrease of entropy in a second system.
 
  • #4
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If we consider a system to undergo an irreversible process from state 1 to state 2 and a reversible process from state 2 to state 2, then through Clausius inequality

(1to2∫dQirrev/T) + (2to1∫dQrev/T) ≤ 0
(1to2∫dQirrev/T) + s1 - s2 ≤ 0
s2-s1 ≥ (1to2∫dQirrev/T)
Δs ≥ (1to2∫dQirrev/T)

Does this mean that the entropy change for a reversible process is greater than that of an irreversible process? I'm convinced I am wrong because my notes say otherwise but isn't Δs the entropy change of a reversible process and (1to2∫dQirrev/T) the entropy change of an irreversible process?
1to2∫dQirrev/T does not represent the entropy change for the irreversible path. In an irreversible process, the temperature of the system is not spatially uniform, and varies from location to location within the system. So what is the value of T that is supposed to be used in this integration? It is supposed to be the temperature at the interface between the system and surroundings where the heat transfer is occurring (not the average temperature of the system). So what the Clausius inequality really says is that, for an irreversible process, the change in entropy for the system is greater than the integral of dQB/TB, where dQB represents an increment of heat transferred to the system across its boundary and TB represents the absolute boundary temperature at which this heat transfer is occurring.

Some people (myself included) like to interpret (1to2∫dQirrev/T) as the amount of entropy transferred to the system from the surroundings across the system boundary. This, plus the entropy generated within the system (as a result of irreversibilities) is equal to the change in entropy of the system.
 

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