Internal energy or enthelpy are minimized in isoentropic processes?

AI Thread Summary
In isoentropic processes, internal energy (U) is minimized at constant volume (V) and entropy (S), while enthalpy (H) is minimized at constant pressure (P) and S. The discussion highlights the relationship between changes in entropy, energy, and work in both reversible and irreversible processes. It emphasizes that an isoentropic process is adiabatic, leading to the conclusion that the heat transfer (δq) is zero in irreversible scenarios. The analysis reveals that assumptions about the nature of isoentropic processes may lead to contradictions, particularly regarding the Clausius inequality. Ultimately, the reasoning presented suggests a clearer understanding of the conditions under which U and H are minimized in these processes.
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Homework Statement



How to demonstrate that U is minimized at constant V and S, while H at constant P and S?

Homework Equations



ΔS universe = ΔS system + ΔS environment ≥ 0
ΔU system = δq reversible + δw reversible = δq irreversible + δw irreversible
ΔS environment = −∫(δq reversible / T)
dU = TdS − PdV = δq + δw

The Attempt at a Solution



ΔS universe = ΔS system + ΔS environment > 0
ΔS universe = ΔS system − ∫ ( δq rev /T ) > 0
ΔS universe = ΔS system − ∫( δq irrev/T + δw irrev − δw rev ) > 0
TΔS system − T∫( δq irrev/T + δw irrev − δw rev ) > 0
T∫( δq irrev/T + δw irrev − δw rev ) − TΔS system < 0

If the process is isoentropic, it follows that it's an adiabatic process with δq irrev = 0

T∫( δw irrev − δw rev ) < 0

* If, furthermore, the volume of the system is constant, it follows that irreversible and reversible works are 0, leading to a senseless expression.* If not the volume but the pressure is constant, we get
T∫( δw irrev − δw rev ) < 0
T∫( δq rev − δq irrev ) < 0 and we recover ∫δq irrev = ΔH

Now, since reversible −∫PdV is minimum in reversible process, δ rev is maximum.
It follows that the T∫(positive quantity) < 0
Which is another senseless expression.

I guess the flaw comes from the assumption that an isoentropic process is always adiabatic... but how?
 
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Now I think I have found it: Clausius inequality
dS ≥ δq reversible/T
I assumed that isoentropic process implied adiabatic irreversible process; however, it means that δq rev≤0

So I have that
ΔS universe = ΔS system − ∫ ( δq rev /T ) > 0

If the process is isoentropic
ΔS universe = − ∫ ( δq rev /T ) > 0
∫ ( δq rev /T ) < 0
∫ ( dU − δw rev ) /T < 0

If now I assume that the volume is constant, I get δw = 0
∫ dU / T < 0

Which seems like the inequality I have been looking for.

∫ ( dU − δw rev ) / T < 0
∫ ( dΗ − PdV − VdP − δw rev ) / T < 0

If I assume that pressure is constant, I get

∫ ( dΗ − PdV − δw rev ) / T < 0

Now If I assume mechanical equilibrium

∫ dΗ / T < 0

Now I guess my reasoning is okay? Someone answer please.
 
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