Internal energy or enthelpy are minimized in isoentropic processes?

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SUMMARY

This discussion focuses on demonstrating that internal energy (U) is minimized at constant volume (V) and entropy (S), while enthalpy (H) is minimized at constant pressure (P) and entropy (S) in isoentropic processes. The key equations utilized include the Clausius inequality and the relationships between changes in entropy, internal energy, and enthalpy. The participant identifies a critical flaw in assuming that isoentropic processes are always adiabatic, leading to a deeper understanding of the conditions under which U and H are minimized.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically isoentropic processes
  • Familiarity with the Clausius inequality and its implications
  • Knowledge of the relationships between internal energy (U), enthalpy (H), and entropy (S)
  • Basic calculus for evaluating integrals in thermodynamic equations
NEXT STEPS
  • Study the implications of the Clausius inequality in thermodynamics
  • Learn about isoentropic processes and their characteristics in detail
  • Explore the derivation of the relationships between U, H, and S in various thermodynamic processes
  • Investigate the conditions under which adiabatic and irreversible processes occur
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying physical chemistry, mechanical engineering, or any field that involves energy transformations and thermodynamic 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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