# Homework Help: Internal energy or enthalpy are minimized in isoentropic processes?

1. May 13, 2012

### 5LAY3R95

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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?

2. May 14, 2012

### 5LAY3R95

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