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.
5LAY3R95
Messages
2
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top