Enthelpy (H=U+PV) What is P, really?

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SUMMARY

The discussion centers on the interpretation of the enthalpy change equation, ΔH = ΔU + P * ΔV, particularly the meaning of pressure (P) within this context. Participants clarify that P represents system pressure, which is assumed constant in this equation, making it less applicable for non-equilibrium systems like engines or refrigerators. They emphasize that for varying conditions, ΔH can be calculated using more complex integrals, such as ΔH = ΔU + ∫P dV + ∫V dP or ΔH = ∫C_P dT. The nuances of pressure's role in enthalpy changes are critically examined, highlighting the importance of understanding system conditions.

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  • Understanding of thermodynamic concepts, specifically enthalpy and internal energy.
  • Familiarity with the first law of thermodynamics and its equations.
  • Knowledge of calculus, particularly integrals, as they apply to thermodynamic equations.
  • Basic principles of pressure in thermodynamic systems.
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  • Study the derivation and applications of the enthalpy equation in various thermodynamic processes.
  • Learn about the implications of constant vs. variable pressure in thermodynamic systems.
  • Explore the integral forms of enthalpy change, specifically ΔH = ∫P dV and ΔH = ∫C_P dT.
  • Investigate real-world applications of enthalpy in engines and refrigeration cycles.
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InTuoVultu
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I'm studying for the Physics GRE and going over my thermo notes.
Ok, so for enthalpy change

(delta)H = (delta) U + P * (delta) V

H is enthalpy, U is just proportional to T
What pressure are they talking about?
If the volume and temperature are changing, then the internal pressure is definitely changing, but in this equation it seems to be constant through out the process. So I assume that this is supposed to mean the pressure of the surroundings.
But this is tricky.
If you have a box of gas in space, the surrounding pressure is zero. So it seems like the only enthalpy change would be due to U. But this doesn't seem right. So let's look at how it expands.
Some latch could get thrown during the process and allow the piston to slide out frictionlessly. Or the piston could be connected to a spring that would get compressed as the gas expanded. Or the piston could just be rusty and release energy as heat when it moves.

I feel like all of these scenarios would have a different enthalpy change, but the equation doesn't seem to account for this. I'm guessing there's something more interesting in the P factor. What does it really mean?
-thanks
 
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The P is system pressure, and the equation you give assumes a constant system pressure (which is not always the case). This form is useful when a system is at equilibrium with the pressure of its surroundings.

(Also, I don't agree that if the volume and temperature of a system are changing, then "the internal pressure is definitely changing.")
 
P is system pressure then?
I know that T and V changing does not imply that P changes. It just usually does when the system is not in mechanical contact with its surroundings.

So I'm guessing that (delta)H is only useful in a setup where the system is held at constant pressure? This means that (delta)H is not very useful when it comes to engines/refrigerators or any system that is not held at constant pressure.
 
Your equation above assumes constant P, so not surprisingly it's only useful for constant-pressure situations. For more general situations, we can calculate \Delta H by

\Delta H=\Delta U+\int P\,dV+\int V\,dP

or

\Delta H=\int C_P\,dT
 
Mapes said:
Your equation above assumes constant P, so not surprisingly it's only useful for constant-pressure situations. For more general situations, we can calculate \Delta H by

\Delta H=\Delta U+\int P\,dV+\int V\,dP

or

\Delta H=\int C_P\,dT


Can u please explain why there is both Delta P * V and Delta V * P?

Would this not take Work into account twice?

When P is constant, the Delta P * V term = 0?

When V is constant, the Delta V * P term =0? I am just guessing
 
ILovePhysics! said:
Can u please explain why there is both Delta P * V and Delta V * P?

Would this not take Work into account twice?

When P is constant, the Delta P * V term = 0?

When V is constant, the Delta V * P term =0? I am just guessing

Work is \int P\,dV (magnitude), not V\Delta P or V\Delta P+P\Delta V, and not necessarily P\Delta V if the pressure isn't constant.

I agree with your guesses.
 

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