Intuition Mismatch: Reconciling ΔH for Ideal Gas Isothermal Expansion

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SUMMARY

The discussion centers on the reconciliation of the change in enthalpy (ΔH) for the isothermal expansion of an ideal gas, specifically addressing the apparent contradiction between ΔH being zero and the heat transfer occurring during the process. It is established that for a reversible isothermal expansion, ΔH=0, as the internal energy remains constant. However, the confusion arises when considering the heat exchange with the surroundings, leading to the conclusion that ΔH=0 applies only to isolated systems, while for closed systems, ΔH may not equal zero due to heat transfer. The key takeaway is that the pressure of the gas does not remain constant during the expansion, which clarifies the relationship between ΔH and heat flow.

PREREQUISITES
  • Understanding of thermodynamic concepts such as enthalpy and internal energy.
  • Familiarity with the ideal gas law and its implications.
  • Knowledge of isothermal and adiabatic processes in thermodynamics.
  • Basic grasp of reversible and irreversible processes in thermodynamic systems.
NEXT STEPS
  • Study the implications of the first law of thermodynamics on closed and isolated systems.
  • Explore the differences between isothermal and adiabatic expansions in detail.
  • Learn about the derivation and application of the equation ΔH=∫CPdT for various thermodynamic processes.
  • Investigate the concept of heat transfer in thermodynamic systems and its impact on enthalpy changes.
USEFUL FOR

This discussion is beneficial for students of thermodynamics, physicists, and engineers who are exploring the principles of heat transfer and enthalpy changes in ideal gas systems.

apb000
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This is probably a trivial problem, but I can't get it straight. It's well known and easy enough to show mathematically that the change in enthalpy for the reversible isothermal expansion of an ideal gas is always zero, ΔH=0. But it's also true that the enthalpy change for a process at constant P is equal to the heat that enters (or leaves) the system, ΔH=qP. Fine.
So if I take a cylinder with a piston and place it in a heat bath, and pull the piston out in a slow, reversible fashion, I can allow that the temperature remains constant and the process is isothermal; it follows that ΔH = 0. Um...but didn't heat cross the boundary of the system? In fact doesn't it HAVE to, to keep the temperature constant? The whole process was expansion against a constant Pressure, so ΔH should equal the heat that entered the cylinder, a quantity definitely not zero.
It seems to me that with ΔT=0 the internal energy is not changing, so q=-w, and here w=P dV. Again, this is a non-zero quantity.
How do I reconcile this?
(Note: Please don't derive for me that ΔH=0 from the ideal gas law. I know, and I'm convinced. But that doesn't explain why the reasoning with the cylinder is wrong.)
 
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Short version: The heat bath cools down.

You're splitting up the problem differently in each case. The "system" is either the cylinder OR the cylinder and the heat bath. If you expand the cylinder but bring heat in from outside, you have to account for the heat LOST outside before you get to a zero-sum.
 
That sounds like it might be the right track to me. Can I get a slightly longer version? I'm taking the system to be just the gas, which I am free to do, and the above reasoning still applies. It sounds like you're saying the solution is that ΔH=0 for an isothermal expansion ONLY if the system is isolated, not merely closed. Is this correct? I've been over the several sources that derive this statement and none of them come right out and say that. Am I right in gleaning that for a closed but not isolated system, ΔH=∫CPdT, and not zero? This is the only thing that makes intuitive sense to me.
But wait. If the system is isolated, that makes it an adiabatic expansion, not an isothermal one. So which one does ΔH=0 apply to?
This is such a simple physical problem, I'm surprised no one has been able to connect the pieces to the theory in a straightforward way.
 
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The paradox, if I've got it right (and I probably haven't), is dependent on the pressure being constant:

"But it's also true that the enthalpy change for a process at constant P is equal to the heat that enters (or leaves) the system, ΔH=qP."

But I can't see how the pressure CAN be constant for the isothermal expansion you describe. Surely, at constant temperature, if V goes up, p must go down?
 
Okay, I think the issue is that I'm imagining the expansion is against a constant pressure, which is perfectly reasonably physically, but the pressure in the gas will not be constant. So I can't take the enthalpy to be equal to the heat flow. Simply, ΔH ≠qP in this situation, and ΔH = 0 even though the there is heat flow across the cylinder.
Is that right?
 
That's my reading of it. ΔT = 0 and Δ(PV) = 0, so ΔH = 0.
 
Okay that makes sense. Thanks.
 

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