Is Heat Exchange Equal to Enthalpy Change in an Isobaric Process?

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Discussion Overview

The discussion revolves around the relationship between heat exchange and enthalpy change in isobaric processes within closed thermodynamic systems. Participants explore whether the equality of heat exchange and enthalpy change can imply constant pressure and seek counterexamples to this assertion.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that in an isobaric process, the heat exchanged equals the change in enthalpy.
  • One participant questions whether the reverse is true: if heat exchange equals enthalpy change, can it be concluded that pressure remains constant?
  • A participant provides a mathematical derivation linking enthalpy change to work done, suggesting that if work is expressed as PdV, then constant pressure follows.
  • Another participant challenges the interpretation of heat and work as true differentials, arguing against the use of delta notation for these quantities.

Areas of Agreement / Disagreement

Participants express differing views on the implications of heat exchange equating to enthalpy change, with no consensus reached on whether this implies constant pressure. The discussion remains unresolved regarding the interpretation of heat and work as differentials.

Contextual Notes

There are limitations in the assumptions made about the definitions of heat and work, as well as the mathematical steps involved in deriving relationships between these quantities.

Carcul
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Consider a closed thermodynamic system capable of exchanging energy in the form of work only as PV work. Under these conditions, for an isobaric process we find that the heat exchanged equals the enthalpy change. Now what about the reverse? For such a system, if for some process the heat exchange equals the change in enthalpy can we conclude that the pressure has remained constant? If not can you find a counter example?
 
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Carcul said:
Consider a closed thermodynamic system capable of exchanging energy in the form of work only as PV work. Under these conditions, for an isobaric process we find that the heat exchanged equals the enthalpy change. Now what about the reverse? For such a system, if for some process the heat exchange equals the change in enthalpy can we conclude that the pressure has remained constant? If not can you find a counter example?
Yes, but only if ∂W = PdV. You can prove this from the definition of enthalpy:

H = U + PV
dH = dU + PdV + VdP

If dH = ∂Q = dU + ∂W then ∂W = PdV + VdP

If ∂W = PdV then VdP = 0 which implies that dP = 0 (isobaric)

AM
 
Thank you very much. But why does ΔH = Q implies dH = δQ?
 
Last edited:
It doesn't.

Neither the heat nor the work exchanged are true differentials, they are actual values so it is wrong talk of delta (of any sort) q or w. Some people prefer capitals, some prefer lower case some use the (as Andrew has done) Greek delta to show this.

But the bottom line is that the heat exchanged is the heat exchanged it is not a small change in the heat exchanged.
 

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