Why is delta-H equal to q at constant pressure?

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Discussion Overview

The discussion revolves around the relationship between heat transfer (q) and change in enthalpy (ΔH) at constant pressure in the context of thermodynamics. Participants explore the implications of the first law of thermodynamics and the definitions of work and enthalpy, seeking to clarify how ΔH relates to q under specific conditions.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • The original poster (OP) expresses confusion about why q at constant pressure equals ΔH, given their understanding of the first law of thermodynamics.
  • Some participants point out that at constant pressure, work (w) is defined as -p times ΔV, leading to a modified first law equation: ΔU = q - pΔV.
  • There is a suggestion to consider how ΔH is defined in terms of ΔU and Δ(pV), and what this means when pressure is constant.
  • One participant argues that ΔH should equal q at all times, questioning the role of work in this relationship.
  • Another participant counters this by stating that work is not always -pΔV and that ΔH is more generally ΔU + Δ(pV), prompting further exploration of these definitions.
  • There is a reference to the multiplication rule from differential calculus to derive the relationship between d(PV) and its components, suggesting a deeper mathematical exploration.
  • One participant expresses satisfaction with their understanding after reviewing the discussion, while another provides clarification on the derivation of d(PV) as a necessary step for understanding ΔH at constant pressure.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relationship between ΔH and q, with some asserting that they should always be equal while others argue that this is conditional on the definitions of work and enthalpy. The discussion remains unresolved regarding the implications of these definitions.

Contextual Notes

Limitations include the potential misunderstanding of the definitions of work and enthalpy, as well as the assumptions made about constant pressure scenarios. The discussion also highlights the need for clarity in the mathematical derivations involved.

DiamondV
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Homework Statement


In my lecture notes(beginner thermodnyamics), we just got introduced to the first law(DeltaU=q+w) and two scenarios. One at constant volume which yields the equation Delta U = qv. I understand that
The second scenario is at Constant pressure and it says that At constant pressure q subcript p = DeltaH. DeltaH being change in enthalpy. I don't understnad that bit at all. How is being at constant pressure tell you that the heat energy is the change in enthalpy?

Homework Equations

The Attempt at a Solution

 
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If the pressure is constant, what is w equal to?
 
Chestermiller said:
If the pressure is constant, what is w equal to?
w = -p times deltaV. note that p is the external pressure
So if pressure is constant, the formula is the same.
 
DiamondV said:
w = -p times deltaV. note that p is the external pressure
So if pressure is constant, the formula is the same.
OK. What do you get if you substitute that into your first law equation?
 
Chestermiller said:
OK. What do you get if you substitute that into your first law equation?
w = -p times deltaV. First law is DeltaU = q+w. So subbing it in would give me:
DeltaU = q-pDeltaV
 
DiamondV said:
w = -p times deltaV. First law is DeltaU = q+w. So subbing it in would give me:
DeltaU = q-pDeltaV
OK. Now how is ΔH defined in terms of ΔU and Δ(pV)?
What does this defining equation reduce to if p is constant?

Chet
 
Chestermiller said:
OK. Now how is ΔH defined in terms of ΔU and Δ(pV)?
What does this defining equation reduce to if p is constant?

Chet
Well DeltaH is change in enthalphy which is change in heat energy, so shouldn't DeltaH always equal to q no matter if its at constant pressure or not since the only other energy transfer is work and that doesn't involve heat. So at all times DeltaH =q? I've just confused myself somehow.
 
DiamondV said:
Well DeltaH is change in enthalphy which is change in heat energy, so shouldn't DeltaH always equal to q no matter if its at constant pressure or not since the only other energy transfer is work and that doesn't involve heat. So at all times DeltaH =q? I've just confused myself somehow.
No, because the work is not always -pΔV, it is more generally the integral of -pdV. And ΔH is not always ΔU+pΔV, it is more generally ΔU+Δ(pV). Try it using that information and see what you get.
 
Remember the multiplication rule from differential calculus: d(f(x)*g(x))/dx =g(x)* df(x)/dx + f(x)* dg(x)/dx. Or, in the differential form (shortcut):
d(f(x)*g(x)) = g(x)* df(x) + f(x)* dg(x). and so, d(p*v) = ?
Now integrate that, given p=constant.
 
  • #10
Mark Harder said:
Remember the multiplication rule from differential calculus: d(f(x)*g(x))/dx =g(x)* df(x)/dx + f(x)* dg(x)/dx. Or, in the differential form (shortcut):
d(f(x)*g(x)) = g(x)* df(x) + f(x)* dg(x). and so, d(p*v) = ?
Now integrate that, given p=constant.
Do you find an error in what DiamondV and I have done so far?

Chet
 
  • #11
Chestermiller said:
Do you find an error in what DiamondV and I have done so far?

Chet
Thanks. I think I've gotten a grasp of it anyways. had my final thermodynamics exam, never seeing it again thank god.
 
  • #12
Chestermiller said:
Do you find an error in what DiamondV and I have done so far?

Chet
No, not at all. Reading over the posts in this thread, I couldn't find an explanation for why d(PV) = PdV + VdP, so I provided one, just in case the OP didn't know that. It's a necessary step in the derivation for delta-H at constant pressure.
Mark
 

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