What Happens to Enthalpy in Constant Volume and Pressure Processes?

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Discussion Overview

The discussion revolves around the behavior of enthalpy in constant volume and constant pressure processes within thermodynamics. Participants explore the definitions and relationships between internal energy and enthalpy, particularly in the context of different types of substances and processes.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Enthalpy is defined as h = u + pv, leading to the relationship dh = du + d(pv) for changes in enthalpy.
  • During a constant pressure process, heat added increases both internal energy and boundary work, resulting in an increase in enthalpy.
  • In a constant volume process, the change in enthalpy is questioned, particularly the term v dp, and whether internal energy change equals enthalpy change.
  • Some participants note that for incompressible substances, changes in enthalpy and internal energy can be considered equal.
  • There is confusion regarding the applicability of property relations for gases in constant volume processes, especially concerning the definitions of cp and cv.
  • It is suggested that while enthalpy can be defined as a state variable, the relationship dh = cp dT is not applicable in constant volume scenarios due to the nature of cp.
  • Participants discuss the First Law of Thermodynamics and its implications for understanding changes in internal energy and heat transfer.

Areas of Agreement / Disagreement

Participants generally agree on the definitions of enthalpy and internal energy, but there is disagreement regarding the applicability of certain relationships in different processes, particularly for gases. The discussion remains unresolved on how to define changes in enthalpy during constant volume processes.

Contextual Notes

Participants express uncertainty about the definitions and relationships between cp and cv, particularly in the context of compressible versus incompressible substances. The applicability of thermodynamic property relations in various processes is also a point of contention.

jason.bourne
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enthalpy is defined as

h = u + pv

so, dh = du + d (pv)

if the process is constant pressure,

dh = du + p dv (change in internal energy + boundary work )

during constant pressure process, if heat is added to the system, it increases its internal energy as well as it does the expansion boundary work. so the net heat added will increase the enthalpy of the system.

if the process is constant volume,

dh = du + v dp (what is this extra term v dp??).

during constant volume process, heat added will just raise the internal energy. so what happens to enthalpy ?
 
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It will equal the change in internal energy because there is no boundary work performed by the system.
 
thanks for the reply Skrambles !

there won't be any boundary work, agreed.

will internal energy change be equal to enthalpy change?

we know for some change in temperature dT,
du = cv dT and
dh = cp dT

and cp and cv are never same.
 
They are indeed equal for an incompressible substance. (k=1)

dh = cp dT applies only when pressure is constant.

du = cv dT applies only when volume is constant.
 
they are equal for incompressible fluids such as water - agreed.

what happens when we consider a gas, assuming compressibility condition holds, in a constant volume process?

"du = cv dT applies only when volume is constant"

so does that mean we can't define change in enthalpy during constant volume process?

this is bit confusing coz the thermodynamics textbooks suggests, property relations can be used irrespective of the processes, reversible or irreversible coz they are state variables and don't depend on the path of integration.
 
jason.bourne said:
so does that mean we can't define change in enthalpy during constant volume process?

No, it just means that you can't use dh = cp dT because of the way cp is defined.

If you can define the state of the system, then you can define its enthalpy because, like you said, it is a state variable.

The change in enthalpy equals the change in internal energy plus boundary work performed during the process. If there is no boundary work, then dh = du and du = cv dT.

You can also use the First Law to get ΔQ = ΔU.

Don't feel bad about getting confused, I struggled for a while with the same concept when I took thermo. I went into a test thinking I could use cp and cv for every process, so I was very surprised when I saw my grade.
 
hahaa...

alright! got it.

thanks for helping me out !
:))
 

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