How Do You Derive Irreversible Adiabatic Work Equations?

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SUMMARY

The discussion focuses on deriving equations for irreversible adiabatic work, contrasting it with reversible adiabatic work. The key equations mentioned include the first law of thermodynamics, \(dU = dq + dW\) (with \(q = 0\) for adiabatic processes), and the work done in reversible processes expressed as \(W = \int P_{ext} dV\). The participants emphasize that the derivation for irreversible processes cannot follow the same path as reversible ones due to differences in internal and external pressures. A specific example provided illustrates that for constant external pressure, the work done is \(W = P_{ext} \Delta V\).

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with adiabatic processes and their characteristics
  • Knowledge of the relationship between pressure, volume, and temperature in gases
  • Ability to perform integration in thermodynamic equations
NEXT STEPS
  • Study the derivation of reversible adiabatic work equations
  • Learn about the implications of the first law of thermodynamics in irreversible processes
  • Explore the concept of internal vs. external pressure in thermodynamic systems
  • Investigate specific examples of irreversible adiabatic processes and their work calculations
USEFUL FOR

Students in thermodynamics, engineers working with gas systems, and anyone interested in understanding the differences between reversible and irreversible adiabatic processes.

speny83
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Homework Statement


So I came down with something and missed some class and am making it up from others notes which arent entirely clear plus the long weekend for turkey day and I am really hoping someone can help me make some sense of this...

I am trying to come up with the derivations for irreversible adiabatic expansions/work, ill try to explain as much as possible and hopefully it will be sufficient for someone to see what I am getting at give me a hand






Homework Equations



The notes begin with adiabatic reversible work
Du=Cv,mdT
dU=dq+dW -->q=0 for adiabatic so dU=dW
pex=Pf=nRT/V
∴nCv,mdT=-nRTdV/V

the notes then integrate each side wrt T & V end then exponentiate each side to arrive at v1T1c=v2T2c where c =Cv,m/R

they then get sloppy to where i can't make anything of it and arrive at Pivi\gammaPfvf\gamma where\gamma=Cpm/Cvm


Then there is a huge note that says DO NOT USE THIS IF WORK IS IRREV and that the instructor wants us to derive similar for irreversible work Hint: (it changes where dU=dw)

The Attempt at a Solution



I tried working on this all morning and didnt make any progress...integrating dV/V to get the ln(vf/vi) is how we were taught to do reversible work and for non-reversible it was just w=-PexΔV

I don't see any way this could turn into similar expressions relate T,P&V

Someone please help! If i see a derivation a couple times they make sense but I can never figure this out solo.

Thank you so much
 
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speny83 said:

Homework Statement


So I came down with something and missed some class and am making it up from others notes which arent entirely clear plus the long weekend for turkey day and I am really hoping someone can help me make some sense of this...

I am trying to come up with the derivations for irreversible adiabatic expansions/work, ill try to explain as much as possible and hopefully it will be sufficient for someone to see what I am getting at give me a hand

Homework Equations



The notes begin with adiabatic reversible work
Du=Cv,mdT
dU=dq+dW -->q=0 for adiabatic so dU=dW


pex=Pf=nRT/V
∴nCv,mdT=-nRTdV/V

the notes then integrate each side wrt T & V end then exponentiate each side to arrive at v1T1c=v2T2c where c =Cv,m/R

they then get sloppy to where i can't make anything of it and arrive at Pivi\gammaPfvf\gamma where\gamma=Cpm/Cvm
The derivation can be found here.

Then there is a huge note that says DO NOT USE THIS IF WORK IS IRREV and that the instructor wants us to derive similar for irreversible work Hint: (it changes where dU=dw)

The Attempt at a Solution



I tried working on this all morning and didnt make any progress...integrating dV/V to get the ln(vf/vi) is how we were taught to do reversible work and for non-reversible it was just w=-PexΔV

I don't see any way this could turn into similar expressions relate T,P&V

Someone please help! If i see a derivation a couple times they make sense but I can never figure this out solo.
You cannot derive a general expression for irreversible processes since they are all different. For reversible processes, the internal pressure is equal to the external pressure so the work done against the external pressure is ∫PextdV = ∫PintdV = ∫nRTdV/V. But if the external pressure is lower than the internal pressure of the gas (irreversible) you can't use this.

For example, if the external pressure is constant, the work done by the gas is W = PextΔV. If you apply the first law, Q = ΔU + W where W is the work done by the gas, then (since Q=0) ΔU = -W so ΔU = nCvΔT = -PextΔV . From that you could work out the relationship between P and V in such a process. But, as I said, you need to know some details about the process before you can relate P and V.

AM
 
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