IRREVERSIBLE adiabatic work

In summary: ENDMENT:In summary, the notes discuss the derivation of adiabatic reversible work using the equation Du = Cv,mdT and the integration of each side with respect to T and V, followed by exponentiation. However, the notes also mention that this method cannot be used for irreversible work. The attempt at a solution includes trying to integrate dV/V and using the equation w = -PexΔV, but the relationship between P and V in an irreversible process cannot be determined without further information.
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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[itex]\gamma[/itex]Pfvf[itex]\gamma[/itex] where[itex]\gamma[/itex]=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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  • #2
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[itex]\gamma[/itex]Pfvf[itex]\gamma[/itex] where[itex]\gamma[/itex]=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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1. What is "IRREVERSIBLE adiabatic work"?

"IRREVERSIBLE adiabatic work" refers to a type of thermodynamic process in which a system undergoes a change in its thermodynamic state without any heat exchange with its surroundings. This type of work is considered irreversible because it cannot be reversed or undone without the introduction of heat from an external source.

2. How is "IRREVERSIBLE adiabatic work" different from "REVERSIBLE adiabatic work"?

The main difference between "IRREVERSIBLE adiabatic work" and "REVERSIBLE adiabatic work" is that in the latter, the process can be reversed without the introduction of heat from an external source. This is because the system undergoes a series of quasi-equilibrium states, meaning that it is always in thermal equilibrium with its surroundings.

3. What is the formula for calculating "IRREVERSIBLE adiabatic work"?

The formula for calculating "IRREVERSIBLE adiabatic work" is W = P(V2 - V1), where W is the work done, P is the pressure, V2 is the final volume, and V1 is the initial volume. This formula applies to a closed system where the pressure remains constant during the process.

4. What are some real-life examples of "IRREVERSIBLE adiabatic work"?

One example of "IRREVERSIBLE adiabatic work" is the expansion of a gas in an insulated container. Another example is the compression of air in a car engine. In both cases, the process is adiabatic and irreversible, as heat cannot be exchanged between the system and its surroundings.

5. What are the implications of "IRREVERSIBLE adiabatic work" in thermodynamics?

One of the main implications of "IRREVERSIBLE adiabatic work" in thermodynamics is the concept of entropy. During an irreversible adiabatic process, the entropy of a closed system increases, meaning that the system becomes more disordered. This is in contrast to a reversible adiabatic process, where the entropy remains constant.

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