Why dU may not equals dW in free expansion?

[kelvin490]

From a video lecture, it is mentioned that "dU≠dW in Joule's free expansion if the process is irreversible and adiabatic"
Mentioned in around 36:00-38:00 in the video:

What I would like to ask is why in this irreversible adiabatic process, dU≠dW? Is it because the W here doesn't include other sort of work?

 

[Thermo]

dU can't be equal to dW if it is irreversible. First and Second rule of thermodynamics which are dU=dQ+dW and dQrev=Tds are for reversible processes. dWir<-PdV and dQir<TdS. However if we combine first and second laws we substitute dU=TdS-PdV which is for reversible and irreversible processes both. Because (U,S,V,T and P) all state variables and the change of those variables not dependent on the way of change. That's why it's not important reversible or irreversible.

 

[Chestermiller]

The reason you are so confused about what this guy is saying is because it is basically incorrect. What he is trying to demonstrate is that dU represents the change in internal energy between two closely neighboring thermodynamic equilibrium states. However, the example he has chosen involves two widely separated thermodynamic equilibrium states. Worse yet, he seems to be using dU and ΔU interchangeably, even though they are not and even though nowhere does he mention ΔU.

You were obviously correct in concluding that, for the irreversible adiabatic free expansion he described, ΔU=0. Of course, there are many reversible paths between these two widely separated equilibrium states. If you determined dU along the segments of any of these paths, the dU for each segment might not be zero, but the integral of dU over each and every entire path would have to be zero.

He is also trying to show that the changes in U, S, and V between two closely neighboring thermodynamic equilibrium states of a system are not independent, but are related to one another by the equation dU = TdS-pdV

Hope this helps.

Chet

 

[kelvin490]

dU can't be equal to dW if it is irreversible. First and Second rule of thermodynamics which are dU=dQ+dW and dQrev=Tds are for reversible processes. dWir<-PdV and dQir<TdS. However if we combine first and second laws we substitute dU=TdS-PdV which is for reversible and irreversible processes both. Because (U,S,V,T and P) all state variables and the change of those variables not dependent on the way of change. That's why it's not important reversible or irreversible.

The reason you are so confused about what this guy is saying is because it is basically incorrect. What he is trying to demonstrate is that dU represents the change in internal energy between two closely neighboring thermodynamic equilibrium states. However, the example he has chosen involves two widely separated thermodynamic equilibrium states. Worse yet, he seems to be using dU and ΔU interchangeably, even though they are not and even though nowhere does he mention ΔU.

You were obviously correct in concluding that, for the irreversible adiabatic free expansion he described, ΔU=0. Of course, there are many reversible paths between these two widely separated equilibrium states. If you determined dU along the segments of any of these paths, the dU for each segment might not be zero, but the integral of dU over each and every entire path would have to be zero.

He is also trying to show that the changes in U, S, and V between two closely neighboring thermodynamic equilibrium states of a system are not independent, but are related to one another by the equation dU = TdS-pdV

Hope this helps.

Chet

Actually I tend to agree with Chestermiller but still wonder whether the reversibility will affect dU in the way Thermo has mentioned. Let's imagine the adiabatic expansion is not free but the pressure decrease very slowly and the gas actually do some work, the dU certainly is not zero.

So I think what the guy in the video was trying to do is to compare the dU in reversible and irreversible processes, but he got the wrong conclusion. Because the final state that the system achieve by a reversible path is different from that by a irreversible path. One cannot say because dU is not zero in a reversible path then it is also not zero if the gas expand freely. First law of thermodynamics must be obey so if dQ and dW are zero then dU must be zero.

 

[Chestermiller]

Actually I tend to agree with Chestermiller but still wonder whether the reversibility will affect dU in the way Thermo has mentioned. Let's imagine the adiabatic expansion is not free but the pressure decrease very slowly and the gas actually do some work, the dU certainly is not zero.

So I think what the guy in the video was trying to do is to compare the dU in reversible and irreversible processes, but he got the wrong conclusion. Because the final state that the system achieve by a reversible path is different from that by a irreversible path. One cannot say because dU is not zero in a reversible path then it is also not zero if the gas expand freely. First law of thermodynamics must be obey so if dQ and dW are zero then dU must be zero.

I totally agree.

Here is a link to my recent article in Physics Forums Insight that may be of interest to you:
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

Chet

 

[Thermo]

"The guy" is me or the one on the video? I didn't understand who you mentioned Professor. I just shared what I know about Thermodynamics basically maybe I was wrong. I will read your article for better understanding. Thanks for sharing.

 

[Chestermiller]

"The guy" is me or the one on the video? I didn't understand who you mentioned Professor. I just shared what I know about Thermodynamics basically maybe I was wrong. I will read your article for better understanding. Thanks for sharing.

I was referring to the guy in the video.