Why dU may not equals dW in free expansion?

Jun 2014
67
0
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: https://www.youtube.com/watch?v=RrVq7Yduz2g

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? From first law if dQ is zero dU should be equal to dW.
 
Apr 2015
1,035
223
Somerset, England
I had a quick look at that lecture, but couldn't get any sound.

Anyway I can't recommend that approach to thermo, something seems wrong to me.

The first law is really a statement of conservation of energy so

Energy stored is always equal to energy in minus energy out.

I don't like the use of differentials for heat exchanged or work done, I think that is misleading.

Adiabatic gas expansions are accompanied by temperature changes as well as volume changes, but dU = q - w in both the cases of revdersible and irreversible change.

It is instructive to calculate the energy changes associated with a real example

say 10L of a monatomic gas expanded from 10 atm to 1 atm starting at 273K
 
Jun 2014
67
0
I had a quick look at that lecture, but couldn't get any sound.

Anyway I can't recommend that approach to thermo, something seems wrong to me.

The first law is really a statement of conservation of energy so

Energy stored is always equal to energy in minus energy out.

I don't like the use of differentials for heat exchanged or work done, I think that is misleading.

Adiabatic gas expansions are accompanied by temperature changes as well as volume changes, but dU = q - w in both the cases of revdersible and irreversible change.

It is instructive to calculate the energy changes associated with a real example

say 10L of a monatomic gas expanded from 10 atm to 1 atm starting at 273K

1. Just copy the link to browser you can have the sound.

2. I don't think your example is necessary

3. 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.
 
Apr 2015
1,035
223
Somerset, England
I don't see the point of watching an hour long video presenting incorrect material, with or without sound.

dU is precisely zero if the change is isothermal. That is the definition of an ideal gas.
 
Jun 2014
67
0
I don't see the point of watching an hour long video presenting incorrect material, with or without sound.

dU is precisely zero if the change is isothermal. That is the definition of an ideal gas.

1. The differential form is very common in all thermodynamics textbook. It is the formal way to express thermodynamic relations, whether you like it or not.

2. The guy in video only got one thing (I have pointed out previously) wrong. All the other content are fine.

3. The focus of the question is not about whether it is ideal gas or whether it is isothermal (although it is), it is about the effect of reversibility on dU.
 
Apr 2015
1,035
223
Somerset, England
1. The differential form is very common in all thermodynamics textbook.
I will put this mis-statement down to English not being your first language.

The erroneous notion that q and w are somehow differential quantities appears in some simple texts, but rarely in advanced ones. Indeed the best ones take great pains to explain why they are not differential quantities.

q and w are not functions of state, indeed they are not even functions of the system itself and can be forced on the system by external agents.
They are almost never able to be treated as though they were quantities that can be handled by the differential or integral calculus.

2. The guy in video only got one thing (I have pointed out previously) wrong. All the other content are fine.
I have no more time to waste on the video

3. The focus of the question is not about whether it is ideal gas or whether it is isothermal (although it is), it is about the effect of reversibility on dU.
You originally stated the expansion was adiabatic.

I offered you consideration of both but since you rejected that it is your loss, not mine.

2. I don't think your example is necessary
 
Jun 2014
67
0
You originally stated the expansion was adiabatic.

I offered you consideration of both but since you rejected that it is your loss, not mine.

It is adiabatic, free expansion. Since dQ is zero, dU can only equal to dW which is also zero. As I am quite sure that guy in video was wrong to say dU is not zero, it should be no problem now.
 
Apr 2015
1,035
223
Somerset, England
It is quite right to say that an adiabatic free expansion is constant internal energy.

However please note that the entropy changes during such an expansion.

You should also carefully distinguish between isentropic processes such as 'throttling' and 'free expansion'.