# Is an adiabatic process possible for an open system?

1. Jun 12, 2015

### gajagaja

Hello all!

Think of an adiabatic process (can be either reversible or irreversible - doesn't matter). dQ = 0. Assume only, mechanical work is done on/by the system. So by 1st law of thermodynamics, dU =dW.

Now, everything is clear to me when I am dealing with an adiabatic process in a closed system, i.e, there is both: (a) no mass transfer (dM = 0), and (b) no heat transfer (dQ = 0).

But, let us say, I imagine an adiabatic process in an open system (where only mechanical work is done), i.e, dQ = 0, but dM NOT EQUAL to 0.

In other words, I am making dM to not be zero, by either doing some mechanical work or letting the system do some mechanical work (using the dW term).

But now, the second law of thermdynamics tells us that, mechanical work, is never possible without some amount of it getting lost as heat (Am I right about this understanding of the second law - or is this wrong??)

Therefore, part of dW is also used to generate a non-zero dQ term. But I started off with the assumption that the process was adiabatic, so I can not have a non-zero dQ term.

So now, I contradict myself, if I assume that, mass transfer also happens in an adiabatic process. But mass transfer happens in all open systems, so overall am I right to say that, adiabatic processes are not possible in open systems???? Or have I reached this contradiction, only due to a poor understanding of the second law??

I would thank you all for your help and useful discussion on this!

Regards,

Gajagaja

2. Jun 12, 2015

### Staff: Mentor

Have you studied the open system version of the first law of thermodynamics yet? If so, please state it for us.

Also, have you considered adiabatic reversible expansion of an ideal gas in a closed system (say a cylinder with a piston)? How does that square with your understanding of the second law of thermodynamics? (If it doesn't, then your understanding must be wrong.)

Chet

Last edited: Jun 12, 2015
3. Jun 12, 2015

### gajagaja

Hello Chet,

I was not familiar with the first law of thermodynamics for open systems. But now, I read about it in the following link:

http://www.learnengineering.org/2013/03/frist-law-of-thermodynamics-open-system.html

And, per this page, there are two additional terms corresponding to the K.E and P.E of the stuff which undergoes mass-transfer.

So now, I think it makes sense to me. By doing some mechanical work, some of the dW gets converted to either the K.E or P.E corresponding to whatever mass which is getting transferred. So since the work has been converted into P.E and K.E, there is no violation of the second law of thermodynamics. So dQ term still remains zero.

Can you please tell me, if this line of reasoning makes sense?

Regards,

Gajagaja.

4. Jun 12, 2015

### gajagaja

Oh and yes - I do have a problem with all kinds of adiabatic processes and their compatability with the second law of thermodynamics.

For example, in any adiabatic process, dQ = 0 (by definition). So, in a closed system, dU = dW.

But now again, due to second law of thermodynamics, mechanical work, is never possible without some amount of it getting lost as heat (Am I right about this understanding of the second law - or is this wrong??)

So again, dQ is NOT EQUAL TO ZERO here!

So, per my understanding, all adiabatic processes should violate the second law of thermodynamics!!!

But I know, this is definitely not the case, and it is bizarre to even think of the second law getting violated - So, I guess I am definitely wrong in my thinking, somewhere. I am not able to find where. Can you guys help me get where I am going wrong here??

Regards,

Gajagaja.

5. Jun 12, 2015

### Staff: Mentor

No, not really. Even if the changes in KE and/or PE are both close to zero, both the First Law and the 2nd Law are satisfied. To see this with regard to the second law, you need to consider the answer to my question about the adiabatic reversible expansion of an ideal gas in a closed system. Is the 2nd Law satisfied for that system, even though Q = 0 while work is done?

To get a better understanding of the 2nd law, it may be helpful for you to read my recent Physics Forums Insight article on Entropy and the 2nd Law at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/.

Chet