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Old May 5th 2009, 03:31 PM   #1
C.E
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Increasing a the volume of a gas

A gas has its volume instantaneously tripled by expansion into a vacuum. The
system remains adiabatic at all times.
(i) Is this process reversible or irreversible? Explain your choice.

(ii) Calculate the change in internal energy and entropy, for this process.

I didn't know how to answer the first bit of this question, could somebody please help? For the second part I though there would be no change in internal energy (as the system is adiabatic) is this correct?
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Old May 5th 2009, 04:37 PM   #2
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Originally Posted by C.E View Post
A gas has its volume instantaneously tripled by expansion into a vacuum. The
system remains adiabatic at all times.
(i) Is this process reversible or irreversible? Explain your choice.
(ii) Calculate the change in internal energy and entropy, for this process.

I didn't know how to answer the first bit of this question, could somebody please help? For the second part I though there would be no change in internal energy (as the system is adiabatic) is this correct?
(i) I've read in a book (molecular physics by Isaac Kikoin) that this process is irreversible. (It's related to entropy that increases instead of being a constant). But I didn't cover it yet at university so I can't enter in details.
It would have been reversible if the gas suffered infinitely small changes in volume. (I'm pretty sure you have to show this using entropy formula)
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Old May 23rd 2009, 04:59 PM   #3
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Ok, I think I have now have an answer.

(i). No, by considering work arguments.

(ii).$\displaystyle \Delta u = 0$

$\displaystyle \Delta S= nRlog(3)$ (seems a bit odd that I have an unknown, n= number of moles, in my final answer).

Anyway what do you think?

Last edited by C.E; May 23rd 2009 at 05:04 PM.
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Old Jul 17th 2009, 05:38 AM   #4
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Originally Posted by C.E View Post
Ok, I think I have now have an answer.

(i). No, by considering work arguments.

(ii).$\displaystyle \Delta u = 0$

$\displaystyle \Delta S= nRlog(3)$ (seems a bit odd that I have an unknown, n= number of moles, in my final answer).

Anyway what do you think?
I agree. No work is done on the gas and no heat flows into the gas so the internal energy of the gas can't change. This implies that the temperature of the gas doesn't change in the process. Since there is no process which can be done to place the system back into its original state the rapid expansion is irriversible.
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Old Jul 17th 2009, 02:33 PM   #5
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-i. adiabatic détente is working with mechanical failure of the initial internal energy.
If compressed gas back to the position initiala.Procesul is reversible.
-ii. in an adiabatic transformation, internal energy change and entropy is constant.
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