Is it possible to have a gas expans

[25362]

Is it possible to have a gas expansion process that is adiabatic and isothermic at the same time ?

 

[quark]

Let us see.

for adiabatic process
dQ = dU + PdV

and dQ = 0, dU can be written as CvdT

But for isothermal process dT = 0, so the equation becomes

PdV = 0 which shows dV = 0 and there is no expansion.

But may I know what is the idea?

Regards,

 

[hacksaw]

my first answer is not commonly, but with gases that are capable of heating on expansion, at lower pressures there is a transition to cooling on expansion. Somewhere between the two, it is conceivable that such a process is possible.

 

[25362]

To quark, the question is essentially theoretical. It was raised because of the many queries of late, concerning J-T expansion effects, and from the very fact that I'm learning a lot from the experts in this forum and am looking forward to seeing their comments, yours included.

As I see it, this is the case of a gas initially in a rigid, insulated container that is allowed to freely (i.e., irreversibly) expand into an evacuated, insulated chamber (a Joule-type expansion).

No work is done on an external body, dW=0.
The expansion is adiabatic (a pre-condition), therefore dQ=0.
From the 1st Law, as you rightly say, for an ideal gas, dU=0. But U depends solely on T. Meaning that dU = 0 reflects the fact that for an ideal gas, C_vdT = 0, or simply dT = 0.

Thus, the temperature stays constant, and we have an isothermic, adiabatic expansion. If the final volume is double the original, P₂ = 0.5 P₁ and V₂=2 V₁.

As in a throttling process dW=dH=0.
dH=integral of C_pdT=0, meaning dT=0.

Entropy, S, should increase simply because of the irreversibility of the process, dS=R ln(V₂/V₁)=-R ln(P₂/P₁) = 1.99 ln 2 = 1.38 cal/(mol.K).

The key word, as I see it, is irreversibility.

I hope not having uselessly spent your time, and that my view of the process is correct.

 

[25362]

To hacksaw, you are right. I should have added another 'key' word: gas "ideality". Thanks.

Could we say, then, that gas ideality is determined by the constancy of temperature on an adiabatic throttling (isoenthalpic) process ?

 

[quark]

25362!

First of all I should thank you. I don't know the usefulness but I really spent all these nights studying thermodynamics after my first post and I enjoyed it.

Coming to your question, Hacksaw made a good point. There may a possibility and for real gases undergoing Joule Thompson expansion temperature remains constant. (straight from my TD text book)

But my thoughts are, if such a process occurs, there can be one more adiabatic process that can be drawn from the initial state point. (this one is not isothermic) So the two adiabatics cut at one place and subsequently prove Classius statement of second law as wrong. I am yet to establish this and posted here just to keep this thread alive. I am thinking of writing it to the author of my TD textbook.

Regards,

 

[25362]

To Quark, please don't forget the magic word "irreversibility".