Re: Joule Thompson Effect and Isenthalpic Expansion
- From: renai <bensegev@xxxxxxxxxxx>
- Date: Sat, 21 Jul 2007 10:07:30 -0700
On Jul 20, 6:29 am, andy everett <vze2q...@xxxxxxxxxxx> wrote:
renai wrote:
In learning about the Joule Thompson Effect, I'm told to consider a
throttling process, where no PV work is done by the gas on its
surroundings and no heat is transferred between the surroundings and
the gas contained within. Sources also cite this process as
isenthalpic. This confuses me.
If H = U+PV, then
dH = dU + PdV + VdP
= Q + W + PdV + VdP
= Q - PdV + PdV + VdP
^^^
no work done, set W = 0?
= Q + VdP
If there is no heat transfer then I understand that we have:
dH = vdP.
Now if throttling is said to be isenthalpic then wouldn't that mean
that dP would have to equal zero for dH to be zero. And if that is
so, how can it be since I would think that dP would be negative for
any expansion since the area over which the total force of the gas
molecules is acting over increases during an expansion and P = F/A.
Can anyone explain this to me?
Cheers,
Renai.
I would have to relearn all this (that assumes i learned it in the first
place). But a little googling (Joule Thompson Effect) i found this:
(http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html
Which might get you farther.- Hide quoted text -
- Show quoted text -
Thanks a lot, it actually helped a lot. It explained everything. I
think I could've also got their even using my equation by just doing
the following:
As per my last line:
dH = VdP
Using the ideal gas equation of state PV = nRT:
P = nRT/V
dP = -nRT/V^2dV
Now subbing back in I get:
dH = V(-nRT)/V^2dV which simplifies to:
dH = -nRT/VdV
but -nRT/V = P so we have:
dH = -PdV and that is expansion work which is zero in this case so we
have:
dH = 0.
I also realize know I could've done this much easier by simply taking
the initial equation and differentiating as follows:
dH = dU + d(PV) and for an ideal gas:
= dU + d (nRT) and since temperature doesn't change for the
ideal gas, nRT won't change, and the internal energy of an ideal gas
only changes with temperature then dU won't change either and then we
emmediately get:
dH = 0.
The cool thing about the article was that it explained that Joule
actually did different experiments which I now understand.
The situation I'm describing is just "free" expansion from a two bulb
system where one gas expands into the other evacuated bulb.
Throttling is different. I'm still trying to work through the latter
half of the paper on it and it's brought up other questions but at
least it's helped me with this.
Thank You,
Renai
.
- References:
- Joule Thompson Effect and Isenthalpic Expansion
- From: renai
- Re: Joule Thompson Effect and Isenthalpic Expansion
- From: andy everett
- Joule Thompson Effect and Isenthalpic Expansion
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