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| Landon |
 Posted: Fri Feb 09, 2007 1:52 pm |
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On definitions,
When speaking of ΔH, we are referring to the change in the heat of a
system that occurs at constant pressure. When speaking of qp, we are
referring to the heat transfer. These two values are equal.
ΔH = qp
However, let us consider an adiabatic change. In an adiabatic change,
there is no heat. Yet, in a reversible, adiabatic expansion, there is
a change in enthalpy. Our simple definition of ΔH as “change in heat
at constant pressure” creates confusion because the enthalpy actually
increases in a process where there is no heat. Is there a more
encompassing definition for ΔH? |
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| Ian Gay |
| Posted: Fri Feb 09, 2007 1:52 pm |
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"Landon" <landon.waldner@gmail.com> wrote in
news:1171048679.903133.71130@k78g2000cwa.googlegroups.com:
[quote:cb20349c93]Actually, things make a lot more sense to me now with this
definition: ΔH = ΔE + PΔV + VΔP.
[/quote:cb20349c93]
It's not a definition, and it's not quite correct.
The definition is H = E + PV.
so delta H = delta E + delta(PV) =
delta E + P delta V + V delta P + (delta P)(delta V)
The last term can be ignored for infinitesimal changes, but not for
finite ones.
[quote:cb20349c93]
When I visualize these processes, I often think of a boundary
transfer by heat (q) which is followed by a change in enthalpy.
That is why when I think about an adiabatic process, I can’t
understand why there is a change in enthalpy when there is no
change in heat. But it makes more sense to think of enthalpy as
ΔH = ΔE + PΔV + VΔP because even if there is no heat (as in an
adiabatic process), there is still a change in enthalpy. This
seems like a more abstract definition, but I think that this
definition works better for me as I find it more general than the
one above. Thanks for the help!
[/quote:cb20349c93]
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| Madalch |
| Posted: Fri Feb 09, 2007 2:18 pm |
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On Feb 9, 9:52 am, "Landon" <landon.wald...@gmail.com> wrote:
[quote:159ddaa750]On definitions,
...Is there a more encompassing definition for ΔH?
[/quote:159ddaa750]
ΔH is the change in enthalpy, and enthalpy is defined as the sum of
the internal energy (E or U, depending on the text you're using) and
the product of pressure times volume.
So H = E + PV, and ΔH = ΔE + PΔV + VΔP.
At constant pressure, ΔP = 0, so
ΔH = ΔE + PΔV.
Since ΔE = q + w, and expansion work is -PΔV,
ΔH = q + w - w = q (at constant pressure).
Is that what you were looking for? |
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| Landon |
| Posted: Fri Feb 09, 2007 3:17 pm |
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Guest
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Actually, things make a lot more sense to me now with this definition:
ΔH = ΔE + PΔV + VΔP.
When I visualize these processes, I often think of a boundary transfer
by heat (q) which is followed by a change in enthalpy. That is why
when I think about an adiabatic process, I can’t understand why there
is a change in enthalpy when there is no change in heat. But it makes
more sense to think of enthalpy as ΔH = ΔE + PΔV + VΔP because even if
there is no heat (as in an adiabatic process), there is still a change
in enthalpy. This seems like a more abstract definition, but I think
that this definition works better for me as I find it more general
than the one above. Thanks for the help! |
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| Fred Kasner |
| Posted: Sat Feb 10, 2007 7:21 pm |
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Landon wrote:
[quote:f2bbc90d0c]On definitions,
When speaking of ΔH, we are referring to the change in the heat of a
system that occurs at constant pressure. When speaking of qp, we are
referring to the heat transfer. These two values are equal.
ΔH = qp
However, let us consider an adiabatic change. In an adiabatic change,
there is no heat. Yet, in a reversible, adiabatic expansion, there is
a change in enthalpy. Our simple definition of ΔH as “change in heat
at constant pressure” creates confusion because the enthalpy actually
increases in a process where there is no heat. Is there a more
encompassing definition for ΔH?
H=E+pV[/quote:f2bbc90d0c]
FK |
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