Why is enthalpy a function of Temperature and Pressure?


by tkdiscoverer
Tags: enthalpy, pressure, thermal physics, thermodynamics
tkdiscoverer
tkdiscoverer is offline
#1
Apr28-12, 10:48 PM
P: 2
right now, I'm following the MIT thermodynamics video lecture.

I've gone thru

dU = dw + dq
(d for "is path dependent")

and

H = U + pV

But why is enthalpy a function of temperature and pressure?
is it because pV = nRT and thus, V = nRT/p, so we only need p and T to get V?

dH = ( [itex]\delta[/itex] H/ [itex]\delta[/itex] )TdT + ( [itex]\delta[/itex] H/ [itex]\delta[/itex] p)dp

but why not:
dH = ([itex]\delta[/itex]H/[itex]\delta[/itex] T)dT + ([itex]\delta[/itex]H/[itex]\delta[/itex]V)dV ???

Thanks! : D
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mSSM
mSSM is offline
#2
Apr29-12, 03:52 AM
P: 24
You have (I won't write the path-dependent parts explicitly; just assume they are there :D ):
[tex]
\mathrm{d}E = T\mathrm{d}S - P\mathrm{d}V
[/tex]

We know that [itex]\mathrm{d}Q = T\mathrm{d}T[/itex], so that for a process at constant pressure we can rewrite the quantity of heat as the differential:
[tex]
\mathrm{d}Q = \mathrm{d}E + P\mathrm{d}V = \mathrm{d}(E + PV) = \mathrm{d}W
[/tex]

of some quantity
[tex]W = E+PV[/tex]

If you now want to find the total differential of the heat function itself (now no longer assuming constant pressure), you can easily get:
[tex]
\mathrm{d}W = T\mathrm{d}S + V\mathrm{d}P
[/tex]

For your notation: look at the definition of your [itex]U[/itex]. When you make the total differential of the heat function, you have to write explicitly the terms for the energy, where [itex]-P\mathrm{d}V[/itex] cancels with one of the differential you get from [itex]\mathrm{d}(PV)[/itex] in the heat function.


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