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Why is enthalpy a function of Temperature and Pressure? 
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#1
Apr2812, 10:48 PM

P: 2

right now, I'm following the MIT thermodynamics video lecture.
I've gone thru dU = ( 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 


#2
Apr2912, 03:52 AM

P: 31

You have (I won't write the pathdependent 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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