Enthalpy of an ideal gas


by fluidistic
Tags: enthalpy, ideal
fluidistic
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May7-13, 12:42 PM
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I've read on Wikipedia that the enthalpy of an ideal gas does not depend on pressure (http://en.wikipedia.org/wiki/Enthalpy:
Quote Quote by WikiTheGreat
Enthalpies of ideal gases and incompressible solids and liquids do not depend on pressure, unlike entropy and Gibbs energy.
).
However mathematically I get that it's false. So I'm wondering weather I'm making some error(s) which seems likely or the wiki article is wrong.
Here's my work:
I consider a monoatomic ideal gas.
From the equations of state ##PV=nRT## and ##U=\frac{3nRT}{2}##, one can obtain the fundamental equation ##S(U,V,n)=nS_0+nR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ]##. I've done it myself and it can be found in Callen's book.
Now using the definition of the enthalpy, I get that ##H=U+PV## where the independent variables of H are S, P and n. So that ##H(S,P,n)##. In order to get the enthalpy I must get ##U(S,P,n)## and ##V(S,P,n)##.
From the equations of state, I get that ##U=\frac{2PV}{3}##. Plugging that into the enthalpy, I reach that ##H=\frac{5}{3}PV##. So if I can find ##V(S,P,n)## I'm done.
I take the exponential in both sides of the fundamental equation to get ##e^S=\exp \{ nS_0+nR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] \}##
##\Rightarrow e^S = e^{nS_0} \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] ^{nR}##
##\Rightarrow \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] ^{nR} =e^{S-nS_0}##
##\Rightarrow \left ( \frac{U}{U_0} \right ) ^{3nR/2} \left ( \frac{V}{V_0} \right ) ^{nR} \left ( \frac{n}{n_0} \right ) ^{-5nR/2} =e^{S-S_0n}##
##\Rightarrow \left ( \frac{2PV}{3U_0} \right )^{3nR/2} \left ( \frac{V}{V_0} \right ) ^{nR} n^{-5nR/2} =e^{S-nS_0} ##
##\Rightarrow \left ( \frac{2}{3U_0} \right ) ^{3nR/2} \cdot \frac{1}{V_0 ^{nR}} \cdot n ^{-5nR/2 } \cdot V^{5nR/2} \cdot P ^{3nR/2} =e^{S-nS_0}##
##\Rightarrow V(S,P,n) = e^{\frac{2(S-nS_0)}{5nR}}P^{-3/5}nV_0^{2/5} \left ( \frac{2}{3U_0} \right )^{3/5}##
Which makes [tex]H(S,P,n)= e^{\frac{2(S-nS_0)}{5nR}}P^{2/5}nV_0^{2/5} \left ( \frac{2}{3U_0} \right )^{3/5}= c_1 \cdot e^{c_2(S-nS_0)}P^{2/5}n[/tex] where there's a dependence of the enthalpy on the pressure and ##c_1## and ##c_2## are positive constants.
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DrClaude
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May7-13, 01:57 PM
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Quote Quote by fluidistic View Post
From the equations of state, I get that ##U=\frac{2PV}{3}##.
This should be ##U=\frac{3PV}{2}##.
fluidistic
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May7-13, 02:27 PM
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Quote Quote by DrClaude View Post
This should be ##U=\frac{3PV}{2}##.
True. Also in my last expression for H(S,P,n) I forgot to multiply by that constant. Nevertheless this doesn't change the general expression I reached, namely ##H(S,P,n)=c_1 \cdot e^{c_2(S-nS_0)}P^{2/5}n##. The dependence of H on P remains.

qbert
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May7-13, 03:37 PM
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Enthalpy of an ideal gas


Let me simplify your claim:

claim 0. the enthalpy of an ideal gas doesn't depend on the pressure.
but from the second law dH = TdS + VdP
we have [itex]\frac{\partial H}{\partial P} = V \neq 0[/itex].

All of the extra junk is just working with the integrated forms
of the entropy expression.

Here's a similar line of reasoning for internal energy.
claim 1. the internal energy of an ideal gas doesn't depend on the volume.
but from the second law TdS = dU + pdV
we have [itex]\frac{\partial U}{\partial V} = -p \neq 0[/itex].

So what gives?

The claim isn't about what is the dependence of H on p
when entropy is constant - but when the temperature is held constant.

From your first expressions
[itex] H = U + pV = \frac{3}{2}nR T + nRT = C_p T [/itex],
from whence it's obvious
[tex] \left( \frac{\partial H}{\partial p} \right)_T = 0, [/tex]
and so the enthalpy is independent of the pressure.
fluidistic
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May7-13, 08:22 PM
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Quote Quote by qbert View Post
Let me simplify your claim:

claim 0. the enthalpy of an ideal gas doesn't depend on the pressure.
but from the second law dH = TdS + VdP
we have [itex]\frac{\partial H}{\partial P} = V \neq 0[/itex].

All of the extra junk is just working with the integrated forms
of the entropy expression.

Here's a similar line of reasoning for internal energy.
claim 1. the internal energy of an ideal gas doesn't depend on the volume.
but from the second law TdS = dU + pdV
we have [itex]\frac{\partial U}{\partial V} = -p \neq 0[/itex].

So what gives?

The claim isn't about what is the dependence of H on p
when entropy is constant - but when the temperature is held constant.

From your first expressions
[itex] H = U + pV = \frac{3}{2}nR T + nRT = C_p T [/itex],
from whence it's obvious
[tex] \left( \frac{\partial H}{\partial p} \right)_T = 0, [/tex]
and so the enthalpy is independent of the pressure.
Thanks for the reply.
If I understand well the claim of Wikipedia is not really accurate. They should have added "when the temperature is held constant", which isn't a given beforehand. But indeed, if one fixes T, then H doesn't depend on P. I didn't know this, good to know.
Have I understood you well?
Chestermiller
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May8-13, 12:16 AM
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Quote by WikiTheGreat
Enthalpies of ideal gases and incompressible solids and liquids do not depend on pressure, unlike entropy and Gibbs energy.

This is incorrect. The enthalpies of incompressible solids and liquids depend on pressure:

[tex]dH = C_pdT+(V-T\frac{\partial V}{\partial T}) dP[/tex]

This is a general thermodynamic relationship the applies to any material.


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