- #1

fluidistic

Gold Member

- 3,926

- 262

I've read on Wikipedia that the enthalpy of an ideal gas does not depend on pressure (http://en.wikipedia.org/wiki/Enthalpy:

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.

).WikiTheGreat said: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.

Last edited by a moderator: