# Enthalpy of an ideal gas

1. May 7, 2013

### fluidistic

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 $$H(S,P,n)= e^{\frac{2(S-nS_0)}{5nR}}P^{2/5}nV_0^{2/5} \left ( \frac{2}{3U_0} \right )^{3/5}= c_1 \cdot e^{c_2(S-nS_0)}P^{2/5}n$$ where there's a dependence of the enthalpy on the pressure and $c_1$ and $c_2$ are positive constants.

2. May 7, 2013

### Staff: Mentor

This should be $U=\frac{3PV}{2}$.

3. May 7, 2013

### fluidistic

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.

4. May 7, 2013

### qbert

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 $\frac{\partial H}{\partial P} = V \neq 0$.

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 $\frac{\partial U}{\partial V} = -p \neq 0$.

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
$H = U + pV = \frac{3}{2}nR T + nRT = C_p T$,
from whence it's obvious
$$\left( \frac{\partial H}{\partial p} \right)_T = 0,$$
and so the enthalpy is independent of the pressure.

5. May 7, 2013

### fluidistic

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?

6. May 8, 2013

### Staff: Mentor

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:

$$dH = C_pdT+(V-T\frac{\partial V}{\partial T}) dP$$

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

7. May 18, 2016

### Brett0

As I understood it if the pressure changes then the PV work required to establish the system in the environment changes, which is what enthalpy is, according to the first few lines of said wikipedia article. I too thought the claim that enthalpy is not pressure dependent was a bit strange.

8. May 18, 2016

### Brett0

Another interesting caveat is found in this document:

http://www.physics.usu.edu/torre/3700_Spring_2015/Lectures/06.pdf

Here we see that for an isobaric H=Q assuming only compressional work is done

However in the following pages conditions are given for dF = W irregardless of the type of work.

Any thoughts on why the work done by an enthalpy change is restricted to PV work?

9. May 18, 2016

### Staff: Mentor

This post makes no sense to me. A key physical characteristic of an ideal gas it that it's enthalpy is independent of pressure. I see absolutedly no reason why one would find it strange the enthalpy is not a function of pressure for an ideal gas.

10. May 18, 2016

### Staff: Mentor

These lecture notes make no sense to me.
Who says that the work done when enthalpy changes is restricted to PV work?

11. May 18, 2016

### Zeppos10

Hello you all:
first I assume a closed non-reactive system with n=constant, the system being an ideal gas.
lets assume H(U,p,V)=U+pV by definition and lets assume that for an ideal gas, U is a function of T only, hence H=H(T,p,V). However, for an ideal gas pV=nRT, hence of the variables (T,p,V) only 2 are independent, or H is to be expressed as a function of 2 variables. We do not want to drop T, hence we write either H(T,p) or H(T,V).
But when the enthalpy of an ideal gas is not a function of p (as is the issue at hand), then only H(T,V) would be admissible. yes H(T) is also admissible, if H is independent from volume. Is it ??
I would conclude that there is something rotten in the state of enthalpy. The asssumption that it could be a function of 3 independent variables as suggested by the equation in #1 seems unsustainable.

12. May 18, 2016

### Staff: Mentor

Sure.
You have just verified that, per unit mass, U, H, S, A, and G can be expressed as function of any two of the three intensive variables, p, T, and V, where V is the specific volume. This is true of any single-phase substance of constant concentration, not just for an ideal gas. This is because, from the equation of state, any of the intensive variables p, T, or V can be expressed as a function of the other two.

13. May 19, 2016

### vanhees71

First of all we have for the ideal monatomic gas (easily derivable from the equipartition theorem)
$$U=N k_B T.$$
This is not in the "natural" variables, because due to the 1st and 2nd Law you have
$\mathrm{d} U=T \mathrm{d} S-p \mathrm{d} V.$
So the natural variables for the caloric equation of state in terms of $U$ are $S$ (entropy) and $V$ (volume) rather than $T$, but of course we can express everything in any two independent thermodynamical variables we like. Further we also know the equation of state
$$p V=N k_B T.$$
To get the entropy (within classical thermodynamics, which is a somewhat difficult issue, because entropy cannot be well defined within classical physics, but you get a good approximation for not too low temperatures). We can only give a relation of entropies with respect to some reference state $(T_0,V_0)$.

The idea is to use the equation
$$\Delta S=S-S_0=\int_{T_0}^{T} \mathrm{d} T \frac{C_V}{T}+\int_{V_0}^{V} \mathrm{d} V \left (\frac{\partial S}{\partial V} \right)_{T},$$
where we have used a path from the reference state to an arbitrary state in the $(T,V)$ plane by first going from $T_0$ to $T$ keeping $V$ constant and then from $V_0$ to $V$ keeping $T$ constant.

Now we get the heat capacity at constant volume from the 1st Law above as
$$C_V=\frac{\partial U}{\partial T}=\frac{3}{2} N k_B.$$

To get the 2nd integral we need a Maxwell relation. To get the right on we have to consider the free energy
$$F=U-T S.$$
From the 1st+2nd Law we find
$$\mathrm{d} F=-S \mathrm{d} T-p \mathrm{d} V,$$
and since this is a total differential we must have (using the "natural" independent variables, $F=F(T,V)$
$$\partial_V \partial_T F=-\left (\frac{\partial S}{\partial V} \right)_T=\partial_T \partial_V F=-\left ( \frac{\partial p}{\partial T} \right )_V,$$
i.e., from the EoS
$$\left (\frac{\partial S}{\partial V} \right)_T=\left ( \frac{\partial p}{\partial T} \right )_V=\frac{N k_B}{V}.$$
Plugging this into the integrals for the entropy we get
$$S=N k_B \left [\frac{3}{2} \ln \left (\frac{T}{T_0} \right) + \ln \left (\frac{V}{V_0} \right) \right]=N k_B \ln \left(\frac{V T^{3/2}}{V_0 T_0^{3/2}} \right).$$
Now the denominator in the logarithm can only be a function of the fixed particle number $N$ and from the extensivity of entropy it follows that it must be proportional to $N$. Thus we find
$$S=N k_B \ln \left (\frac{V T^{3/2}}{\alpha N} \right),$$
where $\alpha$ is a constant.

The natural variables for the entropy are found by using again the 1st+2nd Law:
$$\mathrm{d} S=\frac{1}{T} \mathrm{d} U-\frac{p}{T} \mathrm{d} V.$$
So we should give the entropy as $S=S(U,V)$. With the caloric EoS we get
$$S=N k_B \ln \left [\left(\frac{2 U}{3 k_B N} \right)^{3/2} \cdot \frac{V}{\alpha N} \right ].$$
Now we can derive all thermodynamical potentials with their natural dependence. You can, e.g., solve the previous equation for $U=U(S,V)$. Then you have for the enthalpy
$$H=U+p V,$$
for which with the 1st+2nd Law you get
$$\mathrm{d} H = T \mathrm{d} S+V \mathrm{d} p,$$
So you just use the definition of $H$. Then you just need to express $V$ in terms of $S$ and $p$ to get the natural dependence $H=H(S,p)$.

Last edited: May 19, 2016