# Entropy change of a real gas

1. Dec 5, 2011

### chagular

For a real gas (non-ideal gas) in a reversible process, the way to calculate $\Delta{S}$ should also be independent to path simply because entropy is a state function.

However, I got strange solution while taking different path.

Here is the condition:

1. Equation of state (EOS) for the real gas could be arbitrary (but actually not ideal)
2. The constant volume heat capacity is only the function of temperature or a constant, i.e., $C_V=f(T)$ only.
3. State changes from $A(T_1,V_1)$ to $B(T_2,V_2)$.

As illustrated (attached figure). Basically, there're two simple paths for calculation.
(1) The black route: A-D-B
(2) The red route: A-C-B

The total differential of entropy as function of $T$ and $V$ is:
$dS=\frac{C_V}{T}dT+\left(\frac{\partial P}{\partial T}\right)_VdV$.

Both entropy change in constant-V process A-C and D-B are identical:
$\Delta{S}=\int^{T_2}_{T_1}\frac{C_V}{T}\,dT$.

But I can't prove that the rest two constant T process paths have same contribution as encountering a real gas:
(1) for A-D (in black route)
$\Delta{S_{AD}}=\int^{V_2}_{V_1} \left.\left(\frac{\partial P}{\partial T}\right)_V\right|_{T_1}\,dV$

(2) for C-B (in red route)
$\Delta{S_{CB}}=\int^{V_2}_{V_1} \left.\left(\frac{\partial P}{\partial T}\right)_V\right|_{T_2}\,dV$

It's obvious that $\Delta{S_{AD}}$ is not identical to $\Delta{S_{CB}}$ if $\left(\frac{\partial P}{\partial T}\right)_V$ is function of both V and T. Such a situation could happen for almost all real gas.

The result obtained above is opposed to the fundamental knowledge: entropy is a state function. Regardless of the integration difficulty that might encounter as we could always use numerical integration especially when analytical EOS is available. What is wrong with my derivation?

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2. Dec 6, 2011

### I like Serena

Welcome to PF, chagular!

Where did you get this formula?
$$dS=\frac{C_V}{T}dT+\left(\frac{\partial P}{\partial T}\right)_VdV$$
I can't find it and I doubt it is true in general.

3. Dec 6, 2011

### DrDu

Consider the Maxwell equation:$\partial S/\partial V|_T=\partial p/\partial T|_V$, which is also obvious from the expression you gave for dS.

4. Dec 6, 2011

### chagular

To I like Serena,

The equation is derived with the only assumption: reversible process.

As equation of state connects P, V and T like $f(P,V,T)=0$, the thermodynamical property such as entropy can be expressed (derived) in terms of any two of them.

Now choose T and V, hence $S=S(T,V)$.
The total differential form is then $dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$ ...... eqn. (1)

By fundamental definitions, we have:
$U\equiv Q+W \Rightarrow dU=TdS-PdV$ ...... eqn. (2)
$H\equiv U+PV \Rightarrow dH=dU+d(PV)=TdS+VdP$
$A\equiv U+TS \Rightarrow dA=dU-d(TS)=-SdT-PdV$ ...... eqn. (3)

From eqn. (2): $\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V$
Also by definition: $\left(\frac{\partial U}{\partial T}\right)_V \equiv C_V$

Therefore, $\left(\frac{\partial S}{\partial T}\right)_V = \frac{C_V}{T}$ ...... eqn. (4)

As $A$ is an exact function: $\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$ (Maxwell relationship) ...... eqn. (5)

Substitute eqn. (4) and (5) into (1):
$dS=\frac{C_V}{T}dT+\left(\frac{\partial P}{\partial T}\right)_V dV$

This formula should be valid for any gases operating under reversible process since no further assumption is made.

5. Dec 6, 2011

### DrDu

The formula for dS is ok. If you know the Maxwell relations, I don't understand why you don't use them to evaluate Delta S_AD=S(T_1,V_2)-S(T_1,V_1) and S_BC=S(T_2,V_2)-S(T_1,V_1).
Or to formulate it differently: The Maxwell relation puts a very stringent restriction on how p may vary with T.

6. Dec 6, 2011

### chagular

Since EOS is expressed in terms of P, V and T (but not U, G, H, A or S), therefore changing it to the differential of S or other non-mesurable variables is not possible for calculation. Besides, what I'm doing is exactly taking the advange of Maxwell relationships to transform dS into mesurable variables.

7. Dec 6, 2011

### Andrew Mason

You can take any path between the two states so long as it is a reversible path. Neither the black nor red routes are reversible. To find the change in entropy you have to calculate ∫dQ/T over a reversible path between A and B.

Entropy is defined as ∫dQrev/T. Q is not a state function so there is no reason why, in general, ∫dQ/T should be a state function. It so happens, though, that ∫dQ/T over the reversible path between two states is the same regardless of which reversible path you take. So, ∫dQrev/T depends only on the beginning and end states. That is why it can be considered a state function.

AM

8. Dec 6, 2011

### DrDu

Then I don't understand your problem: Do you fear that there is any real material that doesn't obey Maxwells equations?

9. Dec 6, 2011

### chagular

I'm little confused about what transformation should I do to correctly calculate the entropy change if an analytical EOS (e.g. van der Waals equation, cubic equation, Virial equation, etc.) is available? Do you mean Maxwell's relationship fail while facing a real gas?

10. Dec 6, 2011

### DrDu

No, I don't think they fail. Did you try to calculate dp/dT for a VdW gas?

11. Dec 6, 2011

### chagular

For a comprehensive EOS like van der Waals equation, $\left(\frac{\partial P}{\partial T}\right)_V$ is a constant. That's the benefit as choosing vdw eqn as vdw constants seem to be 'fixed' constants not changing with the change of P, V and T.

However if we treat other gases described by Virial equation and a lot of other equations of state listed in wiki: Equation of state

In this way, $\left(\frac{\partial P}{\partial T}\right)_V$ is no longer a constant but changes with T, which causes the difference to calculations. (They do have same start and end states but different paths.)

12. Dec 6, 2011

### qbert

So clearly you need to check that for your case you have (dp/dT)V doesn't depend on T.

let's see if we can work out why.

1 TdS = dU + p dV
2. dU = CV(T) dT
3. (dp/dT)V = (dS/dV)T

Then
(d/dT)V[(dp/dT)V]= (d/dT)V[(dS/dV)T] = (d/dV)T[(dS/dT)V]
= (d/dV)T[ (dU/dT)V / T] =(d/dV)T[ Cv(T)/ T]
= 0.

13. Dec 6, 2011

### Andrew Mason

This is not correct.

A constant V process is not reversible. To calculate the change in entropy between states of equal volume but different temperature you have to connect the beginning and end points with a combination of isothermal and adiabatic paths. Reversible heat flow has to occur isothermally.

AM

14. Dec 7, 2011

### DrDu

No, you can connect the system to a heat bath whose temperature is slowly increasing so that the system and bath are always (arbitrary nearly) in equilibrium. The heat transfer between the bath and the system is then reversible.

In fact, the whole assumption of reversibility is superfluous for the calculation.
Reversibility is a concept which refers to the actual path the system takes to go from A to B.
However, what counts in thermodynamics is only the end points A and B. Calculation of the entropy change via a path in state space is pure mathematics and has nothing to do with the actual path of the system.

15. Dec 7, 2011

### DrDu

Yes, I came to the same conclusion. Restricting C_v to be a function of T, only, means p to be a linear function of T, while the dependence on V can be non-trivial.

16. Dec 7, 2011

### Andrew Mason

Are you saying that $\int_{1}^{2} dQ/T$ should be the same regardless of the path taken? That seems to be what chagular is saying.

AM

17. Dec 7, 2011

### DrDu

I say that the entropy change should be the same regardless of the path.
However it is not correct to write this as an integral over dQ/T for the real irreversible path taken.
The classical counterexample is the free expansion of an ideal gas where Q=0 but Delta S >0. As T doesn't change, the entropy change is only due to the integration over dp/dT dV in the space of equilibrium states.
Usually it is best to avoid using Q in thermodynamics as far as possible. This was also the guiding principle in the axiomatization of thermodynamics by Caratheodory.

18. Dec 7, 2011

### Andrew Mason

That is because entropy is defined as ∫dQrev/T.
Or taking ∫dQrev/T over a reversible path between those two states: eg. an isothermal quasi-static expansion (which requires heat flow into the gas).
Caratheodory certainly took into account reversible heat flows in entropy calculations. States that could not be reached by a reversible adiabatic path necessarily required heat flow and entropy change. The definition of change in entropy is still ∫dQrev/T. Any method for determining change in entropy is simply a way of doing what is mathematically equivalent to finding ∫dQrev/T.

AM

19. Dec 8, 2011

### DrDu

That's all ok and we mean the same thing. However I wanted to stress that dQ_rev is some very special kind of heat. Specifically, dQ_rev=dU+pdV, so that it can be expressed completely in terms of state variables.

Caratheodory never used heat explicitly but replaces it by the difference of work necessary to to from one state to the other adiabatically (but in general via an irreversible path) and the actual diabatic work (W). He shows that the adiabatic work is a function of state (U actually), so in fact he defines heat as Q=U-W. I think that's quite reasonable as it reduces to consider mechanical variables only. That this is possible was shown long before by Thompson (Count Rumford).