# Thermodynamics, Van der Waals gas problem

1. Jul 18, 2013

### fluidistic

1. The problem statement, all variables and given/known data
Hi guys, I'm stuck on 2 parts for now of the following problem that appeared in an exam I totally bombed:
Given the equation $s=s_0+R\ln \left [ C(v-b)\left (u + \frac{a}{v} \right )^{5/2} \right ]$,
1)Calculate $u(v,T)$
2)Calculate f in its natural variables (the molar Helmholtz equation).
3)Find $P(v,T)$
4)Calculate $C_p$ and $C_v$
5)Show that this system doesn't satisfy the criteria of stability when $RT > \frac{2a(v-b)^2}{v^3}$.
6)Give a way to obtain $v_l$ and $v_s$, the liquid and solid molar volumes. Don't do the algebra but give a way that ensures to find them, with as much details as possible.

2. Relevant equations
(1)$\frac{1}{T} = \left ( \frac{\partial s}{\partial u} \right ) _v$
(2)$F=U-TS \Rightarrow f=u-Ts$
(3)$P=-\left ( \frac{\partial f}{\partial v} \right ) _T$
(4)$C_v= \left ( \frac{\partial u}{\partial T} \right ) _v$
(5)$-\left ( \frac{\partial P}{\partial v}\right ) T >0$ (stability criterium)
3. The attempt at a solution
For part 1), I used eq. (1) to reach that $u(v,T)=\frac{5RT}{2}-\frac{a}{v}$.
For part 2), I used eq. (2) to reach $f(v,T)=\frac{5RT}{2}-\frac{a}{v}-Ts_0-RT \ln \left [ C(v-b) \left ( \frac{5RT}{2} \right ) ^{5/2} \right ]$.
For part 3), I used eq. (3) to reach that $P(v,T)=\frac{RT}{C(v-b)}-\frac{a}{v^2}$.
For part 4), I used eq. (4) to reach that $C_v=\frac{5R}{2}$.
However I don't really know how to find $C_P$.
Here's a try: $C_P =C_v+T \left ( \frac{\partial s}{\partial v} \right ) _T \left ( \frac{\partial v}{\partial T} \right ) _P$. Then I applied Maxwell's equation on $\left ( \frac{\partial s}{\partial v} \right ) _T$ to obtain $C_P =C_v+T \left ( \frac{\partial p}{\partial T} \right ) _v \left ( \frac{\partial v}{\partial T} \right ) _P$.
Then a cyclic relation to obtain $C_P =C_v-T \left ( \frac{\partial p}{\partial v} \right ) _T \left ( \frac{\partial v}{\partial T} \right ) ^2 _P$, that I rewrote as $C_P=C_v- \frac{T \left ( \frac{\partial P}{\partial v} \right ) _T }{\left ( \frac{\partial T}{\partial v} \right ) ^2 _P}$.
Where I calculated $\left ( \frac{\partial P}{\partial v} \right ) _T$ thanks to eq. (3) and from eq. (3) I also isolated T(v,P) and then performed the partial derivative of it with respect to v by keeping P constant. And then I tried to see if some terms would cancel out but couldn't find any. So I got an enormous expression for $C_P$, I can't believe $C_v$ to be so simple but $C_P$ being that horrible. I must have used a bad approach?
For part 5), I've used eq. (5) and found out the desired expression.
I'll think about part 6. The explanation I gave was wrong because I assumed I could find P(T) on the coexistence liquid-solid. But I should have said how I could obtain such a function and it's true, I have no idea. So I'll think about another way.
So for now, I'd like some help for part 4) on how to get $C_P$. Thank you very much for any help.

2. Jul 19, 2013

### TSny

Answers for parts (1), (2), and (4) look correct to me. There's a small error in your answer for part (3) associated with the constant C.

Your method for (4) looks OK and should get you the answer for CP. It should simplify down to something not too complicated.

I found an answer using your relation $C_P =C_v-T \left ( \frac{\partial p}{\partial v} \right ) _T \left ( \frac{\partial v}{\partial T} \right ) ^2 _P$.

To find $\left ( \frac{\partial v}{\partial T} \right )_P$, you can just take the differential of the equation for P, let dP = 0, and rearrange for dV/dT.

I don't know how to approach part (6).

Last edited: Jul 19, 2013
3. Jul 19, 2013

### fluidistic

I see, thanks. So if I'm not wrong, I shouldn't have any "C" constant for my expression for P(v,T).
Using what you said, I reached that $\left ( \frac{\partial v}{\partial T} \right ) _P=\frac{Rv^3(v-b)}{RTv^3-2a(v-b)^2}$.
If I didn't make any mistake, I get that $C_P=\frac{5R}{2}+\frac{TR^2v^3}{RTV^3-2a(v-b)}$ which is less horrible than I thought.
For part 6) I think I made a mistake in my first post. It should be $v_g$ and $v_l$, gaseous and liquid phases, not the solid one. What I have in mind right now might be to equal the chemical potentials in both phases, maybe I could get an equation that relates $v_l$ to $v_g$. However I'd need another equation that relates both variables in order to solve for them.

4. Jul 19, 2013

### TSny

That's what I got, too. [EDIT: Actually, I think the $(v-b)$ in the denominator should be squared.]

OK, liquid and gas makes more sense. Still not sure what to do here. The question doesn't state how to express the molar volumes (in terms of P and T?). For the gas phase, you already have an equation of state which you could use to solve for $v_g$.

As you say, the equality of chemical potentials will lead to one relation (Clapeyron equation) which involves the difference $v_g - v_l$. But the Clapeyron equation contains dP/dT along the coexistent curve of the liquid and gas as well as the heat of vaporization. Seems like too many unknowns.

The parameter b in the Van der Waals equation is essentially the volume of a molecule. [Edit: b is roughly the same as Avogadro's number times the volume of a molecule.] So, maybe you could use that to come up with a very rough estimate of the molar volume of the liquid.

As you can see, I really don't have a good approach to this question. I don't want to mislead you.

Last edited: Jul 19, 2013
5. Jul 19, 2013

### fluidistic

I've tried to equal the chemical potentials but didn't reach Clausius-Clapeyron equation as far as I know.
I used the fact that $\mu = \left ( \frac{\partial F}{\partial n} \right ) _{T,V}$.
I don't have F, but I have f. In order to get F, I multiplied f by n and I changed v by V/n. I reached $F(T,V,n)=\frac{5nRT}{2}-\frac{n^2a}{V}-Tns_0 -RT \ln \left [ C \left ( \frac{V}{n} -b \right ) \left ( \frac{5RT}{2} \right ) ^{5/2} \right ]$.
After I equaled the potentials I got the equation $$\frac{2a}{v_l}+RT \{ \ln \left [ C(v_l-b) \left ( \frac{5RT}{2} \right ) ^{5/2} \right ] +\frac{v_l}{v_l-b} \}=\frac{2a}{v_g}+RT \{ \ln \left [ C(v_g-b) \left ( \frac{5RT}{2} \right ) ^{5/2} \right ] +\frac{v_g}{v_g-b} \}$$. I don't even think it's possible to isolate $v_l$ in terms of $v_g$.

6. Jul 19, 2013

### TSny

OK, scratch everything I said regarding question (6). I had to dig into one of my texts. You were correct in proposing setting the Gibbs potentials (chemical potentials) for the liquid and gas equal to one another (at liquid-vapor equillibrium). Sorry to have mentioned the Clapeyron equation - forget it.

Consider an isotherm of the Van der Waal's equation for a temperature below the critical temperature, such as the blue isotherm in the picture. Equality of the Gibbs potentials at points A and B can be used to find a relation between the pressure P (at A and B), the molar volume of the liquid $v_l$ at point A, and the molar volume of the gas $v_g$ at point B. Details are probably shown in most texts.

You can also use your equation of state at point A to relate P to $v_l$ and again at point B to relate P to $v_g$.

Thus, you have three equations for the unknowns P, $v_l$, and $v_g$. Temperature T is presumed to be given. Looks like it would be hard to actually solve the equations.

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7. Jul 19, 2013

### TSny

OK, I think you have a typo where you left out a factor of n in front of the logarithm term. But it looks like you included it when you did your calculation.

OK, this looks good! But you can simplify it by using log(ab) = log(a) + log(b) to eliminate C and the 5RT/2 terms. Your result is equivalent to what I get if I follow a somewhat different way of expressing equality of the Gibbs potential in the two phases as given in my text. I like your way.

You also have the equality $P_l = P_g$. Using the equation of state (i.e., your expression for P(V,T)), you get another equation relating $v_l$ and $v_g$. So, in principle you could solve separately for $v_l$ and $v_g$ in terms of T.

Last edited: Jul 19, 2013
8. Jul 19, 2013

### fluidistic

Yes I made a typo when writing in latex here for the F function.
I see. I didn't realize that the pressures were equal in the coexistence liquid-gas. So this makes up for the 2nd equation.
I've simplified the argument in the logarithm for the expression of the equated potentials.
The 2 equations I must solve are then: $$2a \left ( \frac{1}{v_l} -\frac{1}{v_g} \right ) +RT \ln \left ( \frac{v_l-b}{v_g-b} \right ) +RT \left ( \frac{v_l}{v_l-b} - \frac{v_g}{v_g-b} \right ) =0$$ (equated potentials)
and $$RT \left [ \frac{1}{v_l-b}-\frac{1}{v_g-b} \right ] +a \left ( \frac{1}{v_g^2} - \frac{1}{v_l^2} \right )=0$$ (equated pressures)
That is, if I made no mistake. I guess that's why they said to give a way to find $v_l$ and $v_g$ but not actually solve the equations explicitly.
So problem solved! Thank you very much for your assistance.

9. Jul 19, 2013

### Staff: Mentor

dH = TdS+VdP

So
$$C_p=\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P$$

You already showed that $u = \frac{5RT}{2}-\frac{a}{v}$

If you substitute this into the original equation for s, you get
$$s=s_0+R\ln{\left[C(v-b)(\frac{5RT}{2})^\frac{5}{2}\right]}$$
so
$$\left(\frac{\partial S}{\partial T}\right)_P=\frac{5R}{2T}+\frac{R}{(v-b)}\left(\frac{\partial v}{\partial T}\right)_P$$
So,
$$C_p=\frac{5R}{2}+\frac{RT}{(v-b)}\left(\frac{\partial v}{\partial T}\right)_P$$

10. Jul 19, 2013

### TSny

Yes, that's a nice way.

Last edited: Jul 19, 2013
11. Jul 19, 2013

### fluidistic

Hmm I don't really understand what the enthalpy has to do with this problem.
Also I don't understand the step where you passed from the expression of s(v,T) to the partial derivative of s with respect to T with P fixed.

12. Jul 19, 2013

### Staff: Mentor

The enthalpy has everything to do with the heat capacity at constant pressure because the heat capacity at constant pressure is defined as the partial derivative of the enthalpy with respect to temperature at constant pressure.

Are you saying that you don't understand how I took the partial derivative of s(v,T) with respect to T at constant P? You first expand the ln term as the sum of three ln terms.

13. Jul 19, 2013

### fluidistic

Thank you I get it now Chestermiller. I also reach your result.