Chapter 3- Answers to selected exercises

1. The chemical potential of a simple fluid of a single component is given by the expression

$$\mu =\mu _{o}\left( T\right) +k_{B}T\ln \frac{p}{p_{o}\left( T\right) },$$

where $T$ is the temperature, $p$ is the pressure, $k_{B}$ is the Boltzmann constant, and the functions $\mu _{o}\left( T\right)$ and $p_{o}\left( T\right)$ are well behaved. Show that this system obeys Boyle's law, $pV=Nk_{B}T$. Obtain an expression for the specific heat at constant pressure. What are the expressions for the thermal compressibility, the specific heat at constant volume, and the thermal expansion coefficient? Obtain the density of Helmholtz free energy, $f=f\left( T,v\right)$. Hint: Note that $\mu \left( T,p\right) =g\left( T,p\right)$. Thus

$$v=\left( \frac{\partial \mu }{\partial p}\right) _{T}=\frac{k_{B}T}{p},$$

which is the expression of Boyle's law, and

$$-s=\left( \frac{\partial \mu }{\partial T}\right) _{p}=\frac{d\mu... ...p_{o}\left( T\right) }-\frac{k_{B}T}{p_{o}\left( T\right) }\frac{dp_{o}}{dT},$$

from which we obtain the specific heat at constant pressure. All other expressions are straightforward. In particular,

$$f=g-pv.$$

2. Consider a pure fluid of one component. Show that

$$\left( \frac{\partial c_{V}}{\partial v}\right) _{T}=T\left( \frac{\partial ^{2}p}{\partial T^{2}}\right) _{v}.$$

Use this result to show that the specific heat of an ideal gas does not depend on volume. Show that

$$\left( \frac{\partial \alpha }{\partial p}\right) _{T,N}=-\left( \frac{ \partial \kappa _{T}}{\partial T}\right) _{p,N}.$$

3. Consider a pure fluid characterized by the grand thermodynamic potential

$$\Phi =Vf_{o}\left( T\right) \exp \left( \frac{\mu }{k_{B}T}\right) ,$$

where $f_{o}\left( T\right)$ is a well-behaved function. Write the equations of state in this thermodynamic representation. Obtain an expression for the internal energy as a function of $T$, $V$, and $N$. Obtain an expression for the Helmholtz free energy of this system. Calculate the thermodynamic derivatives $\kappa _{T}$ and $\alpha$ as a function of temperature and pressure. From Euler's relation, we have

$$p=-\frac{\Phi }{V}=-f_{o}\left( T\right) \exp \left( \frac{\mu }{k_{B}T} \right) .$$

Thus, we can write

$$\mu =k_{B}T\ln \frac{p}{-f_{o}\left( T\right) },$$

which is identical to the expression for the chemical potential in exercise 1, if we make $\mu _{o}=0$ and $p_{o}\left( T\right) =-f_{o}\left( T\right)$ . Therefore, we have Boyle's law and the usual expressions for $\kappa _{T}$ and $\alpha$.

4. Obtain an expression for the Helmholtz free energy per particle, $f=f\left( T,v\right)$, of a pure system given by the equations of state

$$u=\frac{3}{2}pv p=avT^{4},$$

where $a$ is a constant. Note that we can write the following equations of state in the entropy representation

$$\frac{1}{T}=\left( \frac{3a}{2}\right) ^{1/4}v^{1/2}u^{-1/4} \frac{p}{T}=\frac{2}{3}\left( \frac{3a}{2}\right) ^{1/4}v^{-1/2}u^{3/4},$$

that lead to the fundamental equation

$$s=\frac{4}{3}\left( \frac{3a}{2}\right) ^{1/4}v^{1/2}u^{3/4}+c,$$

where $c$ is a constant. The Helmholtz free energy is given by

$$f\left( T,v\right) =u-Ts=\frac{3a}{2}v^{2}T^{4}-T\left[ \frac{4}{... ...{2}\right) ^{1/4}v^{1/2}\left( \frac{3a}{2}v^{2}T^{4}\right) ^{3/4}+c\right] .$$

5. Obtain an expression for the Gibbs free energy per particle, $g=g\left( T,p\right)$, for a pure system given by the fundamental equation

$$\left( \frac{S}{N}-c\right) ^{4}=a\frac{VU^{2}}{N^{3}},$$

where $a$ and $c$ are constants. From the fundamental equation

$$s=a^{1/4}v^{1/4}u^{1/2}+c,$$

we write the equations of state

$$\frac{1}{T}=\left( \frac{\partial s}{\partial u}\right) _{v}=\frac{1}{2} a^{1/4}v^{1/4}u^{-1/2} \frac{p}{T}=\left( \frac{ \partial s}{\partial v}\right) _{u}=\frac{1}{4}a^{1/4}v^{-3/4}u^{1/2}.$$

The Gibbs free energy per particle is given by the Legendre transformation

$$-\frac{g}{T}=s-\frac{1}{T}u-\frac{p}{T}v,$$

where $u$ and $v$ come from the equations of state.

6. Consider an elastic ribbon of length $L$ under a tension $f$. In a quasi-static process, we can write

$$dU=TdS+fdL+\mu dN.$$

Suppose that the tension is increased very quickly, from $f$ to $f+\Delta f$ , keeping the temperature $T$ fixed. Obtain an expression for the change of entropy just after reaching equilibrium. What is the change of entropy per mole for an elastic ribbon that behaves according to the equation of state $L/N=cf/T$, where $c$ is a constant? Using the Gibbs representation, we have the Maxwell relation

$$\left( \frac{\partial S}{\partial f}\right) _{T}=\left( \frac{\partial L}{ \partial T}\right) _{f}.$$

The equation of state, $L/N=cf/T$, leads to the result

$$\frac{\Delta S}{N}=-\frac{cf}{T^{2}}\Delta f.$$

7. A magnetic compound behaves according to the Curie law, $m=CH/T$, where $C$ is a constant, $H$ is the applied magnetic field, $m$ is the magnetization per particle (with corrections due to presumed surface effects), and $T$ is temperature. In a quasi-static process, we have

$$du=Tds+Hdm,$$

where $u=u\left( s,m\right)$ plays the role of an internal energy. For an infinitesimal adiabatic process, show that we can write

$$\Delta T=\frac{CH}{c_{H}T}\Delta H,$$

where $c_{H}$ is the specific heat at constant magnetic field.

8. From stability arguments, show that the enthalpy of a pure fluid is a convex function of entropy and a concave function of pressure.

*9. Show that the entropy per mole of a pure fluid, $s=s\left( u,v\right)$, is a concave function of its variables. Note that we have to analyze the sign of the quadratic form

$$d^{2}s=\frac{1}{2}\frac{\partial ^{2}s}{\partial u^{2}}\left( du\... ... v}dudv+\frac{1}{2}\frac{\partial ^{2}s}{\partial v^{2}}\left( dv\right) ^{2}.$$

This quadratic form can be written in the matrix notation

$$d^{2}s=\frac{1}{2}\left( \begin{array}{cc} du & dv \end{array... ...{array} \right) \left( \begin{array}{c} du \\ dv \end{array} \right) .$$

The eigenvalues of the $2\times 2$ matrix are the roots of the quadratic equation

$$\lambda ^{2}-\left[ \frac{\partial ^{2}s}{\partial u^{2}}+\frac{\... ...rtial v^{2}}-\left[ \frac{\partial ^{2}s}{\partial u\partial v}\right] ^{2}=0.$$

For a concave function, the eigenvalues are negative, that is,

$$\frac{\partial ^{2}s}{\partial u^{2}}\frac{\partial ^{2}s}{\partial v^{2}}- \left[ \frac{\partial ^{2}s}{\partial u\partial v}\right] ^{2}>0$$

and

$$\frac{\partial ^{2}s}{\partial u^{2}}+\frac{\partial ^{2}s}{\partial v^{2}} >0.$$

Now it is straightforward to relate these derivatives of the entropy with positive physical quantities (as the compressibilities and the specific heats).