Chapter 4

Prove that the slope of the sublimation curve of a pure substance at the triple point must be greater than that of the vaporization curve at the triple point.

Consider the vaporization curve for liquid mercury. The latent heat of vaporization, \(L\) (in \(\mathrm{J} / \mathrm{mol}\) ) varies slowly with pressure, but has significant variation with temperature and can be written \(L=(7724-0.9768 T) R\), where \(R\) is the gas constant and \(T\) is measured in kelvin. It is known that at atmospheric pressure ( \(P_{\mathrm{am}}=1.013 \times 10^{5} \mathrm{~Pa}\) ) mercury begins to vaporize at temperature \(T=\) \(630 \mathrm{~K}\). (a) Plot the vaporization curve for mercury between temperatures \(T=500 \mathrm{~K}\) and \(T=650 \mathrm{~K}\). (b) At what pressure does mercury begin to vaporize when \(T=530 \mathrm{~K}\) ? (Hint: The volume of the liquid mercury can be neglected relative to that of the vapor and the vapor can be treated as an ideal gas.)

Two phases of solid carbon are called graphite and diamond. At standard temperature \(\left(T_{0}=298 \mathrm{~K}\right)\) and standard pressure \(\left(P_{0}=1.0\right.\) bar \()\), the difference in the molar Gibbs free energy for these two phases is \(\Delta g=g_{\mathrm{G}}-g_{\mathrm{D}}=-2.9 \mathrm{~kJ} / \mathrm{mol}\), so graphine is the stable phase at standard temperature and pressure (STP). At STP, the difference in molar volume is \(\Delta v=v_{\mathrm{G}}-v_{\mathrm{D}}=\) \(1.9 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{mol}\), and the difference in molar entropy is \(\Delta s=s_{\mathrm{G}}-s_{\mathrm{D}}=3.4 \mathrm{~J} /(\mathrm{K} \mathrm{mol}) .\) (a) If temperature is held fixed at \(T=T_{0}=298 \mathrm{~K}\), estimate the pressure at which a phase transition occurs and diamond becomes the most stable form of the crystal. (b) At temperature \(T=398 \mathrm{~K}\), at approximately what pressure does the phase transition from graphine to diamond occur?

One kilogram of superheated steam, at temperature \(t=350^{\circ} \mathrm{C}\), pressure \(P=100\) bar, and specific entropy \(s=5949 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), is expanded reversibly and adiabatically to form wet steam at \(t=\) \(200^{\circ} \mathrm{C}\) and pressure \(P=15.55\) bar. The specific entropy of water vapor and liquid water on the coexistence curve at \(t=200{ }^{\circ} \mathrm{C}\) are \(\mathrm{s}_{\mathrm{g}}=6.428 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) and \(s_{1}=2.331 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), respectively. The specific enthalpy of water vapor (gas) and liquid water on the coexistence curve at \(t=200^{\circ} \mathrm{C}\) are \(h_{\mathrm{g}}=\) \(2791 \mathrm{~kJ} / \mathrm{kg}\) and \(h_{1}=852.4 \mathrm{~kJ} / \mathrm{kg}\). (a) What is the specific enthalpy of the wet steam at \(t=200^{\circ} \mathrm{C}\) ? (b) What fraction of the wet steam is liquid water?

Consider a monatomic fluid along its liquid-gas coexistence curve. Compute the rate of change of chemical potential along the coexistence curve \((\mathrm{d} \mu / \mathrm{d} T)\) owex, where \(\mu\) is the chemical potential and \(T\) is the temperature. Express your answer in terms of \(s_{1}, v_{1}\) and \(s_{\mathrm{g}}, v_{\mathrm{g}}\), which are the molar entropy and molar volume of the liquid and gas, respectively.

A system in its solid phase has a Helmholtz free energy per mole, \(a_{s}=B / T v^{3}\) and in its liquid phase it has a Helmholtz free energy per mole \(a_{1}=A / T v^{2}\), where \(\mathrm{A}\) and \(\mathrm{B}\) are constants, \(v\) is the volume per mole, and \(T\) is the temperature. (a) Compute the molar Gibbs free energy density of the liquid and solid phases. (b) How are the molar volumes, \(v\), of the liquid and solid related at the liquidsolid phase transition? (c) What is the slope of the coexistence curve in the \(P-T\) plane?

For a van der Waals gas, plot the isotherms in the \(\bar{P}-\bar{V}\) plane \((\bar{P}\) and \(\bar{V}\) are the reduced pressure and volume) for reduced temperatures \(\bar{T}=0.5, \bar{T}=1.0\), and \(\bar{T}=1.5\). For \(\bar{T}=0.5\), is \(\bar{P}=0.1\) the equilibrium pressure of the liquid-gas coexistence region?

Consider a binary mixture composed of two types of particles, A and B. For this system the fundamental equation for the Gibbs free energy is \(G=\mathrm{n}_{\mathrm{A}} \mu_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}} \mu_{\mathrm{B}}\), the combined first and second laws are \(\mathrm{d} G=-S \mathrm{~d} T+V \mathrm{~d} P+\mu_{A} \mathrm{~d} \mathrm{n}_{\mathrm{A}}+\mu_{\mathrm{B}} \mathrm{d} \mathrm{n}_{\mathrm{B}}\) (S is the total entropy and \(V\) is the total volume of the system), and the chemical potentials \(\mu_{\mathrm{A}}\) and \(\mu_{\mathrm{B}}\) are intensive so that \(\mu_{\mathrm{A}}=\mu_{\mathrm{A}}(P, T\), \(x_{\mathrm{A}}\) ) and \(\mu_{\mathrm{B}}=\mu_{\mathrm{B}}\left(P, T, x_{\mathrm{A}}\right)\) where \(x_{\mathrm{A}}\) is the mole fraction of \(\mathrm{A}\). Use these facts to derive the relations $$ s \mathrm{~d} T-v \mathrm{~d} P+\sum_{\alpha=\mathrm{A}, \mathrm{B}} x_{\alpha} \mathrm{d} \mu_{\alpha}=0 $$ and $$ \sum_{\alpha=A \cdot B} x_{\alpha}\left(\mathrm{d} \mu_{a}+s_{\alpha} \mathrm{d} T-v_{\alpha} \mathrm{d} P\right)=0 $$ \(\alpha=A, B\) and \(\beta=A, B\),

Assume that two vessels of liquid \(\mathrm{He}^{4}\), connected by a very narrow capillary, are maintained at constant temperature; that is, vessel A is held at temperature \(T_{A}\), and vessel B is held at temperature \(T_{\mathrm{B}}\). If an amount of mass, \(\Delta M\), is transferred reversibly from vessel A to vessel B, how much heat must flow out of (into) each vessel? Assume that \(T_{\mathrm{A}}>T_{\mathrm{B}}\).

The molar free energy of a spin system can be written $$ \begin{aligned} \phi(T, H)=\phi_{0}(T) &-\frac{1}{2} J m^{2} \\ &+\frac{1}{2} k_{\mathrm{B}} T[(1+m) \ln (1+m)+(1-m) \ln (1-m)]-m H \end{aligned} $$ where \(J\) is the interaction strength, \(m\) is the net magnetization per mole, \(\phi_{0}(T)\) is the molar free energy in the absence of a net magnetization, \(H\) is an applied magnetic field, \(k_{\mathrm{B}}\) is Boltzmann's constant, and \(T\) is the temperature. (a) Compute the critical temperature (called the Curie temperature). (b) Compute the linear magnetic susceptibility of this system. (Hint: Only consider temperatures in the neighborhood of the critical point where \(m\) is small.)

A liquid crystal is composed of molecules which are elongated (and often have flat segments). It behaves like a liquid because the locations of the center- of-mass of the molecules have no long-range order. It behaves like a crystal because the orientation of the molecules does have long-range order. The order parameter for a liquid crystal is given by the dyatic \(S=\eta(n n-1 / 3 I)\), where \(\boldsymbol{n}\) is a unit vector (called the director) which gives the average direction of alignment of the molecules. The free energy of the liquid crystal can be written. $$ \phi=\phi_{0}+\frac{1}{2} A S_{i j} S_{i j}-\frac{1}{3} B S_{i j} s_{j k} S_{k i}+\frac{1}{4} C S_{i j} S_{i j} S_{k l} S_{k l} $$ where \(A=A_{0}\left(T-T^{*}\right), A_{0}, B\) and \(C\) are constants, \(I\) is the unit tensor so \(\hat{x}_{i} \cdot I \cdot \hat{x}_{i}=\hat{x}_{i}\) \(\delta_{i i}, S_{i i}=\hat{x}_{i} \cdot S \cdot \hat{x}_{i, \text { and the summation is over repeated indices. The quantities are the unit vectors }}\) \(\hat{x}_{1}=\hat{x}, \hat{x}_{2}=\hat{y}\), and \(\hat{x}_{3}=\hat{z}\). (a) Perform the summations in the expression for \(\Phi\) and write \(\Phi\) in terms of \(\eta, A, B, C\). (b) Compute the critical temperature \(T_{\mathrm{c}}\) at which the transition from isotropic liquid to liquid crystal takes place, and compute the magnitude of the order parameter \(\eta\) at the critical temperature. (c) Compute the difference in entropy between the isotropic liquid \((\eta=0)\) and the liquid crystal at the critical temperature.

The equation of state of a gas is given by the Berthelot equation \(\left(P+a / T v^{2}\right)(v-b)=R T\). (a) Find values of the critical temperature \(T_{c}\), the critical molar volume \(v_{c}\), and the critical pressure \(P_{c}\), in terms of \(a, b\), and \(R\). (b) Does the Berthelot equation satisfy the law of corresponding states? (c) Find the critical exponents \(\beta, \delta\), and \(\gamma\) from the Berthelot equation.

A mixture of particles A and B have a molar Gibbs free energy of the form $$ g=x_{\mathrm{A}} \mu_{\mathrm{A}}^{\circ}(P, T)+x_{\mathrm{B}} \mu_{\mathrm{B}}^{\circ}(P, T)+R T x_{\mathrm{A}} \ln x_{\mathrm{A}}+R T x_{\mathrm{B}} \ln x_{\mathrm{B}}+\lambda x_{\mathrm{A}} x_{\mathrm{B}} $$ where \(\mu_{\mathrm{A}}^{\circ}(P, T)\) and \(\mu_{\mathrm{B}}^{\circ}(P, T)\) are the chemical potentials of pure A and pure B, respectively, at pressure \(P\) and temperature \(T, R\) is the gas constant, \(x_{A}\) and \(x_{\mathrm{B}}\) are the mole fractions of A and B, respectively, and \(\lambda\) measures the strength of coupling between \(\mathrm{A}\) and \(\mathrm{B}\). In terms of dimensionless parameters, \(\bar{g}=g / \lambda, \bar{\mu}_{A}^{\circ}(P, T)=\mu_{\mathrm{A}}^{\circ}(P, T) / \lambda, \bar{\mu}_{\mathrm{B}}^{0}(P, T)=\mu_{\mathrm{B}}^{0}(P, T) / \lambda\), and \(\tau=R T / \lambda\), the molar Gibbs free energy takes the form. $$ \bar{g}=x_{\mathrm{A}} \bar{\mu}_{\mathrm{A}}^{\circ}(P, T)+x_{\mathrm{B}} \bar{\mu}_{\mathrm{B}}^{\circ}(P, T)+\tau x_{\mathrm{A}} \ln x_{\mathrm{A}}+\tau x_{\mathrm{B}} \ln x_{\mathrm{B}}+x_{\mathrm{A}} x_{\mathrm{B}} $$ Assume that \(\overline{\mu_{\mathrm{B}}}=0.45\) and \(\bar{\mu}_{\mathrm{A}}=0.40\). (a) Find the critical temperature \(\tau_{c}\) at which phase separation occurs and plot the curve separating the chemically stable from unstable regions in the \(\tau-x_{\mathrm{A}}\) plane. (b) For \(\tau=1 / 2.6\), find equilibrium values of \(x_{\mathrm{A}}\) on the coexistence curve. (c) For \(\tau=1 / 3.6\), find equilibrium values of \(x_{\mathrm{A}}\) on the coexistence curve. (d) On the same plot as in (a), plot (sketch) the coexistence curve. You can estimate its location based on your results in (b) and (c).

Compute the equilibrium vapor pressure of a monomolecular gas in equilibrium with, a spherical droplet of liquid of the same substance, as a function of the radius \(R\) of the droplet and for fixed temperature. Assume the gas phase is well described by the ideal gas equation of state and the liquid can be assumed to be incompressible. Use the fact that for mechanical equilibrium \(P_{1}-P_{\mathrm{g}}=\) \(2 \sigma / R\), where \(P_{1}\left(P_{g}\right)\) is the pressure of the liquid (gas) and \(\sigma\) is the surface tension.