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Chapter 4: Problem 8

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.

Short Answer

Expert verified
The rate of change of chemical potential along the coexistence curve is \frac{v_g - v_l}{s_g - s_l}.

Step by step solution

01

Understand the Coexistence Curve

Along the coexistence curve of a liquid and gas, the chemical potential \( \mu \) of the liquid phase is equal to the chemical potential of the gas phase: \[ \mu_l = \mu_g \] where the subscripts \( l \) and \( g \) denote the liquid and gas phases, respectively.
02

Express Chemical Potential Differentials

Since \( \mu_l= \mu_g \, \) we also have their differentials equal along the coexistence curve: \[ d\mu_l= d\mu_g \]
03

Use Gibbs-Duhem Relation

Using the Gibbs-Duhem relation, we express the differential of chemical potential: \[ d \mu = -s \, dT + v \, dP \] where \( s \) is the molar entropy and \( v \) is the molar volume.
04

Relate Changes in Chemical Potential to Temperature and Pressure

Therefore, for the liquid phase: \[ d \mu_l = -s_l \,, dT + v_l \, dP \] and for the gas phase: \[ d \mu_g = -s_g \,, dT + v_g \, dP \]
05

Set Differential Changes Equal

Since \ d\mu_l = d \mu_g \,, we equate the equations from Step 4: \[ -s_l \,, dT + v_l \, dP = -s_g \,, dT + v_g \, dP \]
06

Factor and Rearrange Equation

Rearranging the terms to isolate \ dT/dP \,, we get: \[ (s_g - s_l) \,, dT = (v_g - v_l) \,, dP \] Or \[ \frac{dT}{dP} = \frac{v_g - v_l}{s_g - s_l} \]
07

Express the Rate of Change

Finally, the required rate of change of the chemical potential along the coexistence curve is: \[ \frac{d\mu}{dT} = \frac{v_g - v_l}{s_g - s_l} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coexistence Curve
In a system where a liquid and its vapor coexist, the coexistence curve represents the set of conditions (temperature and pressure) where these two phases are in equilibrium. At any point along this curve, the chemical potential (\( \mu \)) of the liquid phase is equal to that of the gas phase (\( \mu_l = \mu_g \)). This equality ensures that there is no net mass transfer between the liquid and vapor phases under equilibrium conditions.
To analyze how chemical potential changes along this curve, it's crucial to understand that phase equilibrium is maintained and both phases share the same pressure and temperature.

Gibbs-Duhem Relation
The Gibbs-Duhem relation is a thermodynamic equation that relates changes in the chemical potential of a component to other thermodynamic variables like temperature and pressure. Mathematically, it is expressed as:
\[ d\mu = -s \ dT + v \ dP \]
where:
  • \( s \) represents molar entropy
  • \(v \) represents molar volume
This relation indicates that any infinitesimal change in the chemical potential is due to contributions from changes in temperature and pressure. In our context, the Gibbs-Duhem relation helps us express differentials of the chemical potential for the liquid and gas phases along the coexistence curve, facilitating the calculation of \( \frac{d\mu}{dT} \).

Molar Entropy
Molar entropy (\(s\)) is a measure of the amount of disorder or randomness in a system per mole of substance. Entropy increases with temperature as molecules move more vigorously and occupy more states. For a system along its liquid-gas coexistence curve, we denote the molar entropy of the liquid as \( s_l \) and that of the gas as \( s_g \). Since gases generally have higher entropy due to their more disordered state compared to liquids, \( s_g \ > s_l \). These entropies are critical for solving for the rate of change of chemical potential as they directly appear in the Gibbs-Duhem relation.

Molar Volume
Molar volume (\(v\)) is the volume occupied by one mole of a substance. In the context of our coexistence curve problem, it is represented as \( v_l \) for the liquid phase and \( v_g \) for the gas phase. Gas molecules are much farther apart than liquid molecules, so the molar volume of the gas is much larger than that of the liquid (\( v_g \ > v_l \)). Understanding the molar volumes and their differences is essential because they affect how pressure changes influence the chemical potential. Based on the Gibbs-Duhem relation, expressions involving \( s_g \), \( s_l \), \( v_g \), and \( v_l \) allow us to understand the rate of temperature and pressure changes along the coexistence curve.

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Most popular questions from this chapter

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 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?

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 \(\beta=A, B\),

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.
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