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?

Short Answer

Expert verified
The molar Gibbs free energy for solid is \(G_s = \frac{B}{Tv^{3}} + Pv\), for liquid is \(G_l = \frac{A}{Tv^{2}} + Pv\). At phase transition, \(v_s = v_l\). The slope of the coexistence curve is given by the Clapeyron equation.

Step by step solution

01

Identify Given Information

Given for solid phase: Helmholtz free energy per mole, \(a_{s}=\frac{B}{Tv^{3}}\). Given for liquid phase: Helmholtz free energy per mole, \(a_{1}=\frac{A}{Tv^{2}}\). Constants: \(A\) and \(B\), Volume per mole: \(v\), Temperature: \(T\).
02

Compute Molar Gibbs Free Energy for Solid Phase

Use the relation for Gibbs free energy: \(G = A + Pv\). For the solid phase, compute \(G_{s} = a_{s} + Pv = \frac{B}{Tv^{3}} + Pv\).
03

Compute Molar Gibbs Free Energy for Liquid Phase

Similarly, use the relation for the liquid phase: \(G_{1} = a_{1} + Pv = \frac{A}{Tv^{2}} + Pv\).
04

Relate Molar Volumes at Phase Transition

At the phase transition, the Gibbs free energies of solid and liquid phases are equal: \(\frac{B}{Tv^{3}} + Pv = \frac{A}{Tv^{2}} + Pv\). Simplify this relation to find the relationship between \(v_{s}\) and \(v_{l}\).
05

Determine Slope of Coexistence Curve in P-T Plane

Use the Clapeyron equation, \(\frac{dP}{dT} = \frac{\Delta S}{\Delta V}\). Find \(\Delta S\) and \(\Delta V\) using the relations \(S = -\left(\frac{\partial a}{\partial T}\right)_{v}\) and volumes at phase transition respectively.

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Key Concepts

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

Helmholtz Free Energy
Helmholtz free energy, denoted as F, is a thermodynamic potential that measures the useful work obtainable from a system at constant temperature and volume. It's particularly useful in phase transitions. For the solid phase in the given exercise, the Helmholtz free energy per mole is represented as \(a_{s} = \frac{B}{Tv^{3}}\). Similarly, for the liquid phase, it is represented as \(a_{1} = \frac{A}{Tv^{2}}\). The equations suggest that as temperature (T) increases, the Helmholtz free energy decreases, making transitions feasible.

The relation between Helmholtz free energy and Gibbs free energy, G, helps in phase transition analysis: \(G = A + Pv\), where \(P\) is pressure and \(v\) is volume per mole. Knowing this aids in calculating the Gibbs free energies for different phases, as HFE directly influences phase stability at specific temperatures and volumes.
Clapeyron Equation
The Clapeyron equation is vital for understanding phase transitions. It describes the slope of the coexistence curve between two phases in the pressure-temperature (P-T) diagram. In the given exercise, the Clapeyron equation is used to determine the slope at which solid and liquid phases coexist:

\[ \frac{dP}{dT} = \frac{\Delta S}{\Delta V} \]
Here, \(\frac{dP}{dT}\) is the slope of the P-T curve, \(\Delta S\) is the change in entropy between phases, and \(\Delta V\) is the change in volume.
To find \(\Delta S\), we need the relation \(S = -\left(\frac{\partial a}{\partial T}\right)_{v}\). Calculating the partial derivatives of Helmholtz free energy for solid and liquid phases with respect to T helps in obtaining \(\Delta S\). Similarly, \(\Delta V\) can be determined from the molar volumes of solid and liquid phases.
Phase Transition Thermodynamics
Phase transition thermodynamics examines how substances change from one phase to another and the energy changes involved. The exercise highlights how Gibbs Free Energy (G) governs phase transitions. At equilibrium between solid and liquid phases, their Gibbs Free Energies are equal:

\[ \frac{B}{Tv^{3}} + Pv = \frac{A}{Tv^{2}} + Pv \]
This equality helps derive the relationship between molar volumes of the phases.
Understanding these transitions requires knowing:
  • **Enthalpy (H)** - the heat content at constant pressure.
  • **Entropy (S)** - the disorder or randomness in the system.
  • **Temperature (T)** - driving factor which influences the phase stability.
The Clapeyron relation and entropy calculations further deepen our understanding of phase transition thermodynamics.
Molar Volumes
Molar volume, denoted as v, is the volume occupied by one mole of a substance. It significantly influences phase transitions. In the exercise, both the solid and liquid phases have different Helmholtz free energies due to their varying molar volumes:

Solid phase: \(a_{s}=\frac{B}{Tv^{3}}\)
Liquid phase: \(a_{1}=\frac{A}{Tv^{2}}\).
At the phase transition point, the Gibbs free energy equality influences how molar volumes of phases relate:

\[ \frac{B}{Tv^{3}} + Pv = \frac{A}{Tv^{2}} + Pv \]
Simplifying this relation offers insights into how solid and liquid molar volumes are connected during transitions. The change in molar volume (\(\frac{\Delta V}{\Delta T}\)) as a function of temperature helps determine the slope of the coexistence curve using the Clapeyron equation. Understanding molar volumes aids in predicting material behaviors under different pressure and temperature conditions.

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Most popular questions from this chapter

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.

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.

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?

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.)

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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