A very small amount of sugar is dissolved in water, and the solution is in equilibrium with pure ice. The equation of phase equilibrium is $$ g^{\prime}=g^{\prime \prime}+R T \ln (1-x) $$ where \(g^{\prime}\) is the molar Gibbs function of pure ice, \(g^{\prime \prime}\) is the molar Gibbs function of pure water, and \(x\) is the mole fraction of sugar in solution. (a) For an infinitesimal change in \(x\) at constant pressure, show that $$ -s^{\prime} d T=-s^{\prime \prime} d T+R \ln (1-x) d T+R T d \ln (1-x) $$ (b) Substituting for \(R \ln (1-x)\) the value obtained from the equation of phase equilibrium, show that the equation in part (a) reduces to $$ \frac{h^{\prime \prime}-h^{\prime}}{T} d T=R T d \ln (1-x) $$ (c) Taking into account that \(x \ll 1\) and calling \(h^{\prime \prime}-h^{\prime}\) the latent heat of fusion \(l_{F}\), show that the depression of the freezing point is $$ \Delta T=-\frac{R T^{2}}{l_{F}} x $$

Step by step solution

01

Analyze the Given Phase Equilibrium Equation

Starting from the given phase equilibrium equation: \[ g' = g'' + RT \, \ln (1-x) \] Differentiate this equation with respect to temperature, keeping the pressure constant.

02

Differentiate Molar Gibbs Function Terms

To differentiate, use the property that the differential of Gibbs function with respect to temperature is the negative entropy: \[ \frac{dg'}{dT} = -s' \, , \quad \frac{dg''}{dT} = -s'' \] Applying this to the given equation: \[ -s' = -s'' + R \ln(1-x) + RT \frac{d\ln(1-x)}{dT} \]

03

Simplify Expression Using Logarithm Properties

Expand and simplify the logarithmic differentiation. Recall that for small \(x\): \[ \frac{d\ln(1-x)}{dT} = \frac{-x}{1-x} \frac{dx}{dT} \approx -\frac{dx}{dT} \] Putting this back into the differentiated equation: \[ -s' = -s'' + R \ln(1-x) + RT(-\frac{dx}{dT}) \]}

04

Show the Expression from Part (a)

If \(-s' dT\) and \(-s'' dT\) are rearranged: \[ -s' dT = -s'' dT + R \ln(1-x) dT + RT \frac{-dx}{dT} dT \] .Now combine terms to obtain expression: \[ -s' dT = -s'' dT + R \ln(1-x) dT + RT d\ln(1-x) \]}

05

Apply Phase Equilibrium Equation

Substitute the value from phase equilibrium equation, \( R \ln(1-x) = g' - g'' \), into the result of Step 4: \[ -s' dT = -s'' dT + (g' - g'') dT + RT d\ln(1-x) \]}

06

Define Enthalpy Difference and Express Differential

Recollect the relationship \( h = g+ Ts \), then clearly for two phases: \[ h'' - h' = ( g'' + T s'') - ( g' + T s') \] Then: \[ g' - g'' = ( h'' - T s'' ) - ( h' - T s') \] Therefore, substituting this back gives: \[ \frac{h''-h'}{T} dT = R T d\ln (1-x) \]}

07

Approximate for Small Mole Fraction

For small \(x\), \( \ln(1-x) \approx -x \). And the latent heat \( l_F = h'' - h' \): \[ \frac{l_F}{T} dT = RT d(-x) \] Solving for \(dT\): \[ dT = \frac{-R T^2}{l_F} dx \]}

08

Final Expression for Freezing Point Depression

Integrate to get final expression for freezing point depression: \[ \Delta T = \int -\frac{R T^2}{l_F} dx = -\frac{R T^2}{l_F} x \]}

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Key Concepts

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

molar Gibbs function

The molar Gibbs function, denoted as g, is a crucial thermodynamic potential that helps determine the spontaneity of processes. It combines enthalpy (h), temperature (T), and entropy (s) through the relationship: \[ g = h - Ts \] In phase equilibrium, the molar Gibbs function of different phases must be equal. When dealing with ice and water, this means: \[ g' = g'' + RT \, \ln (1-x) \] Here, g' is the molar Gibbs function of pure ice, g'' of pure water, R is the universal gas constant, T is the temperature, and x is the mole fraction of the solute.

latent heat of fusion

The latent heat of fusion ( \( l_F \) ) is the energy required to change a substance from solid to liquid at its melting point, without altering its temperature. It is the difference between the enthalpy of the liquid phase ( \( h'' \) ) and the solid phase ( \( h' \) ): \[ l_F = h'' - h' \] This property plays a critical role in understanding and calculating the temperature change during phase transitions, especially when solutes are present.

freezing point depression

Freezing point depression occurs when a solute is added to a solvent, lowering the temperature at which the solvent freezes. This change in temperature ( \( \Delta T \) ) can be calculated using: \[ \Delta T = -\frac{R T^2}{l_F} x \] Here, R is the gas constant, T is the equilibrium temperature, \( l_F \) is the latent heat of fusion, and x is the mole fraction of the solute. The presence of a solute disrupts the equilibrium between ice and liquid water, leading to a lower freezing point.

phase equilibrium

Phase equilibrium describes a state where multiple phases coexist stably without changing over time. For ice and liquid water, this equilibrium is governed by their Gibbs functions. The equation: \[ g' = g'' + RT \, \ln(1-x) \] ensures that the chemical potential of water is the same in both solid and liquid phases. When different phases are at equilibrium, the temperature, pressure, and chemical potentials are balanced, ensuring no net phase change.

entropy and enthalpy relationships

Entropy (s) and enthalpy (h) are fundamental in understanding phase changes and equilibrium. Their relationship with Gibbs function is given by: \[ g = h - Ts \] This expresses how the Gibbs function incorporates both enthalpy (the heat content) and entropy (the measure of disorder). During phase changes, these quantities dictate the energy balance and the preferred phase at given conditions, influencing properties like the latent heat of fusion and the freezing point.