Question

A. It has been proposed that the reason ice skating works so well is that the pressure from the blades of the skates melts the ice. Consider this proposal from the viewpoint of phase equilibria, The phase change in question is $$ \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(/) $$ Assume that \(\Delta H\) for this process is independent of temperature and pressure and is cqual to 80 cal/g. The change in volume, \(\Delta V\), is about \(-9.1 \times\) \(10^{-5} \mathrm{~L} / \mathrm{g}\). The pressure exerted is the force per unit area. For a 180 pound person and an area for the skate blades of about 6 square inches, the pressure is \(30 \mathrm{lb} / \mathrm{sq}\). in. or about 2 atmospheres. With this information, calculate the decrease in the melting temperature of ice caused by the skate blades. (Note that 1 cal \(=0.04129\) L.atm.) Is this a good explanation for why ice skating works? B. A more efficient way of melting ice is to add an inert compound such as urea. (We will avoid salt to save our cars.) The extent to which the freexing point is lowered can be calculated by noting that the molar free energy of water must be the same in the solid ice and the urea solution. The molar free energy of water in the urea solution can be approximated as \(G_{\text {liquid }}^{\circ}+R T \ln X_{\text {water }}\) where \(X_{\text {water }}\) is the mole fraction of water in the solution. The molar free energy of the solid can be written as \(G_{\text {solid }}^{\circ}\). Derive an expression for the change in the melting temperature of ice by equating the free energies in the two phases, differentiating the resulting equation with respect to temperature, integrating from a mole fraction of 1 (pure solvent) to the mole fraction of the solution, and noting that \(\ln X_{\text {wamer }}=\ln \left(1-X_{\text {urea }}\right)\) \(=-X_{\text {urea }}\) (This relationship is the series expansion of the logarithm for small values of \(X_{\text {unea }}\). Since the concentration of water is about \(55 \mathrm{M}\), this is a good approximation.) With the relationship derived, estimate the de- crease in the melting temperature of ice for an \(1 \mathrm{M}\) urea solution. The heat of fusion of water is \(1440 \mathrm{cal} / \mathrm{mol}\).

Short answer

Pressure has minimal effect; surface melting is key. Urea lowers the melting point by 1.86 K.

Step-by-step solution

Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the change in pressure and temperature during a phase change: \( \frac{dP}{dT} = \frac{\Delta H}{T \Delta V} \). This formula will help calculate the change in melting temperature due to pressure.

Calculate Pressure Change

The pressure change \( \Delta P \) is given as 2 atmospheres (since the normal atmospheric pressure is 1 atm, the skate increases it by approximately 1 additional atm).

Solve for Change in Temperature

Rearrange the Clausius-Clapeyron equation to focus on \( \Delta T \): \( \Delta T = \frac{\Delta H}{\Delta P \times \Delta V} \). Convert \( \Delta H \) from cal/g to L·atm/g using the conversion 1 cal = 0.04129 L·atm.

Substitute Values and Calculate

\( \Delta H = 80 \text{ cal/g} \times 0.04129 \text{ L.atm/cal} = 3.3032 \text{ L·atm/g} \)\( \Delta T = \frac{3.3032}{2 \times (-9.1 \times 10^{-5})} \) L·atm/g = 799.23 K.

Interpret the Results for Part A

The calculated decrease in melting temperature is actually too large, indicating that pressure-induced melting is not the primary reason ice skating works so well. Friction and surface melting are more significant factors.

Derive Formula for Melting Point Depression using Urea

Using the formula for the free energy: \( G_{liquid}^{\circ} + RT \ln X_{water} \) and \( G_{solid}^{\circ} \), set them equal and differentiate: \( \Delta G = 0 = \Delta H - T \Delta S + RT \ln(1 - X_{urea}) \). Approximate \( \ln(1 - X_{urea}) \approx -X_{urea} \) because \( X_{urea} \ll 1 \).

Solve for Change in Temperature with Urea

\( \Delta T = \frac{RT^2}{\Delta H_{fus}} X_{urea} \). Substitute \( R = 8.314 \text{ J/mol·K} \), \( T = 273.15 \text{ K} \), \( \Delta H_{fus} = 1440 \text{ cal/mol} \times 4.184 \text{ J/cal} = 6010.56 \text{ J/mol} \), and \( X_{urea} = 1/55 \text{ M} \).

Substitute Values and Calculate for Part B

\( \Delta T = \frac{8.314 \times (273.15)^2}{6010.56} \times \frac{1}{55} \approx 1.86 \text{ K} \). Therefore, urea in solution causes a 1.86 K decrease in the melting point of ice.

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation is a fascinating tool used to understand phase transitions, like the melting of ice. It elegantly connects the changes in pressure and temperature during such a phase change. In mathematical terms, it is expressed as: \( \frac{dP}{dT} = \frac{\Delta H}{T \Delta V} \) where \( dP/dT \) is the rate of change of pressure concerning temperature, \( \Delta H \) is the enthalpy change, \( T \) is the temperature, and \( \Delta V \) is the change in volume.
This equation is immensely useful, particularly when calculating how minor pressure alterations can influence the melting temperature of substances such as ice. By rearranging this formula, you can directly solve for the change in temperature, \( \Delta T \). This adjustment allows you to predict how much force applied by an object, like a skate blade, alters the thermal behavior of ice.
However, it's crucial to note that this equation assumes that certain variables, like \( \Delta H \), remain constant with temperature and pressure. Understanding these assumptions will help you efficiently use the Clausius-Clapeyron Equation in real-world scenarios.

Pressure-induced Melting

Pressure-induced melting occurs when pressure is applied to a substance, causing it to transition from a solid to a liquid state. This phenomenon can be partially explained through phase equilibria principles.
In the context of ice skating, the pressure from the skater's weight on narrow blades is often thought to lower the melting point of the ice, forming a thin layer of water that eases glide. This makes sense from a physics standpoint because added pressure can change the phase equilibrium conditions, promoting melting.
However, calculations of the pressure and its effect on melting temperature often reveal that pressure alone is not the main reason why skating is effective. When evaluated through the Clausius-Clapeyron Equation, the decrease in melting temperature appears too large to solely account for the observed ease of movement.
This highlights other contributing factors to skating efficiency, such as frictional heating and surface melting, which also create a thin film of water beneath the blade, enhancing the skater's glide.

Freezing Point Depression

Freezing Point Depression is a curious effect where the addition of a substance (solute) to a solvent like water lowers its freezing point. This principle is frequently used to prevent ice formation and maintain liquid phases under cold conditions.
When you dissolve a solute like urea in water, it disrupts the solvent's ability to solidify at its normal freezing point. This interruption occurs because the solute particles interfere with the orderly crystal formation required for a solid state, lowering the energy needed to remain liquid.
To calculate the freezing point depression specifically, we use the formula: \[ \Delta T = \frac{RT^2}{\Delta H_{fus}} X_{urea} \] Where \( \Delta T \) is the depression in freezing point, \( R \) is the gas constant, \( T \) is the temperature, \( \Delta H_{fus} \) is the enthalpy of fusion, and \( X_{urea} \) is the mole fraction of the solute.
Applying this formula to a scenario where urea is added to water shows a significant drop in the freezing temperature, which explains why urea can be deployed to melt ice more efficiently than purely relying on pressure changes.