The melting point of a fictional substance X is 225^∘ C at 10.0 atm. If the density of the solid

step 1 We should know that the density of a solid is less than the density of a liquid. So this means t

The melting point of a fictional substance X is 225^∘ C at...

The melting point of a fictional substance $X$ is $225^{\circ} \mathrm{C}$ at 10.0 atm. If the density of the solid phase of $\mathrm{X}$ is 2.67 $\mathrm{g} / \mathrm{cm}^{3}$ and the density of the liquid phase is 2.78 $\mathrm{g} / \mathrm{cm}^{3}$ at 10.0 atm, predict whether the normal melting point of $\mathrm{X}$ will be less than, equal to, or greater than $225^{\circ} \mathrm{C}$ . Explain.

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Key Concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation describes the relationship between pressure and temperature during a phase change and provides the slope of the coexistence curve in a phase diagram. It relates the change in pressure with change in temperature to the latent heat of transformation and the change in specific volume between the two phases, indicating how the phase boundary shifts under different pressures.

Specific Volume and Density Relationship

Specific volume is the volume occupied per unit mass and is inversely related to density. Understanding this relationship is key in phase transition problems, as a change in density between two phases implies a corresponding change in specific volume, which directly affects the thermodynamic parameters associated with phase equilibria.

Pressure Dependence of Melting Point

The melting point of a substance can change with pressure. When the substance transitions from solid to liquid, any resulting change in specific volume (due to differences in density) determines whether the melting point increases or decreases as pressure is varied. A decrease in volume upon melting, for instance, results in a negative slope on the phase boundary, indicating that increased pressure lowers the melting temperature.

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