Consider the following data for xenon: $$ \begin{array}{ll}{\text { Triple point: }} & {-121^{\circ} \mathrm{C}, 280 \text { torr }} \\ {\text { Normal melting point: }} & {-112^{\circ} \mathrm{C}} \\ {\text { Normal boiling point: }} & {-107^{\circ} \mathrm{C}}\end{array} $$ Which is more dense, \(\operatorname{Xe}(s)\) or \(\operatorname{Xe}(l) ?\) How do the melting point and boiling point of xenon depend on pressure?

We know that the phase diagram is a representation of the phases (states) of a substance as a function of temperature and pressure. The three common states of a substance are solid, liquid, and gas, which can be distinguished by their densities. For xenon, we have the following data: - Triple point: \(-121 °C\) and \(280 \,\text{torr}\) - Normal melting point: \(-112 °C\) - Normal boiling point: \(-107 °C\) The triple point represents the unique set of conditions (temperature and pressure) where all three phases (solid, liquid, and gas) coexist in equilibrium. In this problem, we are interested in understanding the density differences between solid and liquid xenon.

To determine which state is denser, we need to compare the melting point of solid xenon to that of liquid xenon. Here, we are given the normal melting point, which is the temperature at which the solid state becomes the liquid state under 1 atmospheric pressure (760 torr). At the normal melting point, we can compare the density of solid and liquid xenon: - If the temperature is higher than the melting point, the substance tends to be in liquid state - If the temperature is lower than the melting point, the substance tends to be in solid state Since the normal melting point of xenon is \(-112 °C\), and since the triple point is at \(-121 °C\), we can conclude that \(Xe(s)\) exists below the normal melting point, while \(Xe(l)\) exists above the normal melting point. However, the triple point is at a lower pressure than atmospheric pressure. Under such a low pressure, we can expect that \(Xe(l)\) will have a lower density than \(Xe(s)\), as it is still somewhat close to its gaseous phase. So, the denser state between solid and liquid xenon is \(Xe(s)\), i.e., solid xenon.

In general, the melting point of a substance increases with an increase in pressure, and the boiling point of a substance decreases with an increase in pressure. The provided data gives boiling points and melting points at "normal" atmospheric pressure. To understand how these properties change with pressure, we need to look at the phase diagram or have additional data points at different pressures. However, without further data or a detailed phase diagram for xenon, we cannot specifically determine the exact dependence of melting and boiling points on pressure. Generally, the melting point and boiling point of xenon will depend on pressure, but the given data is not enough to establish the exact relationship. In conclusion, solid xenon (\(Xe(s)\)) is denser than liquid xenon (\(Xe(l)\)). The melting and boiling points of xenon do depend on pressure, but further data would be necessary to establish the exact relationship between these properties and pressure.