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Oxygen has \(T_{\mathrm{t}}=54.3 \mathrm{~K}, P_{\mathrm{t}}=1.14 \mathrm{~mm} \mathrm{Hg}, T_{\mathrm{c}}=154.6 \mathrm{~K}\), and \(P_{c}=49.77 \mathrm{~atm} .\) The density of the liquid is \(1.14 \mathrm{~g} / \mathrm{cm}^{3}\), and the density of the solid is \(1.33 \mathrm{~g} / \mathrm{cm}^{3}\). Sketch a phase diagram for oxygen, and label all points of interest.

Short Answer

Expert verified
Plot triple and critical points, draw phase boundaries, and label the phases.

Step by step solution

01

Understand the Phase Diagram

A phase diagram is a graphical representation showing the phases of a substance at different temperatures and pressures. Key points to include are the triple point (where solid, liquid, and gas coexist), critical point (highest temperature where liquid and gas can coexist), and normal boiling and melting points at standard atmospheric pressure.
02

Plot the Triple Point

The triple point is given as \(T_{\mathrm{t}}=54.3 \; \mathrm{K}\) and \(P_{\mathrm{t}}=1.14 \; \mathrm{mmHg}\). Convert the pressure into the same units as the critical pressure if needed. Place this point on the diagram carefully.
03

Plot the Critical Point

The critical point is \(T_{\mathrm{c}}=154.6 \; \mathrm{K}\) and \(P_{c}=49.77 \; \mathrm{atm}\). Make sure this is placed at the upper end of the pressure scale, signifying the high pressure and temperature beyond which oxygen cannot exist as a liquid.
04

Draw the Phase Boundaries

Draw lines to connect these points, ensuring the lines denote the boundaries between solid, liquid, and gaseous phases. The line from the triple point to lower pressure and temperature represents sublimation, to higher pressure represents melting, and to higher temperature represents boiling.
05

Label Key Areas

Label the solid, liquid, and gas regions on the diagram. Mark the lines carefully and label the points: Triple Point (TP), Critical Point (CP), and any other standard equilibrium lines like normal boiling and melting points if known or applicable.

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Key Concepts

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

Triple point
The triple point of a substance is a fascinating concept in thermodynamics. It is the unique set of conditions at which the three phases of matter—solid, liquid, and gas—can all exist in equilibrium. For oxygen, this occurs at a temperature of 54.3 K and a pressure of 1.14 mmHg. Here, the different phases transition seamlessly, allowing you literally to visualize the process of phase change at this specific point.

Understanding the triple point is crucial because it serves as a fixed reference in phase diagrams. It highlights the interconnected nature of different states of matter, providing a clear insight into how temperature and pressure work together to affect phase transitions in substances like oxygen.
Critical point
The critical point represents another extraordinary aspect of phase diagrams. This is the point of no return for a substance. At this stage, with oxygen specifically at a critical temperature of 154.6 K and pressure of 49.77 atm, the liquid and gas phases merge into a single fluid phase called a supercritical fluid.

The critical point signifies the limit where surface tension between liquid and gas phases disappears. Beyond this point, the distinct differences between liquid and gas are lost, highlighting the versatile nature of substances under extreme conditions. Understanding this concept helps in the study of fluid dynamics and offers critical insights into industrial processes that utilize supercritical fluids.
Sublimation
Sublimation is a direct phase transition from a solid to a gas without passing through the liquid state. In the context of oxygen, sublimation occurs along the line from the triple point towards lower pressures and temperatures on the phase diagram.

This concept is particularly useful in understanding how substances can change state in environments with exceptionally low pressures. For everyday examples, think of dry ice (solid carbon dioxide) sublimating at room temperature. Understanding sublimation supports the study of atmospheric processes and is especially relevant in the science of cryogenics and materials processing.
Phase boundaries
Phase boundaries in a phase diagram are lines or curves that denote the conditions under which a substance exists in different phases. These boundaries are shaped by thermodynamics and dictate the transition between solid, liquid, and gas phases.

In the phase diagram of oxygen, these boundaries include the
  • solid-liquid boundary (melting or freezing line)
  • liquid-gas boundary (boiling line)
  • solid-gas boundary (sublimation line)

The phase boundaries meet at points like the triple point and extend to areas like the critical point. They enable us to predict how a substance behaves under varying temperature and pressure, which is invaluable for both scientific applications and practical industrial processes.

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Most popular questions from this chapter

In Denver, the Mile-High City, water boils at \(95^{\circ} \mathrm{C}\). What is atmospheric pressure in atmospheres in Denver? \(\Delta H_{\text {vap }}\) for \(\mathrm{H}_{2} \mathrm{O}\) is \(40.67 \mathrm{~kJ} / \mathrm{mol}\)

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How much energy in kilojoules is released when \(25.0 \mathrm{~g}\) of ethanol vapor at \(93.0{ }^{\circ} \mathrm{C}\) is cooled to \(-11.0{ }^{\circ} \mathrm{C} ?\) Ethanol has \(\mathrm{mp}=-114.1{ }^{\circ} \mathrm{C}, \mathrm{bp}=78.3^{\circ} \mathrm{C}, \Delta H_{\mathrm{vap}}=38.56 \mathrm{~kJ} / \mathrm{mol}\), and \(\Delta H_{\text {fusion }}=4.93 \mathrm{~kJ} / \mathrm{mol}\). The molar heat capacity is \(112.3\) \(\mathrm{J} /(\mathrm{K} \cdot \mathrm{mol})\) for the liquid and \(65.6 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\) for the vapor.

Dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), is an organic solvent used for removing caffeine from coffee beans. The following table gives the vapor pressure of dichloromethane at various temperatures. Fill in the rest of the table, and use the data to plot curves of \(P_{\text {vap }}\) versus \(T\) and \(\ln P_{\text {vap }}\) versus \(1 / T\). $$ \begin{array}{lccc} \hline \text { Temp (K) } & P_{\text {vap }}(\mathrm{mm} \mathrm{Hg}) & \ln P_{\text {vap }} & 1 / T \\ \hline 263 & 80.1 & ? & ? \\ 273 & 133.6 & ? & ? \\ 283 & 213.3 & ? & ? \\ 293 & 329.6 & ? & ? \\ 303 & 495.4 & ? & ? \\ 313 & 724.4 & ? & ? \end{array} $$

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