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Chapter 11: Problem 60

Krypton, \(\mathrm{Kr}\), has a triple point at \(-169^{\circ} \mathrm{C}\) and \(133 \mathrm{mmHg}\) and a critical point at \(-63^{\circ} \mathrm{C}\) and 54 atm. The density of the solid is \(2.8 \mathrm{~g} / \mathrm{cm}^{3},\) and the density of the liquid is \(2.4 \mathrm{~g} / \mathrm{cm}^{3} .\) Sketch a rough phase diagram of krypton. Circle the correct word in each of the following sentences (and explain your answers). a)Solid krypton at \(130 \mathrm{mmHg}\) (melts, sublimes without melting) when the temperature is raised. b)Solid krypton at \(760 \mathrm{mmHg}\) (melts, sublimes without melting) when the temperature is raised.

Short Answer

Expert verified
a) Sublimes without melting; b) Melts.

Step by step solution

01

Understanding the Phase Diagram

A phase diagram plots pressure versus temperature and shows the conditions under which different phases (solid, liquid, gas) of a substance are stable. Key points include the triple point where solid, liquid, and gas phases coexist, and the critical point beyond which the liquid and gas phases are indistinguishable.
02

Plot Key Points on the Phase Diagram

Mark the triple point at \( -169^{\circ} \mathrm{C} \) and \( 133 \mathrm{mmHg} \), ensuring the pressure is in mmHg. Mark the critical point at \( -63^{\circ} \mathrm{C} \) and \( 54 \ \text{atm} \). It might be useful to convert \( 54 \ \text{atm} \) to mmHg: \( 54 \times 760 = 41040 \ \text{mmHg} \).
03

Connect the Points with Phase Boundaries

Draw lines to separate the phases. The line from the triple point to the critical point is the liquid-gas boundary. Below the triple point, the sublimation line separates the solid and gas phases, and the melting line separates the solid and liquid phases.
04

Analyze Solid Krypton at 130 mmHg

Since 130 mmHg is below the triple point pressure (133 mmHg), increasing temperature means passing from solid directly to gas, i.e., sublimation occurs without melting.
05

Analyze Solid Krypton at 760 mmHg

Given 760 mmHg is above the triple point pressure, when temperature is increased, the solid will first melt into a liquid (as the transition from solid to liquid occurs).

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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 is a unique condition on a phase diagram where a substance can exist simultaneously in solid, liquid, and gaseous forms. For krypton, this point is marked at a temperature of \(-169^{\circ} \text{C}\) and a pressure of \(133\, \text{mmHg}\). At this exact point of pressure and temperature:
  • Solid, liquid, and gas phases are in equilibrium.
  • The substance can transition between these phases without a change in pressure or temperature.
This means that at the triple point, krypton can exist as a solid, liquid, and gas all at the same time, constantly changing forms.
Critical Point
The critical point marks the last point at which liquid and gas phases of a substance can coexist. Beyond this point, known as the supercritical region, the distinction between liquid and gas disappears. For krypton, the critical point occurs at \(-63^{\circ} \text{C}\) and \(54\, \text{atm}\).
  • Beyond the critical temperature, gas cannot be liquefied by pressure alone.
  • Liquid and gas become indistinguishable.
Understanding the critical point is crucial when studying substances at high pressures and temperatures, as it defines the limits of phase coexistence.
Phase Transition
Phase transitions on a phase diagram represent shifts between solid, liquid, and gas phases. Understanding these transitions for krypton involves:
  • Sublimation: Direct transition from solid to gas as seen with solid krypton at \(130\, \text{mmHg}\) when temperature is raised.
  • Melting: Transition from solid to liquid, occurring when pressure is above the triple point as with krypton at \(760\, \text{mmHg}\).
Transitions are demonstrated by the boundaries on the phase diagram, such as the sublimation line and the melting line. These lines indicate the specific conditions under which krypton changes phase.
Pressure-Temperature Relationship
In a phase diagram, the relationship between pressure and temperature for a substance like krypton is crucial for understanding phase stability. This relationship is shown by:
  • Lines connecting key points like the triple point and critical point, indicating phase boundaries.
  • A decrease in pressure typically favors the gaseous phase.
  • An increase in pressure tends to favor the liquid or solid phases, depending on the temperature.
By understanding this relationship, you can predict the phase of krypton under different conditions of pressure and temperature, which is essential for practical applications and scientific studies.

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Most popular questions from this chapter

Part 1: a Is it possible to add heat to a pure substance and not observe a temperature change? If so, provide examples. Describe, on a molecular level, what happens to the heat being added to a substance just before and during melting. Do any of these molecular changes cause a change in temperature? Part 2: Consider two pure substances with equal molar masses: substance A, having very strong intermolecular attractions, and substance \(\mathrm{B}\), having relatively weak intermolecular attractions. Draw two separate heating curves for 0.25 -mol samples of substance \(\mathrm{A}\) and substance \(\mathrm{B}\) in going from the solid to the vapor state. You decide on the freezing point and boiling point for each substance, keeping in mind the information provided in this problem. Here is some additional information for constructing the curves. In both cases, the rate at which you add heat is the same. Prior to heating, both substances are at \(-50^{\circ} \mathrm{C},\) which is below their freezing points. The heat capacities of \(\mathrm{A}\) and \(\mathrm{B}\) are very similar in all states. As you were heating substances \(\mathrm{A}\) and \(\mathrm{B}\), did they melt after equal quantities of heat were added to each substance? Explain how your heating curves support your answer. What were the boiling points you assigned to the substances? Are the boiling points the same? If not, explain how you decided to display them on your curves. c According to your heating curves, which substance reached the boiling point first? Justify your answer. Is the quantity of heat added to melt substance \(\mathrm{A}\) at its melting point the same as the quantity of heat required to convert all of substance \(\mathrm{A}\) to a gas at its boiling point? Should these quantities be equal? Explain.

The halogens form a series of compounds with each other, which are called interhalogens. Examples are bromine chloride \((\mathrm{BrCl})\), iodine bromide \((\mathrm{IBr})\), bromine fluoride \((\mathrm{BrF})\) and chlorine fluoride (ClF). Which compound is expected to have the highest boiling point at any given pressure? Explain.

Identify the phase transition occurring in each of the following. The water level in an aquarium tank falls continuously (the tank has no leak). A mixture of scrambled eggs placed in a cold vacuum chamber slowly turns to a powdery solid. Chlorine gas is passed into a very cold test tube where it turns to a yellow liquid. When carbon dioxide gas under pressure exits from a small orifice, it turns to a white "snow." Molten lava from a volcano cools and turns to solid rock.

Calculate the percent of volume that is actually occupied by spheres in a face-centered cubic lattice of identical spheres. You can do this by first relating the radius of a sphere, \(r,\) to the length of an edge of a unit cell, \(l .\) (Note that the spheres do not touch along an edge but do touch along the diagonal of a face.) Then calculate the volume of a unit cell in terms of \(r\). The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere \(\left(4 \pi r^{3} / 3\right)\)

Compare the quantities of heat required for each of the following pairings. Justify your answers without performing calculations. Evaporating a \(1.00-\mathrm{g}\) sample of liquid water at \(100^{\circ} \mathrm{C}\) or subliming a \(1.00 \mathrm{~g}\) sample of water at \(-10^{\circ} \mathrm{C} ?\) Condensing a \(0.5-\mathrm{g}\) sample of water vapor at \(60^{\circ} \mathrm{C}\) or melting a \(0.5-\mathrm{g}\) sample of water at \(0^{\circ} \mathrm{C}\). c) Vaporizing a \(50-\mathrm{g}\) sample of liquid water placed into an oven set at \(200^{\circ} \mathrm{C}\) or vaporizing a \(100-\mathrm{g}\) sample of liquid water placed into an oven set at \(100^{\circ} \mathrm{C}\). Both water samples are at \(100^{\circ} \mathrm{C}\) when they are placed in the ovens.
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