The phase diagram of a hypothetical substance is (a) Estimate the normal boiling point and freezing point of the substance. (b) What is the physical state of the substance under the following conditions: (i) \(T=150 \mathrm{~K}, P=0.2 \mathrm{~atm}\) (ii) \(T=100 \mathrm{~K}, P=0.8 \mathrm{~atm},(\mathrm{iii}) T=300 \mathrm{~K}, P=1.0 \mathrm{~atm} ?\) (c) What is the triple point of the substance? [Section 11.6\(]\)

Short Answer

Expert verified
The normal boiling point and freezing point can be estimated by locating the intersection points of the melting and vaporization curves with the 1 atm pressure line on the phase diagram. The physical state of the substance under the given conditions can be determined by locating the corresponding points on the phase diagram and identifying the phase: (i) T=150 K, P=0.2 atm; (ii) T=100 K, P=0.8 atm; (iii) T=300 K, P=1.0 atm. The triple point is found by locating the point on the phase diagram where all three phase lines meet.

Step by step solution

01

Estimate the normal boiling point and freezing point

On a phase diagram, the line separating the solid and liquid phases is the melting curve. The point where this curve intersects the line indicating atmospheric pressure (1 atm) represents the normal freezing point. Similarly, the line separating the liquid and gas phases is the vaporization curve, and its intersection with the atmospheric pressure line represents the normal boiling point. To estimate these points, examine the given phase diagram and locate the intersection points of the melting and vaporization curves with the 1 atm pressure line.
02

Determine the physical state under the given conditions

For each of the given conditions, locate the point corresponding to the temperature (T) and pressure (P) on the phase diagram. Then, identify the phase (solid, liquid, or gas) the substance is in at that point. (i) T=150 K, P=0.2 atm (ii) T=100 K, P=0.8 atm (iii) T=300 K, P=1.0 atm
03

Find the triple point of the substance

The triple point is a unique point on a phase diagram where all three phases (solid, liquid, and gas) coexist in equilibrium. To find it, locate the point where the solid-liquid, liquid-gas, and solid-gas lines all meet on the phase diagram.

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Most popular questions from this chapter

At standard temperature and pressure the molar volume of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are \(22.06 \mathrm{~L}\) and \(22.40 \mathrm{~L},\) respectively (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) \(\mathrm{On}\) cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are \(2.02 \mathrm{~g} / \mathrm{cm}^{3}\) and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\).

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For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

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