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Chapter 12: Problem 60

Why is the triple point of water (ice-liquid-vapor) a better fixed point for establishing a thermometric scale than either the melting point of ice or the boiling point of water?

Short Answer

Expert verified
The triple point of water is a better fixed point for establishing a thermometric scale than the melting point of ice or the boiling point of water because at the triple point, all three phases of water co-exist in equilibrium. This state is well-defined and can be accurately reproduced in a controlled laboratory setting, making it more reliable and accurate. In contrast, the melting and boiling points can be influenced by impurities or atmospheric changes.

Step by step solution

01

Understand the Concepts

A triple point of a substance is the temperature and pressure at which the substance can co-exist in three phases - solid, liquid, and gas. In the case of water, this is 0.01 degrees Celsius at standard pressure. Thermometric scales are established at fixed points, which are physical situations where a simple change can occur.
02

Compare with other Points

Both the melting point of ice and the boiling point of water have been traditionally used as reference points in many thermometric scales. However, these points can be influenced by impurities or atmospheric changes, leading to slight inaccuracies.
03

Explain the advantage of the Triple Point

The triple point of water, however, represents a unique state where all three phases co-exist in equilibrium. The conditions for this state are well-defined and can be reproduced accurately in a controlled laboratory setting, making it more reliable and accurate as a fixed point for establishing a thermometric scale.

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Most popular questions from this chapter

Are the fullerenes network covalent solids? What makes them different from diamond and graphite? It has been shown that carbon can form chains in which every other carbon atom is bonded to the next carbon atom by a triple bond. Is this allotrope of carbon a network covalent solid? Explain.

Trace the phase changes that occur as a sample of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) originally at \(1.00 \mathrm{mmHg}\) and \(-0.10^{\circ} \mathrm{C},\) is compressed at constant temperature until the pressure reaches 100 atm.

Sketch a plausible phase diagram for hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) from the following data: triple point \(\left(2.0^{\circ} \mathrm{C}\right)\) and \(3.4 \mathrm{mm} \mathrm{Hg}\) ), the normal melting point \(\left(2^{\circ} \mathrm{C}\right),\) the normal boiling point \(\left(113.5^{\circ} \mathrm{C}\right),\) and the critical point \(\left(380^{\circ} \mathrm{C} \text { and } 145 \mathrm{atm}\right) .\) The density of the liquid is less than that of the solid. Label significant data points on this diagram. Are there any features of the diagram that remain uncertain? Explain.

Potassium chloride has the same crystal structure as NaCl. Careful measurement of the internuclear distance between \(K^{+}\) and \(C l^{-}\) ions gave a value of 314.54 pm. The density of \(\mathrm{KCl}\) is \(1.9893 \mathrm{g} / \mathrm{cm}^{3}\). Use these data to evaluate the Avogadro constant, \(N_{\mathrm{A}}\)

Briefly describe each of the following phenomena or methods: (a) capillary action; (b) polymorphism; (c) sublimation; (d) supercooling; (e) determining the freezing point of a liquid from a cooling curve.
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