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Chapter 23: Problem 2

Sketch the phase diagram for \(\mathrm{I}_{2}\) given the following data: triple point, \(113^{\circ} \mathrm{C}\) and 0.12atm; critical point, \(512^{\circ} \mathrm{C}\) and 116 atm; normal melting point, \(114^{\circ} \mathrm{C} ;\) normal boiling point, \(184^{\circ} \mathrm{C} ;\) and density of liquid \(>\) density of solid.

Short Answer

Expert verified
Sketch involves notable points: triple (113掳C, 0.12 atm), critical (512掳C, 116 atm), and phase boundaries with marked slope due to density.

Step by step solution

01

Understand the Elements of the Phase Diagram

A phase diagram represents the state of a substance (solid, liquid, or gas) at various temperatures and pressures. Key points on the phase diagram include the triple point, critical point, normal melting point, and normal boiling point.
02

Plot the Triple Point

The triple point is given as 113掳C and 0.12 atm. Plot this point on a pressure-temperature graph, marking the intersection where solid, liquid, and gas phases coexist.
03

Plot the Critical Point

The critical point is at 512掳C and 116 atm, signifying the end point of the gas-liquid boundary. Mark this point on the graph to define the gas region.
04

Indicate the Normal Melting Point

The normal melting point is 114掳C at 1 atm, slightly above the triple point temperature. Place a point at this coordinate, highlighting the transition from solid to liquid phase.
05

Indicate the Normal Boiling Point

Normal boiling occurs at 184掳C and 1 atm. Position this point on the graph to depict the shift from liquid to gas under atmospheric pressure.
06

Sketch the Phase Boundaries

Connect the triple point to the normal melting point vertically to form the solid-liquid line. Extend this to the critical point to depict the liquid-gas boundary. The solid-gas line runs from the triple point to lower temperatures and pressures. Ensure the boundaries reflect the density information: since liquid density > solid density, expect liquid to have a steeper slope.

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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 unique condition on a phase diagram where the three phases 鈥� solid, liquid, and gas 鈥� coexist in equilibrium. In the case of iodine (\( \mathrm{I}_2 \)), this occurs at a temperature of 113掳C and a pressure of 0.12 atm. At this junction, any change in pressure or temperature will shift the substance into either the solid, liquid, or gaseous state, thereby breaking the equilibrium. This point is crucial for defining the specific characteristics of a material's phase behavior. It's a key reference for understanding the conditions under which different phases appear and transition.
Critical Point
The critical point represents the highest temperature and pressure at which a liquid and its vapor can coexist. Above this point, the distinction between liquid and gas phases ceases to exist. For iodine (\( \mathrm{I}_2 \)), the critical point is at 512掳C and 116 atm. This marks the endpoint of the liquid-vapor boundary on the phase diagram. Beyond this point, iodine enters a supercritical fluid state, possessing properties of both liquid and gas. Understanding the critical point helps in determining the limits of phase separation and the behavior of the substance under extreme conditions.
Normal Melting Point
The normal melting point is the temperature at which a substance transitions from solid to liquid under standard atmospheric pressure (1 atm). For iodine (\( \mathrm{I}_2 \)), the normal melting point is 114掳C. It is essential to note that this point is slightly above the triple point temperature. Understanding the normal melting point involves recognizing the conditions under which a solid becomes a liquid without a change in pressure. This helps in predicting the behavior of substances under everyday conditions and designing processes that require precise temperature control.
Normal Boiling Point
The normal boiling point is defined as the temperature at which a substance converts from liquid to gas at 1 atm pressure. In the case of iodine (\( \mathrm{I}_2 \)), this occurs at 184掳C. The normal boiling point is a critical aspect for industrial and laboratory applications, dictating the conditions needed for a substance to evaporate. It helps in identifying the energy required to induce a phase shift from liquid to gas under normal atmospheric pressure. Knowledge of the normal boiling point facilitates understanding evaporation rates and the heating processes necessary for vaporization.

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Most popular questions from this chapter

Use the following data for methanol at one atm to plot \(\bar{G}\) versus \(T\) around the normal boiling point \((337.668 \mathrm{K}) .\) What is the value of \(\Delta_{\mathrm{vap}} \bar{H} ?\) $$\begin{array}{lcc}T / \mathrm{K} & \bar{H} / \mathrm{kJ} \cdot \mathrm{mol}^{-1} & \bar{S} / \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \\\\\hline 240 & 4.7183 & 112.259 \\\280 & 7.7071 & 123.870 \\\300 & 9.3082 & 129.375 \\\320 & 10.9933 & 134.756\\\330 & 11.8671 & 137.412 \\\337.668 & 12.5509 & 139.437 \\\337.668 & 47.8100 & 243.856 \\\350 & 48.5113 & 245.937 \\\360 &49.0631 & 247.492 \\\380 & 50.1458 & 250.419 \\\400 & 51.2257 & 253.189\end{array}$$

When we refer to the equilibrium vapor pressure of a liquid, we tacitly assume that some of the liquid has evaporated into a vacuum and that equilibrium is then achieved. Suppose, however, that we are able by some means to exert an additional pressure on the surface of the liquid. One way to do this is to introduce an insoluble, inert gas into the space above the liquid. In this problem, we will investigate how the equilibrium vapor pressure of a liquid depends upon the total pressure exerted on it. Consider a liquid and a vapor in equilibrium with each other, so that \(\mu^{\prime}=\mu^{8}\). Show that \(\bar{V}^{1} d P^{1}=\bar{V}^{g} d P^{8}\) because the two phases are at the same temperature. Assuming that the vapor may be treated as an ideal gas and that \(\bar{V}^{1}\) does not vary appreciably with pressure, show that \(\ln \frac{P^{\S}\left(\text { at } P^{1}=P\right)}{P^{8}\left(\text { at } P^{1}=0\right)}=\frac{\bar{V}^{1} P^{1}}{R T}\) Use this equation to calculate the vapor pressure of water at a total pressure of 10.0 atm at \(25^{\circ} \mathrm{C} .\) Take \(P^{\mathrm{g}}\left(\text { at } P^{1}=0\right)=0.0313 \mathrm{atm}\).

Determine the value of \(d T / d P\) for water at its normal boiling point of \(373.15 \mathrm{K}\) given that the molar enthalpy of vaporization is \(40.65 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), and the densities of the liquid and vapor are \(0.9584 \mathrm{g} \cdot \mathrm{mL}^{-1}\) and \(0.6010 \mathrm{g} \cdot \mathrm{L}^{-1}\), respectively. Estimate the boiling point of water at 2 atm.

Fit the following vapor pressure data of ice to an equation of the form \(\ln P=-\frac{a}{T}+b \ln T+c T\) where \(T\) is temperature in kelvins. Use your result to determine the molar enthalpy of sublimation of ice at \(0^{\circ} \mathrm{C}\). $$\begin{array}{cccc}t /^{\circ} \mathrm{C} & P / \text { torr } & t /^{\circ} \mathrm{C} & P / \text { torr } \\\\\hline-10.0 & 1.950 & -4.8 &3.065 \\\\-9.6 & 2.021 & -4.4 & 3.171 \\\\-9.2 & 2.093 & -4.0 & 3.280 \\\\-8.8 & 2.168 & -3.6 & 3.393 \\\\-8.4 & 2.246 & -3.2 & 3.509 \\\\-8.0 &2.326 & -2.8 & 3.360 \\\\-7.6 & 2.408 & -2.4 & 3.753 \\\\-7.2 & 2.493 & -2.0 & 3.880 \\\\-6.8 & 2.581 & -1.6 & 4.012 \\\\-6.4 & 2.672 & -1.2 &4.147 \\\\-6.0 & 2.765 & -0.8 & 4.287 \\\\-5.6 & 2.862 & -0.4 & 4.431 \\\\-5.2 & 2.962 & 0.0 & 4.579\end{array}$$

In this problem, we will show that the vapor pressure of a droplet is not the same as the vapor pressure of a relatively large body of liquid. Consider a spherical droplet of liquid of radius \(r\) in equilibrium with a vapor at a pressure \(P,\) and a flat surface of the same liquid in equilibrium with a vapor at a pressure \(P_{0} .\) Show that the change in Gibbs energy for the isothermal transfer of \(d n\) moles of the liquid from the flat surface to the droplet is \(d G=d n R T \ln \frac{P}{P_{0}}\) This change in Gibbs energy is due to the change in surface energy of the droplet (the change in surface energy of the large, flat surface is negligible). Show that \(d n R T \ln \frac{P}{P_{0}}=\gamma d A\) where \(\gamma\) is the surface tension of the liquid and \(d A\) is the change in the surface area of a droplet. Assuming the droplet is spherical, show that \(\begin{aligned} d n &=\frac{4 \pi r^{2} d r}{\bar{V}^{1}} \\ d A &=8 \pi r d r \end{aligned}\) and finally that \(\ln \frac{P}{P_{0}}=\frac{2 \gamma \bar{V}^{1}}{r R T}\) Because the right side is positive, we see that the vapor pressure of a droplet is greater than that of a planar surface. What if \(r \rightarrow \infty ?\)
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