Lecture 40 Coexistence Curves Critic Essay

Submitted By Josh-Elson
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Coexistence Curves
• Under a range of temperature and pressure conditions one can have liquid and vapor (gas) in dynamic equilibrium. Gas phase molecules are entering the liquid at the same rate as liquid phase molecules are evaporating. (No heat needs to be supplied for this process to occur. Why?) The simplest cases involve a pure liquid and a pure gas in equilibrium.

Normal boiling point!

Most volatile liquid. Why?

Molecules of liquid “e” have the strongest intermolecular forces. Why?

Least volatile liquid. Why?



FIGURE 12-18

Vapor pressure curves of several liquids
General Chemistry: Chapter 12

Slide 2 of 61

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Calculations Related to Coexistence
Curves
• One can imagine putting a volatile liquid in an empty container and seeing one of two results: • 1. All of the volatile liquid evaporates. In this case PGas will be less than the equilibrium vapor pressure of the substance/liquid.
• 2. Only some of the liquid eventually evaporates. Here Pgas equals the equilibrium vapor pressure.

Class Example:
• 1. A total of 3.00 g of liquid water was placed in an evacuated (initially) 12.0L container maintained at a temperature of 80.0 0C. Will a liquid/gas equilibrium be established? What piece of data is required to solve this problem? If no liquid/gas equilibrium is established, determine how much additional liquid water must be added to have some liquid in equilibrium with H2O(g) in the container. Class Example – Partial Solution::
• If 3.00 g (0.1666 mol) are placed in an evacuated 12.0L container at a temperature of
80.0 0C (353.2 K) we can use PV=nRT to calculate the P due to the gas if all of the liquid water evaporates. We obtain P=40.8 kPa. We can look up the vapor pressure of water at 80.0 0C as 47.4 kPa. This tells us that all of the liquid water has evaporated. Why?

Class Example – Partial Solution::
• Using PV = nRT again we can calculate how much water vapor would produce a gas pressure of 47.4 kPa. We get 0.1937 mol
(3.49g) of water. We must add a minimum of
0.49g of water to establish a liquid/ gas equilibrium. If we add more liquid water the gas pressure will still be 47.4 kPa, the volume of water will increase steadily and the gas volume will be less than 12.0 L.

Critical Points
• The normal boiling point of water is 100 oC.
At this temperature the vapor pressure of water is exactly 760mm Hg (the normal average atmospheric pressure at sea level). If we heat water in a sealed container all of the steam that is formed is trapped above the liquid water. The additional steam formed increases the pressure of the gas above the liquid and “naturally” the boiling point rises.

Critical Points
• There is a limit to how high we can raise the temperature of a liquid/gas mixture and still have a well-defined two phase equilibrium.
One eventually reaches a temperature/pressure point beyond which there is only one fluid phase. The specific T and P where this occurs is referred to as the critical point of the substance.

The Critical Point




FIGURE 12-22

Attainment of the critical point for benzene

In a sealed container, the meniscus separating a liquid from its vapor is just barely visible at the instant the critical point is reached. At the critical point—the liquid and vapor become indistinguishable.
General Chemistry: Chapter 12

Slide 9 of 61

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Canada Inc.

Critical Points
• The next slide shows critical point data for a number of substances. Do the “permanent” and “non-permanent” gases differ appreciably if one considers intermolecular forces?

General Chemistry: Chapter 12

Slide 11 of 61

Copyright © 2011 Pearson
Canada Inc.

Critical Points
• We don’t need a quantitative understanding of the previous slide. The “permanent” gases are those with very weak intermolecular forces. They have both low boiling points and low critical temperatures. “Nonpermanent” gases have much stronger intermolecular forces and, correspondingly, higher…