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By the end of this section, you will be able to:
  • Define phase transitions and phase transition temperatures
  • Explain the relation between phase transition temperatures and intermolecular attractive forces
  • Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes

We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.

Vaporization and condensation

When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation    . When the rate of condensation becomes equal to the rate of vaporization    , neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium    , the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure    (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in [link] , and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.

Three images are shown and labeled “a,” “b,” and “c.” Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid. The phrase, “Molecules escape surface and form vapor” is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase “Equilibrium reached, vapor pressure determined,” is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, “Vapor pressure.”
In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.

The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
Syamthanda Reply
hey , can you please explain oxidation reaction & redox ?
Boitumelo Reply
hey , can you please explain oxidation reaction and redox ?
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for grade 12 or grade 11?
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the value of V1 and V2
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advantages of electrons in a circuit
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we're do you find electromagnetism past papers
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it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
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how to calculate coefficient of static friction
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how to calculate static friction
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Tumelo
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How to calculate force
Monambi
a structure of a thermocouple used to measure inner temperature
Anna Reply
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
Amahle Reply
How is energy being used in bonding?
Raymond Reply
what is acceleration
Syamthanda Reply
a rate of change in velocity of an object whith respect to time
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Kidist
Acceleration is a rate of change in velocity.
Justice
t =r×f
Khuthadzo
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Precious Reply
hi
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Leago
use fnet method. how many obects are being calculated ?
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Lungile Reply
you use Fnet equals ma , newtoms second law formula
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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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