Occurs when rate of evaporation = rate of condensation. Must have some liquid (or solid for sublimation) present. (Figure 10.39, p 460)
Recall that:
Quantitative variation of vapor pressure with temperature:
Plot (P_{vap} vs. T; upward curve) Figure 10.42, p 462
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Plot (lnP_{vap} vs 1/T; linear with negative slope, T = K) Figure 10.42, p 4462
For the linear plot can find the equation (y = ax + b):
This expression is known as the Clausius-Clapeyron Equation.We can use this equation to find useful information such as the boiling points of liquids at different elevations (and thus pressures).
Example: Find the boiling point of water at 10,000 ft elevation if the atmospheric pressure is 508.4 mmHg. H_{vap} = 4.39 x 10^{4} J/mol.
How do we solve this? If we take the difference between the two situations we get:
ln P_{1} - ln P_{2} = -H_{vap}/R (1/T_{1} - 1/T_{2}) + b - b
reaarranging and recalling that log a - log b = loga/b
ln P_{1}/ P_{2} = H_{vap}/R (1/T_{2} - 1/T_{1})
and
1/T_{2} - 1/T_{1} = (R/H_{vap}) (ln P_{1}/ P_{2})
putting in numbers
1/T_{2} = (8.315 JK^{-1}mol^{-1}/ 4.39 x 10^{4} J/mol) ln (760 mmHg / 508.4 mmHg) + 1/373.15 K
1/T_{2} = 7.62 x 10^{-5} + 2.68 x10^{-3} = 2.76 x 10^{-3}
T_{2} = 362.7 K = 89.7 °C Notice that we can also use the data from vapor pressures (or boiling points) at two pressures to calculate a value for H_{vap}!
A consideration of vapor pressure etc leads to the behaviors of substances with increasing (or decreasing) temperture (see Fig 10.44, p 464 of Zumdahl 5th ed):
- Properties:
- Hard and rigid - they have virtually no tendency to flow or diffuse.
- Nearly incompressible - need to increase pressure about 1,000,000 times to decrease volume by half.
- Very low thermal coefficients of expansion.
- Crystal lattice
- Melting and freezing points are sharp - all units in the interior of a perfect crystal have the same relationships, and therefore the same bonds. Thus when enough energy is added to break the bonds for one unit, there is enough to break bonds with all, so melting is sudden as all the particles break bonds with each other at same temperature and thus same energy.
- heat of fusion/crystallization (saw last time with liquids)
- Nearly all solids expand when they melt (after all the particles are moving faster). As a consequence, nearly all solids will sink in their liquid forms (water is of course a notable exception - we'll look at why later).
- Structure Determination: So how do we know how atoms are arranged in crystals?
- X-ray diffraction is THE tool for crystal structure determination. It gives a full 3-D picture of how atoms are arranged via the interpretation of diffraction patterns. So what kind of information is obtained, and how do we use it to reconstruct a crystal?
- Xray scattering (Figure 10.10, Zumdahl p 433) and diffraction patterns.
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- Bragg Equation: we can determine the distances between layers in a repeating structure using x-rays. The relevant distances can be seen in Figure 10.11, Zumdahl p 433, resulting in the Bragg equation:
n = 2d sin
- STEM: a newer method with equal resolution is Scanning Tunneling Electron Microscopy. This technique give a highly detailed view of the surface of a crystal (or other object). Distances between surface atoms can be calculated etc.
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© R A Paselk
Last modified 11 April 2011