# Equilibrium Vapor Pressure, cont.

Last time we found the Clausius-Clapeyron Equation described the plot for the variation of vapor pressure with temperature. 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. Hvap = 4.39 x 104 J/mol.

How do we solve this? If we take the difference between the two situations we get:

ln P1 - ln P2 = -Hvap/R (1/T1 - 1/T2) + b - b

reaarranging and recalling that log a - log b = loga/b

ln P1/ P2 = Hvap/R (1/T2 - 1/T1)

and

1/T2 - 1/T1 = (R/Hvap) (ln P1/ P2)

putting in numbers

1/T2 = (8.315 JK-1mol-1/ 4.39 x 104 J/mol) ln (760 mmHg / 508.4 mmHg) + 1/373.15 K

1/T2 = 7.62 x 10-5 + 2.68 x10-3 = 2.76 x 10-3

T2 = 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 Hvap!

## Heating & Cooling Curves

A consideration of vapor pressure etc. leads to the behaviors of substances with increasing (or decreasing) temperature (see Fig 10.42, p 488 of Zumdahl 9th ed):

Note that this curve is completely reversible, that is you can trace it from left to right as a heating curve, or right to left as a cooling curve. Or you can flip the curve around the y-axis to get the cooling curve.

#### Example calculations

How much heat is needed to melt 100.0 g of ice at –10.0°C and get water at 10.0°C?

Cpice = 2.03 J g–1 °C–1, Cpwater = 4.2 J g–1 °C–1, Cpsteam = 2.0 J g–1 °C–1, Hvap = 40.7 kJ/mol, Hfus = 6.02 kJ/mol

This is a four part problem, first need to calculate the heat to raise the temperature of the ice to the melting point, then need to calculate the heat of melting the ice, then calculate the heat needed to raise the liquid to 10.0°C, and finally add the three parts together.

1. (Cpice)(mass)(T) = (2.03 J g–1 °C–1)(100.0 g)(10.0°C) = 2.03 x 103 J = 2.03 kJ

2. (Hfus)(moles water) = (6.02 kJ/mol)(100.0g/18.01g/mol) = 33.45 kJ

3. (Cpwater)(mass)(T) = (4.2 J g–1 °C–1)(100.0 g)(10.0°C) = 4.2 x 103 J = 4.2 kJ

4. Adding the three parts, Hprocess = 2.03 kJ + 33.45 kJ + 4.2 kJ = 39.68 kJ = 39.7 kJ

How much heat is released when 100.0g of 110.0°C steam is condensed to give water at 90.0°C?

For this problem we follow a similar strategy to that above, but backwards.

1. (Cpsteam)(mass)(T) = (2.0 J g–1 °C–1)(100.0 g)(10.0°C) = 2.0 x 103 J = 2.0 kJ

2. (Hvap)(moles water) = (40.7 kJ/mol)(100.0g/18.01g/mol) = 225.99 kJ

3. (Cpwater)(mass)(T) = (4.2 J g–1 °C–1)(100.0 g)(10.0°C) = 4.2 x 103 J = 4.2 kJ

4. Adding the three parts, Hprocess = 2.0 kJ + 225.99 kJ + 4.2 kJ = 232.19 kJ = 232.2 kJ

# Solids and Crystals

Solids: Recall earlier definition - solids have fixed or definite shapes and volumes. By this definition solids are strictly limited to the crystalline solids. (The amorphous (noncrystalline) solids discussed in our text are what we have discussed as supercooled liquids.)
• 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 - due to very open ice structure with each water molecule H-bonded to four others rather than being closs packed to each other.).
• 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, 10.11; Zumdahl p 462) and diffraction patterns.

public domain image via Wikipedia Creative Commons

• 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

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