Help for a medal! In the diagram, the solid eventually becomes a gas. What will happen to bring the substance from a solid to a gas? Energy will be added. The substance will lose pressure. The substance will lose volume. Energy will be taken away. http://takeexam.next.ecollege.com/ec/courses/46122/CRS-SCI102B-3950282/temperature_pic.gif
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Vaporization is the process of converting a liquid into a gas. It is also called evaporation. Since we know that the particles of a gas are moving faster than those of a liquid, an input of energy must be required for a liquid to become a gas. The most common way to add energy to a liquid system is by adding heat. Evaporation As a liquid gains energy, the molecules begin to move around faster. If a molecule is on the surface of the liquid, and has enough energy, it can break free and become a gas molecule. As with anything in chemistry, or life for that matter, there are other factors that determine how easily a molecule can break free from the liquid. We just discussed some of them: intermolecular forces. The stronger the intermolecular forces that are holding a liquid together, the more energy that will be required to pull them apart. What this means in practicle terms is that a liquid with strong intermolecular forces will have to be heated to a higher temperature before it will evaporate. For example: Look at Methane (CH4 M.W. 16 g/mol) and Water (H2O 18 g/mol). Their molecular weights are very similar, but their Heats of Vaporization (how much heat per mole that has to be added to make them evaporate) are very different. Water has a DHvap of 40.7 kJ/mol and Methane has a DHvap of 8.2 kJ/mol. Methane is actually a gas at room temperature because of its low heat of vaporization.
Well, the opposite of vaporization, of course. The process by which a gas changes phases into a liqud. And if you must add energy/heat a liquid to convert it to a gas, then you must remove energy from or cool a gas to convert it into a liquid. Make sense? Yes. What else makes sense is that the amount of energy required to go from gas to liquid phase would be the same as that required to go from liquid to gas, just opposite in sign. So the Heat of Vaporization is the same for both processes, just positive (endogonic/endothermic) for evaporation and negative (exergonic/exothermic) for condensation. Heat of Vaporization Another property that affects the value of the DHvap is the molecular weight or size of the molecule. This is generally because you have more surface (London Force) interactions between larger molecules than small and the density of a larger molecule is generally greater than that of a small molecule. Remember, that to become airborne a molecule must have a lower density than the air molecules around it. So as a general trend, as the MW/size of a molecule increases so will the DHvap. Exceptions in the trend will occur when a small molecule is capable of a stronger intermolecular force, E.g. HF versus HBr. HF hydrogen bonds and thus has a higher DHvap. So now that you know all about what a DHvap is, why the heck do we care about it? Well, here is a real life use for this value that is important to us all: fingernail polish drying times. For fingernail polishes, if you want the polish to dry faster, you use a solvent that has a lower DHvap. For example: Table 2 lits some of the more common solvents used in fingernail polishes. Methyl Acetate is the preferred solvent for most. This is because not only is it fast drying, it is also somewhat water soluble.
Another simple example of how to use the value is determing how much heat it will take to boil water out of a pot? The DHvap for water is 40.7 kJ/mol. If your pot contains 2.2 liters of water and we assume the density of the water is ~1g/mL then the mass of the water in the pot is 2.2 x 103 grams. The molecular weight of water is 18.0 g/mol so this means we have 2.2 x 103 grams/18.0g/mol = 122.2 moles of water. We can now use the 40.7 kJ/mol factor to show that it will take 122.2 mol x 40.7 kJ/mol = 4974.4 kJ of energy to evaporate all of the water in the pot. What if we take it one step further? Let's combine some concepts we learned in CHM1045 with the concept we just discussed and solve the following problem: How much heat energy is required to convert 5.00 kg of ice at 0°C, into 5.00 kg of steam at 100.°C?
What do we need to calculate? First we need convert the ice from solid to liquid using the Enthalpy of Fusion (a value just like the Heat of Vaporization that measures the amount of heat required to convert a substance from solid to liquid) and then we need to heat the material from 0 up to 100 degrees and finally, we need to convert it from liquid to gas using the Heat of Vaporization. So now we know the three steps to take, we need to collect our conversion factors to use: Step 1: Melt the solid ice, for this we need to know the DHfus of the ice which is 335 kJ/kg. So we can simply multiply the number of kg in our Step 2: We need to heat the liquid water from 0 to 100 degrees Celsius. The energy (q) required to do this can be calculated using q= mcDT, where m is the mass of the water, DT is the Change in temperatue and c is the specific heat of water (a CHM1045 concept that descibes how much heat energy a substance can absorb before it rises in temperature one degree). The specific heat of water is 4.19 J/goC . Step 3: Convert the liquid water into vapor at 100 oC. For this we use the Heat of Vaporization as we did above. (Note: If you convert the units of the DHVap from kJ/mol to kJ/kg you get ~2300 kJ/kg)
Vapor pressure is a measurable quantity that exists when a liquid and its vapor are in equilibrium. This is only possible in closed systems. But please note that the earth's atmosphere is considered a closed system. On a much smaller scale this would be when you place any liquid into a sealed container. The molecules at the surface of the liquid would be changing into gas phase molecules and returning to the liquid until an equilibrium was reached. This is what we call a "dynamic equilibrium". As molecules from the liquid move into the gas phase within the container, this increases the pressure above the liquid. A measure of this pressure (minus the normal atmospheric pressure) gives us the vapor pressure of the liquid. The higher this pressure, the more volatile a liquid is said to be. Vapor pressure is also dependent on the temperature of the system. The liquid will still have a heat of vaporization to contend with so the equilibrium vapor pressure will increase at higher temperatures.( And anyone who has ever microwaved their lunch in a closed tuperware container knows what will happen if the pressure or temperature gets too high, don't they?)
You are going away for the weekend and you want to make sure you leave enough water in the cage for your bird so it doesn't die of thirst while you are gone. If your dorm room has an average temperature of 21oC how many liters of water should you put out to be sure some remains for the entire weekend? Well, at 21oC the vapor pressure of water is ~18 mmHg. If we assume your dorm room has a volume of about 5.0 x 104 L, we can use the ideal gas law to determine how much water will evaporate and just make sure more than that is available for the bird. Water Calculation 49.09 moles of water is only about 883 g of water or ~883 mL so if we leave anything in excess of this for the bird it won't evaporate and the bird will be fine! A couple of things to note about the calculation above: 1) If the room were bigger, more water would need to evaporate before the equilibrium vapor pressure was met and 2) if the temperature was higher in the room the equilibrium vapor pressure would be higher as well. Vapor pressure is temperature dependent! So for practical use, we would need to know how the vapor pressure changed with temperature (that way you can leave your bird at someone else's house, yeah! even if their place doesn't have air conditioning). Unfortunately, we quickly see that most vapor pressure relationships with temperature are not linear, which makes predictions difficult if not impossible. Boo!
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