446_01c_test1_soln

446_01c_test1_soln - METR 446 Test # 1 Name: Solution...

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Unformatted text preview: METR 446 Test # 1 Name: Solution September 18, 2001 1) (15 pts) On the axes below draw and label curves representing equilibrium vapor pressure over a pure water droplet and the equilibrium vapor pressure over two solution drops, one that contains twice as much solute as the other. Also, draw a horizontal line corresponding to 0% supersaturation (100% RH). 2) (10 pts) On the axes below, draw a curve showing how you would expect supersaturation to vary with height inside a cloud. Also explain why this curve looks the way it does by discussing the competing sources and sinks of water vapor in a cloud. Radius e hc /e s Cloud base Cloud top Supersaturation (%) | 0% | Peak supersaturation Not many cloud droplets have formed yet, and those that have are still pretty small, so the available surface area for condensation of water vapor is low. The rate at which the saturation vapor pressure decreases due to the decrease in temperature with height exceeds the rate at which condensation can remove water vapor from the air, so the supersaturation increases. Now, the concentration and size of droplets in the cloud has increased to the point that the condensation rate removes water vapor from the air faster than the saturation vapor pressure decreases with decreasing temperature, so the supersaturation decreases. Pure water 1x solute 2x solute 3) (5 pts) In the derivation of Raoults Law (effect of solutes on equilibrium vapor pressure), we looked at the influence of a solute on both condensation and evaporation rates. We found that the equilibrium vapor pressure above a solution is different than that over pure water because...
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446_01c_test1_soln - METR 446 Test # 1 Name: Solution...

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