AOS145_HW02_KEY - AOS 145 Homework#2 3 QUESTIONS(10 points...

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Unformatted text preview: AOS 145 Homework #2 3 QUESTIONS (10 points total) 1.​ a) [1 point] Use the Clausius­Clapeyron equation and the water vapor pressure at 0°C of 6.1129 mbar (1 mbar = 1 hPa = 100 Pa) to calculate the saturation vapor pressure of water at 5, 10, 15, 25, and 30°C. The temperature ​T ​ in the Clausius­Clapeyron is in unit K. So, be sure to convert the temperatures from °C to °K before plugging them in. ∆​Hv​​ = 4.5×10​4​ J mol​­1​ (at 0°C) is the vaporization heat (of water) on a molar basis. b) [1 point] How well do the calculated values compare to the attached list of measured values from the Handbook of Chemistry and Physics attached below? (SVP short for saturation vapor pressure). The reason that your results are slightly different from those given in the Handbook is because a single value of ∆​Hv​​ for all temperatures. ∆​Hv​​ actually varies with temperature. c) [1 point] Make a plot of the water saturation vapor pressure against temperature from 0 to 30°C based on your calculation. 2.​ [4 points] You dissolve 10 g of NaCl in a glass containing 1000 cm​­3​ of water. The glass is in a room with temperature of T = 20°C and a constant relative humidity of RH = 97%. Calculate the volume of the solution that will be left in the glass after several days. The key is to understand that after several days of evaporation, an equilibrium reaches between the solution (with a flat surface) in the glass and the atmosphere. In other words, the water saturation vapor pressure over the flat surface of the solution remaining in the glass should be equal to the ambient water vapor pressure: On the one hand, the water saturation vapor pressure over the flat surface of the solution left in the glass is related to the water saturation vapor pressure over a flat surface of pure water via Raoult's law: or equivalently, On the other hand, by definition, relative humidity in the room is Now it is clear that at equilibrium, X​H2O​ = RH = 0.97 How to calculate X​H2O​? Remember from the example given during lecture that 1000 cm​3​ water is 1000 g and contains 55.5 mol of water molecules. Assume that the final volume of water in the glass is V, then the corresponding water molecules is 55.5 mol × V/1000 = V/18 mol (This is just simple proportionality: since there are 55.5 moles of water molecules in 1000 cm​ water, then how many moles of water molecules are in V volume of water?) Also from the same example during lecture, 1 g NaCl dissolved in water results in 2 × 3​ 0.017 moles of ions (NaCl → Na​+​ + Cl​­​). For 10 g of NaCl, the corresponding number of moles of ions are 10 × 2 × 0.017 moles = 0.34 moles Hence the mol fraction of H​ O in the solution remaining in the glass at equilibrium is 2​ which gives V ≈ 197 cm​3​. 3.​ [3 points] For a drop of pure water to be formed from water vapor (without the help of a condensation nuclei or other particles) it has to accumulate water molecules starting at clusters that contain less than 10 water molecules. The saturation ratio in the atmosphere rarely exceeds 1.01. How likely do you think it is to find a droplet of pure water large enough to exceed the critical radius at this saturation? Support your conclusions with an estimate using the Kelvin equation. First, let's assume an atmospheric temperature of 25°C – a reasonable assumption in the real atmosphere. At ​T ​ = 25°C = 298 K and ​S​ = 1.01, one can easily calculate the critical radius (Keep in mind that 1 J = 1 N m​­1​. The units work out to be ​µ​m). Next, let's figure out the radius of a cluster of 10 water molecules, assuming the cluster assumes a spherical shape. Start with the density of water, 1 g cm​­3​. 1 g of water contains 1g / (18 g mol​­1​) moles of water molecules, which upon multiplication with Avogadro's number (6.022×10​23​ molecules mol​­1​), gives you the total number of water molecules in 1 g of water, 3.34×10​22​ molecules. These molecules occupy a volume of 1 cm​3​ (again, just from the density of water, 1 g cm​­3​) – now one can compute the volume occupied by a single water molecule in liquid phase as 1 cm​3​ divided by 3.34×10​22​ – the resulting volume is 3×10​­23​ cm​3​. Thus the volume of a cluster with 10 water molecules is 3×10​­22​ cm​3​. From the simple relationship of V = (4/3)π​Rp​​ 3​, you can calculate the corresponding radius of the cluster: ​Rp​​ = 0.0042 ​µ​m – far smaller than the required critical radius. The conclusion: it is unlikely to find a droplet of pure water large enough to exceed the critical radius at the saturation ​S​ = 1.01. ...
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