*** Please answer to part (C) *** Problem 3: Osmotic Pressure Seawater has a sal
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*** Please answer to part (C) ***
Problem 3: Osmotic Pressure Seawater has a salinity of 3.5% meaning that if you boil away a kilogram of seawater, there will be 35g of solids (mostly NaCI). When dissolved, the Sodium Chloride dissociates into separate Nat and Cl ions. a) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume that all the dissolved salts in seawater are NaCl. b) Reverse osmosis occurs when you apply a pressure difference greater than the osmotic pressure to a solution separated from a pure solvent by a membrane permeable only to the solvent. In this process, the solvent flows out of the solution and this can be used to desalinate sea water. Use your result from Part (a) to find the minimum work required to desalinate one liter of water. c) It is also possible to desalinate water by evaporation. The water vapor is recondensed and the salt is left behind. Compare your result from part (b) with the amount of heat that would be required to evaporate a liter of water. Assume that the water starts at a temperature of 300 K and the latent heat of vaporization at a pressure of one atmosphere is 40.7kJ/mol. Compare your result with that of Part (b)Explanation / Answer
(c) We are given 1 litre of water. Molar mass of water is 18g. Mass of 1 litre of water= 1000 grams
Hence, number of moles (n) of water in 1 litre is equal to (Mass of 1 litre of water) / (Molar mass of water)
n = (1000/18)= 55.55 moles
Now, given that the latent heat of vapourization (L) = 40.7 kJ/ mol
Hence, amount of heat required to evapourate 1 litre of water Q = nL
Q = nL
Q= (55.55 x 40.7) KJ
Q=2260.88 KJ
Q= 2.2608 MJ
Now compare Q with the work required to desalinate one litre of water, and you will find that the work done in case (b) is lesser than case (c)
I hope I answered the question in a descriptive way.
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