Question

course name is environmental engineering

TABLE 4 Henry's Law Coefficients, KH (mol/L atm) T (C) 10 15 20 25 CO2 0.076425 0.063532 0.053270 0.045463 0.039172 0.033363 0.0021812 0.0019126 0.0016963 0.0015236 0.0013840 0.0012630 6) Using the Henry's Law Coefficient given in the ENV3001 lexthook, pages 63 to 64, and assuming a concentation o molecular oxygen in air of 21% compute the equilbnum concettratice efonypen in water at 20C and at 25℃ mpl. DO. Make a note of the diffecrence and commest on the significance that you sce in this difsereace units of 7) Browse theough section 5 5 (pages 199-209) Based on the knowledge that the Polyseed used in our lab is "Free e nitrifying microorganisms (manufacturer statement), Explais what do you think are the implications of conducting the est as a BDO-day, rather than BOD-Sdays Answer questions 8 a) and b) by exercising reasoning 8) Last week we set up three "seed contral" boetles by adding to boles "3, 14, and粌espectrvely 10-mL 1SmL, and 20. mL of activaned Polyseed and completing the bonle with BOD dilusion water a) İfthe seed is well activated, the relation of depleted osypen to seed volume should be Incit Using the symbols mondi.JDOe dts, and ΔDOed h show how would you figure out the DO consumpon of 5 mL of the scod

Answer 1

Given :

The volume % of O₂ = 21 % = 0.21 = mole fraction for gases

Henry's law coefficient at 20^oC - 0.0013840 $\frac{mol}{L.atm}$

                                     25^oC - 0.0012630 $\frac{mol}{L.atm}$

Solution :

Using Henry's law which states that at a given constant temperature the concentration of the gas dissolved in a given liquid is proportional to the partial pressure of the gas exerted above the liquid we can solve this problem.

Generally, the equation is given by the formula -

                                                         $K_{H} = \frac{p}{C}$

where,

K_H - Henry's law constant

p- partial pressure of the gas

C- moles of gas per liter of solution (Concentration)

But K_H can also be expressed as C/p which is known as the Henry's law volatility constant. This should be taken care while solving problems since Henry's law volatility constant is the inverse of the Henry's law constant. This should be checked by comparing the units, Henry's constant has the units of (L.atm)/ (mol) and Henry's law volatility constant has units (mol)/(L.atm). But both are referred to as Henry's law coefficient.

Looking at the units given, the constant given is Henry's law volatility constant, so therefore, take the inverse of it. Thus, the value becomes

20^oC-

            $722.54 \frac{L.atm}{mol}$

25^o C -

            $791.76 \frac{L.atm}{mol}$

Since, the partial pressure of O₂ is given by p = mole fraction or volume fraction x Total pressure

                                                                  = 0.21 x 1 atm

                                                                  = 0.21 atm

Therefore at 20^o C, the equilibrium concentration is given by C = p/K_{H =} 0.21/722.54 = 2.91 x 10^-4 mol/L

Now to convert it into mg/L multiply the answer by atomic mass of O₂ i.e 32g and multiply by 1000 to convert it mg. Therefore the final answer is 9.3 mg/L at 20^o C.

Similarly for 25^o C the equilibrium concentration is 8.49 mg/L .

The solubility has decreased by 0.81 mg/L, this implies that as the temperature increases the solubility of O₂ decreases in the water.