One of the detrimental effects of the \"thermal pollution\" of water supplies is

Question

One of the detrimental effects of the "thermal pollution" of water supplies is that a rise in temperature reduces the amount of dissolved oxygen available to fish. Using the information in the table of Henry's Law constants (Ku) below, calculate how many litres of water a fish requires at 30.0°c to obtain the same amount of oxygen that t could obtain from 4.71 L of water at 25.o°c TABLE HENRY'S LAW CONSTANTS (KH) IN AQUEOUS SOLUTION (M/Bar) Gas 0.0 C 1.1 x 10 2.5 x 103 1.6 × 10-3 2.5 × 10.3 4.1 × 10 7.7 × 10-2 25 °C N2 0, co Ar He CO2 6.6× 104 1.3 × 10°, 9.5 × 10-4 1.5× 10-3 38× 10-4 3.4 × 10-2 30°C 40 × 10-4 8.8 x 10-4 4.4 × 10-4 1.0× 10-3 38 × 10.4 1.6 × 10-2

Explanation / Answer

find volume of water

T = 30°C V water

for

T = 25°C V water = 4.71 L

then...

P-O2 = 0.21 atm = 0.2127 bar

[O2] = K*P-O2 = (8.8*10^-4) * 0.2127 = 0.000187176 M

this is per liter

total O2 mol = MV = 4.71*0.000187176 = 0.00088159 mol of O2

then.. at T = 30°C

n = 0.00088159

[O2] = K*P-O2

find [O2] at 30°C

[O2] = (4*10^-4)(0.2127 ) = 0.00008508

V = mol/M = 0.00088159/0.00008508

V = 10.361 Liters required!

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