Air is a mixture gases. It contains 20% by volume of O2 gas and 79% by volume of N2 gas at 298K. Water is in quilibrium with air at a pressure of 10 atmospheres. KH for O2 and N2 at 298 K are 3.30x107mm and 6.51x 107 mm respectively. Determine the composition of these gases in water.

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Answers

2015-04-25T09:09:02+05:30
The partial pressure of N2 must be
(0.79 *(760*0.6*23.76)mm hg
and the partial pressure of O2 must be
(0.21*(760-0.6*23.70) mm hg
0
2015-04-25T12:28:34+05:30

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Total pressure of air above water = P = 10 atm.
partial pressure of O2 in air above water  = P1
partial pressure of N2 above water = P2

Volume of air above water = V
Volume of O2 = V1 = 0.20 V
Volume of N2 = V2 = 0.79 V
let n1 moles of Oxygen and n2 moles of N2 be present in air in volume V.
Let n  = total number of moles of gases in air

From Daltons' law of partial pressures and the law of ideal gas mixture:

\frac{V1}{V}= \frac{P1}{P}=\frac{n1}{n}\\\\\frac{V2}{V}=\frac{P2}{P}=\frac{n2}{n}\\\\P1=\frac{V1}{V}*P=0.20*10=2.0\ atm\\P2=P*\frac{V2}{V}=0.79*10=7.9\ atm

            
Henry's law for solubility :    p=k_H\ c
  where  p = partial presssure of the gas above the liquid in equilibrium with solution
     k_H  Henry's constant
     c = molar concentration of the gas in moles/Litre in the solution
     k_H_1=\frac{3.30*10^7}{760}\ =4.34*10^4\ atm-Litre/moles
     k_H_2=\frac{6.51*10^7}{760}\ = 8.565*10^4\ atm-litre/mol

 c1 = concentration of O2 in water
     = \frac{P1}{k_H_1}=\frac{2.0}{4.34*10^4}=4.606*10^{-5} mol/Litre
     = 4.606 * 32 * 10⁻⁵ gm/Litre = 1.474 mg/kg of water
     =  0.1474 % of water by mass  or  8.291 * 10⁻⁵  moles/100 moles of water

c2 = concentration of N2 in water
   = \frac{P2}{k_H_2}=\frac{7.9}{8.565*10^4}=9.22*10^{-5} mol/Litre
   = 2.582 mg/Litre = 0.2582 % of water by mass  = 16.6*  10⁻⁵ moles/100 moles of water

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