Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions

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.



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 0 0

This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

1 5 1
click on thanks button (Azur blue) above pls
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts