Use Henry\'s law and the solubilities given below to calculate the total volume
ID: 637003 • Letter: U
Question
Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 2.0 L of water upon warming from 25 ?C to 50 ?C. Assume that the water is initially saturated with nitrogen and oxygen gas at 25 ?C and a total pressure of 1.0 atm. Assume that the gas bubbles out at a temperature of 50 ?C. The solubility of oxygen gas at 50 ?C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50 ?C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.
Explanation / Answer
Solution:
Step 1:
Calculation of the Henry’s Law constant at 25 oC
We know
Sgas = kHPgas
Where P is the pressure and KH is the Henry's law constant
Therefore,
KH = 1.3 x 10-3 for Oxygen and KH = 6.1 x 10-4 for Nitrogen
Step 2:
Calculation of dissolved O2 at 25 oC
Given that, Partial pressure for O2 = 0.21 atm
Therefore, Solubility of O2 = 1.3 x 10-3 M/atm x 0.21 atm
= 0.273 x 10-3 M
Thus, the dissolved O2 = 0.273 x 10-3 g/L x 2.0 L = 0.546 x 10-3 g
Step 3:
Calculation of dissolved O2 at 50 oC
We know
Sgas = kHPgas
Or KH = = 27.8 x 10 -3 M/atm
Therefore, Solubility of O2 = 27.8 x 10 -3 M/atm x 0.21 atm
= 5.84 x 10-3 M
Thus, the dissolved O2 at 50 oC = 5.84 x 10-3 g/L x 2.0 L = 11.6 x 10-3 g
Therefore, the difference of oxygen is released = (11.6 x 10-3 g - 0.546 x 10-3 g)
= 11.05 x 10-3 g
The amount of O2 in , moles = 11.05 x 10-3 g x 1.0 mol/32 g= 0.36 x 10-3 mol
From ideal gas equation
PV =nRT
V = n RT/P
= (0.36 x 10-3 mol x 0.0821 Latm/mol.K x 323K)/1 atm
= 9.5x10-3 L
Therefore the volume of Oxygen is 9.5x10-3 L
Step 4:
Calculation of dissolved N2 at 25 oC
Given that, Partial pressure for N2 = 0.78 atm
Therefore, Solubility of N2 = 6.1 x 10-4 x 0.78 atm
= 4.76 x 10-4 M
Thus, the dissolved N2 = 4.76 x 10-4 g/L x 2.0 L = 9.5 x 10-4 g
Step 5:
Calculation of dissolved N2 at 50 oC
We know
Sgas = kHPgas
Or KH = 14.6 mg/L/1.0 atm = 14.6 x 10 -3 M/atm
Therefore, Solubility of N2 = 14.6 x 10 -3 M/atm x 0.78 atm
= 11.38 x 10-3 M
Thus, the dissolved N2 at 50 oC = 11.38 x 10-3 g/L x 2.0 L = 22.77 x 10-3 g =2.27 x 10-4 g
Therefore, the difference of Nitrogen is released = (9.5 x 10-4 g - 2.27 x 10-4 g)
= 7.23 x 10-4 g
The amount of Nitrogen in , moles = 7.23 x 10-4 g x 1.0 mol/28 g = 0.25 x 10-4 mol
From ideal gas equation
PV =nRT
V = n RT/P
= (0.25 x 10-4 mol x 0.0821 Latm/mol.K x 323K)/1 atm
= 6.6x10-4 L
Therefore the volume of Nitrogen is 6.6x10-4 L
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.