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At a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:
S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
At 0.890 atm,
S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows
3.30 L x (0.000273 mol / L) = 0.0012012 mol
3.30 L x (0.00024297 mol / L) = 0.001069068 mol
0.0012012 mol - 0.001069068 mol = 0.000132 mol
S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
At 0.890 atm,
S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows
3.30 L x (0.000273 mol / L) = 0.0012012 mol
3.30 L x (0.00024297 mol / L) = 0.001069068 mol
0.0012012 mol - 0.001069068 mol = 0.000132 mol
0.448g of oxygen is released from the container
Data;
- p1 = 1.0atm
- v1 = 3.30L
- p2 = 0.896 atm
- T = 298K
Ideal Gas Equation
Using the Ideal gas equation, let's make the number of moles the subject.
[tex]pv= nRT\\n = \frac{PV}{RT}[/tex]
For n1
[tex]n_1 = \frac{p_1v}{RT}\\ n_1 = \frac{1.0*3.3}{0.082*298}\\ n_1 = 0.135 mol[/tex]
For n2
[tex]n_2 = \frac{p_2v}{RT} \\n_2 = \frac{0.896*3.3}{0.082*298}=0.121mol[/tex]
The amount of oxygen released is the difference between n1 and n2
[tex]n_1 - n_2 = 0.135-0.121=0.014mol[/tex]
The amount of oxygen released in the sealed container is 0.014moles.
We can convert this into grams.
[tex]number of moles = \frac{mass}{molar mass}\\ 0.014= mass /32\\mass = 0.014 * 32\\mass = 0.448g[/tex]
from the calculations above, 0.448g of oxygen is released from the container.
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