Let\'s do a simplified calculation of dissolved gases in blood plasma when scuba

Question

Let's do a simplified calculation of dissolved gases in blood plasma when scuba diving. When you breathe compressed air from a scuba tank, you arc breathing the air at the same pressure as the surrounding water. Assume that the tank has an air composition of 78.8% N_2 and 21.2% O_2- Henry's Law constants for aqueous solutions at 25 degree C: k_H, N_2 = 1600.0 L middot atm middot mol^-1 and k_H, 0_2 = 756.7 L middot atm middot mol^-1. a. Calculate the solubility of N_2, O_2, and total air at sea level (1.00 atm). b. At a depth of 100 ft, the pressure is 4.00 atm. Calculate the solubility of N_2, O_2, and total air at this depth.

Explanation / Answer

K-N2 = 1600 atm/M

K-O2 = 756.7 atm/M

a)

solubility of O2 and N2 at sea level:

P total = 1 atm

so

P-N2 = x-N2*PT = 0.788*1 = 0.788 atm

P-O2 = x-O2*PT = 0.212*1 = 0.212 atm

so:

[O2] = PO2 /H-O*= 0.212 atm / ( 756.7 amt/M) = = 0.0002801 M

[N2] = H-N2*PN2 = 0.788/1600 = 0.0004925 M

b)

at P = 4 atm (assume this is abs pressure)

P-N2 = x-N2*PT = 0.788*4 = 3.152 atm

P-O2 = x-O2*PT = 0.212*4 = 0.848 atm

[O2] = PO2 /H-O*= 0.848 atm / ( 756.7 amt/M) = = 0.0011206M

[N2] = H-N2*PN2 = 3.152 /1600 = 0.00197 M

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