Hi. I'm having some trouble solving this problem. Can you check my work ?
The solubility of N2
in blood at 37 ºC and at a partial pressure of 0.8 atm is 5.6 x 10 -4
mol/L. A diver breathes compressed air with a partial pressure of N2
of 4 atm. Say the volume of blood in his body is 5 L. Determine the quantity of N2
released in litres when the diver comes back to the surface where the partial pressure of N2
is 0.8 atm.
The solution for this problem is 0.28 dm3
. I got a different result.
First, I calculated Henry's constant for N2
at 37 ºC. I got K = 7 x 10 -4
. Then, I determined the solubility of N2
at a partial pressure of 4 atm using the value I got earlier for K. The solubility I got was 2,8 x 10 -3
mol/L. After that, I calculated the nº of moles for each solubility. For a solubility of 5.6 x 10 -4
mol/L the nº of moles is 2,8 x 10 -3
mol and for a solubility of 2,8 x 10 -3 the nº of moles is 0,014 mol. Then I subtracted 0,014 mol by 2,8 x 10-3
mol and got 0,0112 mol. Finally, I applied the equation pV = nRT to determine the value of V using 0,8 atm as the value for pressure. I got V = 0,38 dm3