Hi. I'm having some trouble solving this problem. Can you check my work ?
The solubility of N
2 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 N
2 of 4 atm. Say the volume of blood in his body is 5 L. Determine the quantity of N
2 released in litres when the diver comes back to the surface where the partial pressure of N
2 is 0.8 atm.
The solution for this problem is 0.28 dm
3. I got a different result.
First, I calculated Henry's constant for N
2 at 37 ºC. I got K = 7 x 10
-4 mol dm
-3 atm
-1. Then, I determined the solubility of N
2 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 dm
3.