Question

At ${37}^{\circ }C$ and 0.80 atm partial pressure, the solubility of $N_2$ was found to be $5.6\times {10}^{-4}mol{L}^{-1}$. A deep sea diver breathes compressed air where the partial pressure of $N_2$ is 4.0 atm. The total volume of blood in his body is 5.0 L. After sometime he comes back to the surface of the water where the total pressure of $N_2$ is 0.80 atm. If the volume of $N_2$ that escapes during his return to the surface, at ${37}^{\circ }C$ and 1 atm, is x, find the value of x (two decimal places).

Answer 1

By Henry's Law: $K_H=\frac{p}{C_w}$
Henry's law constant for nitrogen would be:
$K_H=\frac{0.8}{5.6\times {10}^{-4}}$
Solubility of nitrogen at 4 atm $=\frac{4}{K_H}$.
Moles of $N_2$ dissolved at 4 atm in $5L$ blood:
$=5.6\times {10}^{-4}\times \frac{4}{0.8}\times 5=5.6\times 5\times 5\times {10}^{-4}mols$... (i)
Moles of $N_2$ dissolved on return to the surface - 0.8 atm in $5L$ $=5.6\times 5\times {10}^{-4}mols$... (ii)
Moles escaped (i) - (ii): $5.6\times 4\times 5\times {10}^{-4}mols=112\times {10}^{-4}moles$
Vol. of $N_2$ (X) at ${37}^{\circ }C$ escaped at 1 atm (using the ideal gas law) = $112\times {10}^{-4}\times 0.082\times 310=0.2847L$