Question

Air is a mixture of gases that is about78.0percent N2 by volume. When air is at standard pressure and 25.0 degreeC, the N2 component will dissolve in water with a solubility of 4.88 x 10^-4M. What is the value of Henry's law constant for N2 under these conditions? Express the constant numerically inmoles per liter per atmosphere. =6.26?10^-4 mol/(L.atm) Correct Part B As a scuba diver descends under water, thepressure increases. At a total air pressure of 2.73 atm and a temperature of 25.0 degreeC, what is the solubility of N2 in a diver's blood? [Use the value of the Henry's lawconstant calculated in Part A, 6.26 x 10^-4mol/(L.atm). Assume that the composition of the air in the tank is the sameas on land and that all of the dissolved nitrogen remains in theblood. Express your answer numerically inmoles per liter. solubility =mol/L

Answer 1

Answer:

$(a) 6.256\cdot 10^{-4} M/atm; (b) 1.33\cdot 10^{-3} M$

Explanation:

(a) According to the Henry's law, the solubility is equal to the product between the Henry's law constant and the partial pressure of a gas:

$S = k_Hp^o$

Air is at standard pressure:

$p = 1.00 atm[/atm]Nitrogen's pecentage is:[tex]\omega = 0.780$

Therefore, its partial pressure is:

$p^o = \omega p$

Solving for the Henry's constant:

$k_H = S/p^o = \frac{S}{\omega p}$

$k_H = \frac{4.88\cdot 10^-4 M}{0.780\cdot 1.00 atm} = 6.256\cdot 10^{-4} M/atm$

(b) Using the constant we've found in the previous part, we know that:

$S = k_H\omega p$

In this case, the percentage is kept the same, however, the total air pressure is:

$p = 2.73 atm$

Substituting the variables:

$S = 6.256\cdot 10^{-4} M/atm \cdot 0.78\cdot 2.73 atm = 1.33\cdot 10^{-3} M$