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Calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm (a typical value at high altitude). Atmospheric Gas Mole Fraction kH mol/(L*atm) N2 7.81 x 10-1 6.70 x 10-4 O2 2.10 x 10-1 1.30 x 10-3 Ar 9.34 x 10-3 1.40 x 10-3 CO2 3.33 x 10-4 3.50 x 10-2 CH4 2.00 x 10-6 1.40 x 10-3 H2 5.00 x 10-7 7.80 x 10-4

Respuesta :

Answer : The solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]

Explanation :

First we have to calculate the partial pressure of nitrogen.

Formula used :

[tex]p_{N_2}=X_{N_2}\times P_{atm}[/tex]

where,

[tex]p_{N_2}[/tex] = partial pressure of nitrogen = ?

[tex]X_{N_2}[/tex] = mole fraction of nitrogen = [tex]7.81\times 10^{-1}[/tex]

[tex]p_{atm}[/tex] = atmospheric pressure = 0.480 atm

Now put all the given values in the above formula, we get :

[tex]p_{N_2}=7.81\times 10^{-1}\times 0.480 atm[/tex]

[tex]p_{N_2}=0.375atm[/tex]

Now we have to calculate the solubility of nitrogen in water.

Formula used :

[tex]s_{N_2}=p_{N_2}\times K_H[/tex]

where,

[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 0.375 atm

[tex]s_{N_2}[/tex] = solubility of nitrogen in water = ?

[tex]K_H[/tex] = Henry's constant = [tex]6.70\times 10^{-4}mole/L.atm[/tex]

Now put all the given values in the above formula, we get :

[tex]s_{N_2}=0.375atm\times 6.70\times 10^{-4}mole/L.atm[/tex]

[tex]s_{N_2}=2.5125\times 10^{-4}mole/L[/tex]

Therefore, the solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]

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