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

Answer : The solubility of nitrogen in water at an atmospheric pressure will be,

Explanation :

First we have to calculate the partial pressure of nitrogen.

Formula used :

where,

= partial pressure of nitrogen = ?

= mole fraction of nitrogen =

= atmospheric pressure = 0.480 atm

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

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

Formula used :

where,

= partial pressure of nitrogen = 0.375 atm

= solubility of nitrogen in water = ?

= Henry's constant =

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

Therefore, the solubility of nitrogen in water at an atmospheric pressure will be,

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer would be c