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The solubility of gas in a liquid is profoundly affected by pressure. This relat

ID: 920597 • Letter: T

Question

The solubility of gas in a liquid is profoundly affected by pressure. This relation is expressed by Henry's law, which states that the solubility of a gas in a liquid, expressed in moles per liter (or M), at a given temperature is directly proportional to the partial pressure of the gas over the solution. This relation can be expressed mathematically as

solubility=kP

The constant k is characteristic of the specific gas and P is the partial pressure of the gas over the solution usually expressed in atmospheres. The constant k usually has units of moles per liter per atmosphere and is reported at 25 C. This expression can be used to calculate any of the three variables provided that the other two are known for any gas.

As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.69 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood? [Use the value of the Henry's law constant k calculated in Part A, 6.26×104mol/(Latm).]

Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.

Express your answer with the appropriate units

THE ANSWER BELOW IS INCORRECT (computer tells me: Incorrect; Try Again You used the total pressure of air instead of the partial pressure of nitrogen. Air is 78% nitrogen, so N2 only exerts 78% of the total pressure.):

P = 2.69 atm

k = 6.26×104mol/(Latm).

solubility = henry law constant x P

S = K x P

S = 6.26×10^4 x 2.69

S = 1.68 x 10^-3 mol / Litre

solubility = 0.00168 M

Explanation / Answer

we know that

mole fraction = moles / total moles


assuming 100 moles of air

moles of N2 = 78

so

mole fraction of N2 = 78/100

mole fraction of N2 = 0.78

now

we know that

partial pressure = mole fraction x total pressure

so

partial pressure of N2 = 0.78 x 2.69

partial pressure of N2 = 2.0982

now

solubility = k x P

given

henry law constant (k) = 6.26 x 10-4

so

solubility = 6.26 x 10-4 x 2.0982

solubility = 1.31 x 10-3

so

the solubility is 1.31 x 10-3 M