Mole fraction of n2 Mole fraction of o2 Answers given without the use of the giv

Question

Mole fraction of n2

Mole fraction of o2

Answers given without the use of the given temperatures, pressures, and densities are wrong and have already been tried.

A gaseous mixture consists of 76.0 mole percent N2 and 24.0 mole percent O2 (the approximate composition of air). Suppose water is saturated with the gas mixture at 25°C and 1.00 atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of N2 and O2at 25°C and 1.00 atm are 0.0175 g/L H2O and 0.0393 g/L H2O, respectively.

Explanation / Answer

ASWER:

According to Henrys law

pi = kxi  -----1)

pi = partial pressure of a given gaseoussolute i

xi = mole fraction of i dissolved in solvent (water in our case)

k = Henry constant

pO2 = kO2xO2 and pN2 = kN2xN2 ------2)

By definition partial pressure = mole fractio X Total pressure

Hence

pO2 = yO2P and pN2 = yN2P -----3)

P = total pressure

yO2 = mole fraction of oxygen in air and yN2 = = mole fraction of nitrogen in air

Usig eqn 3) in eqn 2)

yO2P = = kO2xO2 and yN2P = kN2xN2

xO2 = yO2P / kO2  and xN2 = yN2P / kN2 -----4)

Mole fraction of O2 in the air (as per given data) 76 / 100 = 0.76

Mole fraction of N2 in the air (as per given data) 24 / 100 = 0.24

inserting the values in eqn 4)

xO2 = 0.76 X P / kO2  and xN2 = 0.24 X P / kN2

xO2 = 0.76 X 760 / 3.3 X 107 and xN2 = 0.24 X 760 / 6.51 X 107

kO2 = 3.3 X 107 and kN2 = 6.51 X 107

xO2 = 0.0000175 and xN2 = 0.0000028

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