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.

Answer 1

ASWER:

According to Henrys law

p_i = kx_i  -----1)

p_i = partial pressure of a given gaseoussolute i

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

k = Henry constant

p_O2 = k_O2x_O2 and p_N2 = k_N2x_N2 ------2)

By definition partial pressure = mole fractio X Total pressure

Hence

p_O2 = y_O2P and p_N2 = y_N2P -----3)

P = total pressure

y_O2 = mole fraction of oxygen in air and y_N2 = = mole fraction of nitrogen in air

Usig eqn 3) in eqn 2)

y_O2P = = k_O2x_O2 and y_N2P = k_N2x_N2

x_O2 = y_O2P / k_O2  and x_N2 = y_N2P / k_N2 -----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)

x_O2 = 0.76 X P / k_O2  and x_N2 = 0.24 X P / k_N2

x_O2 = 0.76 X 760 / 3.3 X 10⁷ and x_N2 = 0.24 X 760 / 6.51 X 10⁷

k_O2 = 3.3 X 10⁷ and k_N2 = 6.51 X 10⁷

x_O2 = 0.0000175 and x_N2 = 0.0000028

.