The mole fraction of gas in solution is related to the partial pressure of the g

Question

The mole fraction of gas in solution is related to the partial pressure of the gas above the solution by P_B = XBkB where k_B is the Henry's Law constant of the solute. Calculate the ratio of solubility of N_2:0_2 in water at 100degreeC, 1 atm, given kB (in atm) = 12.6 x 10^4 for N2 and 7.01 x 10^4 for O2. [You may assume that N2 is 80% and O2 is 20% of air.] Prepare a plot of the partial pressures and the total vapour pressure of the solution as a function of concentration for the CH_3OH-H-2O system at 40degreeC. Use the following table: Does this system (CH_3OH, H_2O, overall) show positive or negative deviations from ideality?) Calculate the activity coefficients of CH3OH and H2O at x_meth = 0.211

Explanation / Answer

a) Given that pressure = 1 atm

K (N2) at 100oC = 12.6×104 atm

K (O2) at 100oC = 7.1×104 atm

Mole fraction of O2 in air = 0.20

Mole fraction of N2 in air = 0.80

Molar Wt of O2 = 31.9988 g/mol

Molar Wt of N2 = 28.0134 g/mol

Therefore Co = [ (1 atm ×0.20)×31.99 g/mol]/7.01×104 atm/(mol/lit)

                       = 6.398/(7.01×104) g/lit

                      Co = 0.911×10-4 g/lit

                     CN = [ (1 atm ×0.80)×28.0134 g/mol]/12.6×104 atm/(mol/lit)

                                = 1.7786×10-4 g/lit

Hence N2:O2 = 1.7786×10-4 g/lit: 0.911×10-4 g/lit

                                = 1.94

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