Equation 1.4-18 from page 23 of BS&L gives us the viscosity of a gas at low density.

> mu[i]:=2.6693e-5*sqrt(M[i]*T)/(sigma[i]^2*Omegamu[i]);

In order to have units work out properly later, we must define the units of  the constant (2.6693e-5) in this equation.

> constunit:=g/(cm*s)/sqrt(g/mol*K)*A^2;

Temperature as given in problem in Kelvin:

> T:=293*K;

Molecular weights of species:

Index numbers: 1 corresponds to CO2; 2 to O2; 3 to N2

> M[1]:=44.010*g/mol; M[2]:=32.000*g/mol; M[3]:=28.016*g/mol;

Use Table B.1 in the back of BS&L to find the Lennard-Jones parameters, which are sigma and (epsilon/K). Input respective sigmas in Angstroms:

> sigma[1]:=3.996*A; sigma[2]:=3.433*A; sigma[3]:=3.681*A;

Now input (epsilon/K) in units of K:

> eps_K[1]:=190*K; eps_K[2]:=113*K; eps_K[3]:=91.5*K;

Now we will find Omegamu from Table B-2 in the back of BS&L. In order to do that, we need to know the values for (K*T/epsilon). So we take the inverse of the Lennard-Jones parameter we just entered and multiply by the temperature. To evaluate this equation, we  can use a loop in Maple - the syntax for which, although long, is pretty self-explanatory!

> for i from 1 to 3 do KTeps[i]:=evalf(T*(1/eps_K[i])); od;

We can find Omegamu from the second column of Table B.2:

> Omegamu[1]:=1.296; Omegamu[2]:=1.081; Omegamu[3]:=1.022;

Finally, we have everything necessary to compute mu for each species. Again, we use a  loop over the 3 species to calculate:

> for i from 1 to 3 do mu[i]=2.6693e-5*constunit*sqrt(M[i]*T)/(sigma[i]^2*Omegamu[i]); od;

These 3 answers are in agreement with those given in the BS&L problem of:
 mu1 = 1462e-7  g/cm/s
 mu2 = 2031e-7  g/cm/s
 mu3 = 1754e-7  g/cm/s