Example 18.2-2 Determination of Diffusivity
>  restart;
>  z2:=z1+17.1*cm;
>  Ptot:=755*mmHg;
>  T:=273.15*K;
From the ceng301 data and the function vappr, we get for the vapor
pressure of CCl4:
vappr(273.15)
ans =
5.5562 <-- 41.67mmHg?? The text gives 33 mmHg.
>  vapP:=5.5562*kPa;
>  mmHg:=kPa*101.325/760;xA1:=vapP/Ptot;
>  pB1:=(1-xA1)*Ptot;pB2:=Ptot;
>  S:=0.82*cm^2;evaprA:=.00208*cm^3/hr;
>  SGCCl4:=1.595; From Felder & Rousseau, Table B.2
>  rhoA:=SGCCl4*g/cm^3;
>  MWA:=153.82*g/mol; From Appendix E.1 of BSL
>  hr:=3600*second;NA:=evaprA*rhoA/MWA/S;
>  c:=Ptot/Rgas/T;
>  Rgas:=82.0578*cm^3*atm/mol/K;
>  atm:=101.325*kPa;c;
>  DAB:=NA*(z2-z1)/c/log(1/(1-xA1)); From eq. 18.2-14 using xA2=0.

> 
The book found a higher value due to the lower vapor pressure of CCl4 at 0C.