pH2 = 66600;   
pO2 = 
      33300;
p = pH2 + pO2;
T = 298;   
nAv = 
      6.022*10^23;
mH2 = 0.002/nAv;   
mO2 = 
      0.032/nAv;
kB = 1.38065*10^-23;   
dH2 = 
      2.72*10^-10;
dO2 = 3.60*10^-10;
sigmaH2 = π*dH2^2;
sigmaO2 = π*dO2^2;

cMeanH2 = (((((8 kB T)/(π mH2))))^0.5)   
cMeanO2 = (((((8 kB T)/(π mO2))))^0.5)

(Subscript[Z, WH2]) = 
      (pH2/(((2  π mH2 kB T))^0.5))  
(Subscript[Z, WO2]) = (pO2/(((2  π mO2 kB T))^0.5))

nH2 = pH2/kB/T  
nO2 = pO2/kB/T

dH2O2 = 0.5*(dH2 + dO2);
μ = mH2*mO2/(mH2 + mO2);
(Subscript[Z, H2O2]) = 
      nH2*nO2*dH2O2^2*(((8 π kB T/μ))^0.5)/
            nAv/1000

OK, so the answer to part (c) is close enough.

(Subscript[Z, H2O2]) = 
      nO2*dH2O2^2*(((((π kB T)/μ)))^0.5)  
(Subscript[Z, O2H2]) = nH2*dH2O2^2*(((((π kB T)/μ)))^0.5)

These answers are different from the provided solution.. Why? (Possible hint: Note that the relevant ratios are the same!?) Please double-check each one of these 'solutions' and be prepared to discuss them in class.

9.258/3.287
4.63/1.6435

LRR3@psu.edu (updated 10/25/2004, 11 pm)