Answer:
D=1.0x10^-5 m^2/s
Explanation:
the data given by the exercise are as follows:
T=308 K
d=1 cm
PA=0.195 atm
pL=0.85 g/cm^3
The expression for binary gas-phase diffusion coefficient is equal to:
[tex]D=\frac{\frac{pL*yB}{MA(\frac{t1^{2}-t2^{2} }{2} } }{c(yA1-yA2}[/tex]
[tex]yB=\frac{1-0.805}{ln(\frac{1}{0.805}) }=0.89[/tex]
[tex]C=\frac{P}{RT}=\frac{1}{82.06*308}=3.95x10^{-5} mol/cm^{2}[/tex]
substituting the values in the diffusion equation:
[tex]D=9.6x10^{-6} m^{2}/s[/tex]
from Appendix J-1 from Welty:
298 K, D=9.62x10^-5 m^2/s
At 308 K, we have the following:
[tex]D=9.62x10^{-5}(\frac{308}{298})^{3/2}=1.0x10^{-5}m^{2}/s[/tex]