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Sagot :
Solution :
[tex]$\text{Helium and nitrogen}$[/tex] gases are contained in a conduit [tex]$7 \ mm$[/tex] is diameter and [tex]$0.08 \ m$[/tex] long at 317 K (44°C) and a uniform constant pressure of 1 atm.
Given :
Diameter, D = 7 mm
L = 0.1 m
T = 317 K
[tex]$P_{A1}=0.075 \ atm $[/tex]
[tex]$P_{A2}=0.03 \ atm $[/tex]
P = 1 atm
From, table
[tex]$D_{AB}= 0.687 \times 10^{-4} \ m/s$[/tex]
We know :
[tex]$J_{A}^* = D_{AB} \frac{d_{CA}}{dz}$[/tex]
[tex]$J_A^*=\frac{(0.687 \times 10^{-4})(0.075-0.03)(\frac{101.32}{1 \ atm}) }{8.319 \times 298 \times 0.10}$[/tex]
= [tex]$1.26 \times 10^{-6} \ kgmol/m^r s$[/tex]
[tex]$P_{B1} = P-P_{A1}$[/tex]
= 1 - 0.075
= 0.925 atm
[tex]$P_{B2} = P-P_{A2}$[/tex]
= 1 - 0.03
= 0.97 atm
[tex]$J_B^*=D_{AB}\frac{(P_{B1} \times P_{B2})}{RT( \Delta z)}$[/tex]
[tex]$=\frac{0.687 \times 10^{-4}(0.925-0.97)(\frac{101.32}{1 \ atm})}{8.314 \times 298 \times 0.1}$[/tex]
[tex]$=-1.26 \times 10^{-6} \ kg \ mol /m^r s$[/tex]
Partial pressure of helium [tex]$=\frac{0.075+0.03}{2}$[/tex]
= 0.0525 atm
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