At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
