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Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.

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We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."

Answers:

  • Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex]
  • Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]

From the question we are told

The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s

 

A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube

Therefore, the mass flow rate of CCI_4 at point A

=  [tex]5.86*10^{-13} kg/s[/tex]

B) From Fick's law

[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]

Then,

[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]

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