Helium and air are contained in a conduit 7 mm in diameter and 0.08 m long at 44 deg C and 1 atm. The partial pressure of helium at one end

Question

Helium and air are contained in a conduit 7 mm in diameter and 0.08 m long at 44 deg C and 1 atm. The partial pressure of helium at one end of the tube is 0.075 atm and at the other end is 0.03 atm. Calculate the following for steady state equimolar counter diffusion. (a) Molar flux of He, (b) Molar flux of air, and (c) Partial pressure of helium at half way point of the conduit.

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Calantha 5 years 2021-08-23T09:21:31+00:00 1 Answers 265 views 0

Answers ( )

  1. Eirian
    0
    2021-08-23T09:23:22+00:00
    Reply

    Solution :

    $\text{Helium and nitrogen}$ gases are contained in a conduit $7 \ mm$ is diameter and $0.08 \ m$ 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

    $P_{A1}=0.075 \ atm $

    $P_{A2}=0.03 \ atm $

    P = 1 atm

    From, table

    $D_{AB}= 0.687 \times 10^{-4} \ m/s$

    We know :

    $J_{A}^* = D_{AB} \frac{d_{CA}}{dz}$

    $J_A^*=\frac{(0.687 \times 10^{-4})(0.075-0.03)(\frac{101.32}{1 \ atm}) }{8.319 \times 298 \times 0.10}$

        = $1.26 \times 10^{-6} \ kgmol/m^r s$

    $P_{B1} = P-P_{A1}$

          = 1 – 0.075

          = 0.925 atm

    $P_{B2} = P-P_{A2}$

          = 1 – 0.03

          = 0.97 atm

    $J_B^*=D_{AB}\frac{(P_{B1} \times P_{B2})}{RT( \Delta z)}$

        $=\frac{0.687 \times 10^{-4}(0.925-0.97)(\frac{101.32}{1 \ atm})}{8.314 \times 298 \times 0.1}$

        $=-1.26 \times 10^{-6} \ kg \ mol /m^r s$

    Partial pressure of helium  $=\frac{0.075+0.03}{2}$

                                                 = 0.0525 atm

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