The concentration of oxygen at two places

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Reference no: EM13943094

Question - 1 In an oxygen -nitrogen mixture at 10 atmosphere and 25 oc , the concentration of oxygen at two places 0.2 cm apart are 10 and 20 volume percent respectively. The rate of diffusion of oxygen expressed as g /cm² -hr for the case of unicomponent diffusion ( nitrogen is non diffusing ) ,will be value of diffusivity =0.181 cm²/s Solution - Na = Dab *Pt *( Pa1 - Pa2)/Rt *Zp *Pbm Given - Dab = 0.181 cm2/s Pt =10 atm T = 273 +25 =298 ok Z=0.2 cm R = 82.06 atm -cc/g-mol ok Now ,

volume percent = mole percent Pa₁ = 0.2 *10 =2 atm Pa₂ =0.1*10 =1 atm Pa₁ +Pb₁ = Pa₂ +Pb₂ = Pt =10 atm Pb₁ =8 atm , Pb₂ =9 atm Pbm = Pb₂ -Pb₁/ ln Pb₂/Pb₁ = 9-8 /ln 9/8 = 8.49 atm Na = 0.181 *10 *(2-1) /82.06 *298 *0.2 *8.49 = 4.36 *10^-5 g-mol/s -cm² = 4.36 *10^-5*32*3600 =5.022 g /cm²-hr

Question - 1 In an oxygen -nitrogen mixture at 10 atmosphere and 25^oc , the concentration of oxygen at two places 0.2 cm apart are 10 and 20 volume percent respectively. The rate of diffusion of oxygen expressed as g /cm² -hr for the case of unicomponent diffusion ( nitrogen is non diffusing ) ,will be value of diffusivity =0.181 cm2/s Solution - Na = Dab *Pt *( Pa₁ - Pa₂)/Rt *Zp *Pbm Given - Dab = 0.181 cm2/s Pt =10 atm T = 273 +25 =298 ok Z=0.2 cm R = 82.06 atm -cc/g-mol ok Now , volume percent = mole percent Pa₁ = 0.2 *10 =2 atm Pa₂ =0.1*10 =1 atm Pa₁ +Pb₁ = Pa₂ +Pb₂ = Pt =10 atm Pb1 =8 atm , Pb₂ =9 atm Pbm = Pb₂ -Pb₁/ ln Pb₂/Pb₁ = 9-8 /ln 9/8 = 8.49 atm Na = 0.181 *10 *(2-1) /82.06 *298 *0.2 *8.49 = 4.36 *10^-5 g-mol/s -cm² = 4.36 *10^-5*32*3600 =5.022 g /cm²-hr

Reference no: EM13943094

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