Question

Ammonia (A) diffuses through a stagnant layer of air (B), 1cm thick, at 25 ºC and 1 atm total pressure. The partial pressures of ammonia on the two sides of the air layer are: PA0=0.9 atm and PAl=0.1 atm respectively. Air is none diffusing. Calculate the molar flux of ammonia. DAB= 0.214 cm2 /s

Answer 1

Answer:

The value $N_A = 0.192 \ mol \cdot m^{-2} \cdot \ s$

Explanation:

From the question we are told that

     The thickness of the air is $z_2 - z_1 = 1 \ cm =0.01 \ m$

      The temperature is  $T = 25^oc = 25 +273 = 298 \ K$

     The total pressure is  $P_T = 1 atm = 1.01325*10^{5} \ Pa$

      The partial pressure of Ammonia first side is  $P_{AO} = 0.9 \ atm = 0.9 * 1.01325*10^{5} = 91192.5 \ Pa$

      The partial pressure of Ammonia to the second side is $P_{A} = 0.1 \ atm = 0.1 * 1.0325*10^{5} = 10132.5 \ Pa$

 Rate of flow of ammonia is  

          $D_{AB} = 0.214 \ cm/s = \frac{0.214 }{10000} = 2.14 *10^{-5} \ m^2 /s$

Generally the  molar flux of ammonia is mathematically represented as

         $N_A = \frac{D_{AB} * P_T }{RT(z_2 -z_1)} * ln [\frac{P_T - P_{Al}}{P_T - P_{AO}} ]$

Here R is the gas constant with value  

         $R = 8.314 \ m^3 \cdot Pa \cdot mol^{-1} \cdot K$

             $N_A = \frac{2.14 *10^{-5} * 1.01325*10^{5} }{8.314 *298 (0.01)} * ln [\frac{1 - 0.1}{1 - 0.9} ]$

=>           $N_A = 0.192 \ mol \cdot m^{-2} \cdot \ s$