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

The value
`N_A = 0.192 \ mol \cdot m^(-2) \cdot \ s`

Step-by-step 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 = (0.214 )/(10000) = 2.14 *10^(-5) \ m^2 /s`

Generally the molar flux of ammonia is mathematically represented as

`N_A = (D_(AB) * P_T )/(RT(z_2 -z_1)) * ln [(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 = (2.14 *10^(-5) * 1.01325*10^(5) )/(8.314 *298 (0.01)) * ln [(1 - 0.1)/(1 - 0.9) ]`

=>
`N_A = 0.192 \ mol \cdot m^(-2) \cdot \ s`