Question

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contains 50 per cent by volume ammonia at one boundary of the stagnant layer. The ammonia diffusing to the other boundary is quickly absorbed and the concentration is negligible at that plane. The temperature is 295 K and the pressure atmospheric, and under these conditions the diffusivity of ammonia in air is 1.8 x 10~5 m2/s. Estimate the rate of diffusion of ammonia through the layer.

Answer 1

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$, $P_{A1} =y_{A}$, $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  $\frac{kJ}{(kmol)(K)} ,$    T = 298 K   and  $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵$\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴$\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)