Respuesta :
Assuming steady state and one boundary impermeable to water, the flux of HCl is; Nₐ = 2.355 × 10⁻⁶ kmol/m.s²
We are given;
Density of Solution at Point 1; ρ₁ = 1060.7 Kg/m³
Density of Solution at Point 2; ρ₂ = 1030.3 kg/m³
Diffusion coefficient of HCl in water; D_ab = 2.5 × 10⁻⁹ m²/s
Thickness of thin film; Z = 2 mm = 0.002 m
From tables;
Molar mass of HCl; M_a = 36.5 kg/mol
Molar mass of H₂O; M_b = 18 kg/mol
At Point 1;
Let us calculate mole fraction of HCl;
[tex]x_{A1} = \frac{\frac{12}{36.5}}{\frac{12}{36.5} + \frac{88}{18}}[/tex]
⇒ [tex]x_{A1} = 0.0629[/tex]
[tex]x_{B1} = 1 - 0.0629\\\\x_{B1} = 0.9371[/tex]
Average molecular weight here is;
[tex]M_{1} = \frac{{100}}{\frac{12}{36.5} + \frac{88}{18}}[/tex]
⇒ [tex]M_{1} = 19.1662 kg/kmol[/tex]
At Point 2;
Let us calculate mole fraction of HCl;
[tex]x_{A2} = \frac{\frac{6}{36.5}}{\frac{6}{36.5} + \frac{94}{18}}[/tex]
⇒ [tex]x_{A2} = 0.0305[/tex]
[tex]x_{B2} = 1 - 0.0305\\\\x_{B2} = 0.9695[/tex]
Average molecular weight here is;
[tex]M_{2} = \frac{{100}}{\frac{6}{36.5} + \frac{94}{18}}[/tex]
⇒ [tex]M_{2} = 18.5646 kg/kmol[/tex]
Average concentration is;
C_av = ¹/₂[(ρ₁/M₁) + ρ₂/M₂)]
C_av = ¹/₂((1060.7/19.1662) + (1030.3/18.5646))
C_av = 55.42 kmol/m³
x_bm = (x_b2 - x_b1)/(In(x_b2/x_b1))
x_bm = (0.9695 - 0.9371)/(In (0.9695/0.9371))
x_bm = 0.9529
Thus, flux of HCl is;
Nₐ = [D_ab × C_av × (x_a1 - x_a2)]/(x_bm * Z)
Plugging in the relevant values gives;
Nₐ = [2.5 × 10⁻⁹ × 55.42 × (0.0629 - 0.0305)]/(0.9529 × 0.002)
Nₐ = 2.355 × 10⁻⁶ kmol/m.s²
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