Consider a non-reactive binary mixture of two ideal gases A and B in an isothermal horizontal closed tube. Due to the concentration difference at the two ends of the tube, equimolal counterdiffusion occurs along the axial direction. No concentration gradients exist in the radial direction. The profile for partial pressure of component A (\(p_A\)) as a function of axial distance \(z\) is shown in the figure. The partial pressure of component B (\(p_B\)) approaches zero at \(z = 0.8\) m. At a location \(z = z_1\) the partial pressures of the components A and B are equal. The value of \(z_1\) (in m) is ______ (rounded off to two decimal places).
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Total pressure is constant along an equimolal counterdiffusion path; find where the linear p_A profile equals half the total pressure.
Step 1: Physical principle for equimolal counterdiffusion. In equimolal counterdiffusion, the molar flux of A is equal and opposite to the molar flux of B (\(N_A = -N_B\)), so there is no net bulk molar flow. For an ideal gas mixture at constant temperature and steady state, this means the total pressure \(P = p_A + p_B\) is constant at every axial location \(z\), and each partial pressure varies linearly with \(z\), consistent with the straight line shown in the figure.
Step 2: Read the linear profile of \(p_A\) from the figure. The graph shows \(p_A = 0.6\) bar at \(z = 0\) and the line rises linearly, passing through \(p_A = 1.1\) bar at \(z = 0.2\) m and \(p_A = 1.6\) bar at \(z = 0.4\) m. Slope \(= \dfrac{1.6-0.6}{0.4-0} = \dfrac{1.0}{0.4} = 2.5\) bar/m. So the profile is: \[ p_A(z) = 0.6 + 2.5z \quad \text{(bar)} \]
Step 3: Determine the total pressure using the given boundary condition. It is given that \(p_B \to 0\) at \(z = 0.8\) m. Since \(P = p_A+p_B\) is constant along the tube, at \(z=0.8\) m, \(P = p_A(0.8)\). \[ p_A(0.8) = 0.6+2.5(0.8) = 0.6+2.0 = 2.6 \text{ bar} \] So the (constant) total pressure is \(P = 2.6\) bar.
Step 4: Apply the condition \(p_A = p_B\) at \(z = z_1\). Since \(p_A(z_1)+p_B(z_1)=P\) and \(p_A(z_1)=p_B(z_1)\): \[ 2\,p_A(z_1) = P = 2.6 \ \Rightarrow \ p_A(z_1) = 1.3 \text{ bar} \]
Step 5: Solve for \(z_1\) using the linear profile from Step 2. \[ 0.6 + 2.5 z_1 = 1.3 \] \[ 2.5 z_1 = 0.7 \] \[ z_1 = 0.28 \text{ m} \]
