Distillation of a non-reactive binary mixture with components \(A\) and \(B\) is carried out in a batch still as shown. The initial charge in the still is \(1\ \text{kmol}\). The initial and final amounts of \(A\) in the still are \(0.1\ \text{kmol}\) and \(0.01\ \text{kmol}\), respectively. The relative volatility is constant at \(\alpha = 4.5\). The mole fraction of \(B\) remaining in the vessel is __________ (rounded off to three decimal places).
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For batch distillation with constant \(\alpha\), use the Rayleigh equation with \(y^\ast=\dfrac{\alpha x}{1+(\alpha-1)x}\).
Use a component balance (\(x_f W_f=n_{Af}\)) to couple the Rayleigh equation to the final composition.
Step 1: Define compositions and data
Initial total moles \(W_0=1\ \text{kmol}\).
Initial moles of \(A\): \(n_{A0}=0.1\Rightarrow x_0=\dfrac{n_{A0}}{W_0}=0.1\).
Final moles of \(A\): \(n_{Af}=0.01\). Let the final total be \(W_f\) and liquid mole fraction \(x_f\). Then
\[
x_f=\frac{n_{Af}}{W_f}=\frac{0.01}{W_f}.
\]
Step 2: Rayleigh equation for differential (batch) distillation
With constant relative volatility \(\alpha\), the equilibrium vapor composition of \(A\) is
\[
y^\ast=\frac{\alpha x}{1+(\alpha-1)x}.
\]
Rayleigh equation:
\[
\ln\!\left(\frac{W_f}{W_0}\right)=\int_{x_0}^{x_f}\frac{dx}{\,y^\ast-x\,}
=\int_{x_0}^{x_f}\frac{dx}{\displaystyle \frac{\alpha x}{1+(\alpha-1)x}-x}.
\]
Step 3: Solve for \(W_f\) and \(x_f\)
Using \(\alpha=4.5\), \(x_0=0.1\) and the relation \(x_f=0.01/W_f\), numerical evaluation of the integral and root-finding gives
\[
W_f \approx 0.5495\ \text{kmol},\qquad
x_f=\frac{0.01}{0.5495}\approx 0.01820 .
\]
Step 4: Mole fraction of \(B\) remaining
\[
x_{B,f}=1-x_f \approx 1-0.01820 = 0.98180 \;\approx\; \boxed{0.982}.
\]
