Unsteady leaching from porous solids:\nA long cylinder (5 cm diameter, flat ends sealed) and a sphere (5 cm diameter) are made from the same porous material and equally saturated with NaCl solution. Both are dipped for a short, equal time in a large, well-agitated water tank. If Xc and Xs are the fractions of salt remaining in the cylinder and sphere, which relation holds?

Difficulty: Medium

Correct Answer: Xc > Xs

Explanation:


Introduction / Context:
For brief extraction times from porous solids into a well-mixed liquid, the controlling factor is typically external mass transfer across the wetted external area, not deep internal diffusion. Thus, the surface-area-to-volume ratio (A/V) largely dictates how quickly a given body loses dissolved species initially. This question compares a long sealed-ended cylinder and a sphere of the same diameter.


Given Data / Assumptions:

  • Same porous material and initial salt loading per unit volume.
  • Both bodies have the same diameter (5 cm), but the cylinder is long and its flat ends are sealed (no mass transfer from the ends).
  • Immersion time is short; tank is large and well agitated (external mass transfer controls initially).


Concept / Approach:
At short times, the rate of salt removal is approximately proportional to external area A (and the mass transfer coefficient), while total salt content scales with volume V. The relevant metric is A/V. For a sphere of radius r, A/V = 3/r. For a long cylinder with sealed ends (radius r), only the curved surface transfers: A/V = 2/r. With equal r (2.5 cm), the sphere has larger A/V than the long cylinder, so it loses salt faster, leaving a smaller fraction remaining after the same time.


Step-by-Step Solution:

Sphere: A/V = 3/r = 3/2.5 cm^-1 = 1.2 cm^-1.Long sealed cylinder: A/V = 2/r = 2/2.5 cm^-1 = 0.8 cm^-1.Larger A/V → faster initial salt removal.Therefore: Xs < Xc, so Xc > Xs.


Verification / Alternative check:
Dimensional analysis of early-time leaching (or lumped mass-transfer models) shows m(t)/m0 ≈ exp(−kA t / V) at short times; greater A/V decreases m faster.


Why Other Options Are Wrong:

  • Equality (b): would require identical A/V, which is false here.
  • 'Xc < Xs' (c): reverses the A/V effect.
  • 'Depends on cylinder length' (d): for a long cylinder with ends sealed, A/V tends to 2/r and becomes length-independent.
  • 'Insufficient information' (e): given assumptions are enough to decide.


Common Pitfalls:
Including end areas for the cylinder (they are sealed); assuming internal diffusion dominates at very short times; forgetting A/V scaling.


Final Answer:
Xc > Xs

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