The crystal of Problem 1.25 is a sphere \(2 \mathrm{~cm}\) in diameter. It is falling at terminal velocity under the influence of gravity into a big tank of water at \(288 \mathrm{~K}\). The density of the crystal is \(1464 \mathrm{~kg} / \mathrm{m}^{3}\) (Perry and Chilton, 1973).

(a) Estimate the crystal's terminal velocity.

(b) Estimate the rate at which the crystal dissolves and compare it to the answer obtained in Problem 1.25.

Data From Problem 1.25:-

A crystal of Glauber’s salt (Na₂SO₄ ⋅10H₂O) dissolves in a large tank of pure water at 288 K. Estimate the rate at which the crystal dissolves by calculating the flux of Na₂SO₄ from the crystal surface to the bulk solution. Assume that molecular diffusion occurs through a liquid film 0.085 mm thick surrounding the crystal. At the inner side of the film—adjacent to the crystal surface—the solution is saturated with Na₂SO₄, while at the outer side of the film the solution is virtually pure water. The solubility of Glauber’s salt in water at 288 K is 36 g of crystal/100 g of water and the density of the corresponding saturated solution is 1240 kg/m³ (Perry and Chilton, 1973). The diffusivity of Na₂SO₄ in dilute aqueous solution at 288 K can be estimated. The density of pure liquid water at 288 K is 999.8 kg/m3; the viscosity is 1.153 cP.