Let's break down the problem and solve it step by step. ## **Given:** - **Diameter of sphere, \( d \) = 1.25 cm = 0.0125 m** - **Temperature, \( T = 10^\circ C \)** - **The naphthalene is in a stagnant, infinite medium (air)** - **Evaporation is so slow that the sphere's diameter can be considered constant at any instant (quasi-steady state)** - **Find the time required for the sphere to completely sublimate** --- ## **Step 1: Mass transfer from a sphere (quasi-steady state)** The mass transfer rate from a sphere by diffusion in a stagnant medium is given by: \[ \dot{m} = 4 \pi r D_{AB} (C_{A,s} - C_{A,\infty}) \] Where: - \( r \) = radius of sphere (m) - \( D_{AB} \) = diffusivity of naphthalene vapor in air (m²/s) - \( C_{A,s} \) = concentration of naphthalene vapor at the surface (kg/m³) - \( C_{A,\infty} \) = concentration of naphthalene vapor in the bulk (0, since it's a closed room and initially air is pure) --- ## **Step 2: Physical Properties** ### **Radius:** \[ r = \frac{d}{2} = \frac{0.0125}{2} = 0.00625\, \text{m} \] ### **Molecular Diffusivity (\(D_{AB}\)):** At \(10^\circ C\), for naphthalene vapor in air, \( D_{AB} \approx 6 \times 10^{-6} \, \text{m}^2/\text{s} \) (Source: Perry's Chemical Engineers' Handbook, estimate) ### **Vapor Pressure of Naphthalene at \(10^\circ C\):** From vapor pressure tables: - At \(10^\circ C\), \(P_A^* \approx 0.020 \, \text{kPa} = 20 \, \text{Pa}\) ### **Total Pressure (\(P\)):** Assume atmospheric pressure, \(P = 101325 \, \text{Pa}\) ### **Molar concentration at the surface:** \[ C_{A,s} = \frac{P_A^*}{RT} \] Where: - \(R = 8.314\,\text{J/(mol K)}\) - \(T = 10^\circ C = 283.15\,\text{K}\) \[ C_{A,s} = \frac{20}{8.314 \times 283.15} \approx \frac{20}{2355} \approx 0.0085\, \text{mol/m}^3 \] \[ C_{A,\infty} = 0 \] --- ## **Step 3: Mass of sphere** \[ \text{Volume of sphere} = V = \frac{4}{3} \pi r^3 \] \[ V = \frac{4}{3} \pi (0.00625)^3 = \frac{4}{3} \pi \times 2.44 \times 10^{-7} \] \[ V \approx 1.02 \times 10^{-6} \, \text{m}^3 \] \[ \text{Density of naphthalene,}~ \rho = 1,140\, \text{kg/m}^3 \] \[ \text{Mass of sphere,}~ m = \rho V = 1,140 \times 1.02 \times 10^{-6} \approx 1.16 \times 10^{-3} \, \text{kg} \] --- ## **Step 4: Rate of mass loss** First, convert \(C_{A,s}\) to mass concentration: - Molar mass of naphthalene = 128.17 g/mol = 0.12817 kg/mol \[ C_{A,s}~(\text{kg/m}^3) = 0.0085~\text{mol/m}^3 \times 0.12817~\text{kg/mol} \approx 0.00109~\text{kg/m}^3 \] So, \[ \dot{m} = 4\pi r D_{AB} (C_{A,s} - C_{A,\infty}) \] \[ \dot{m} = 4 \pi (0.00625) \times 6 \times 10^{-6} \times (0.00109) \] Calculate step by step: - \(4 \pi \times 0.00625 = 0.07854\) - \(0.07854 \times 6 \times 10^{-6} = 4.7124 \times 10^{-7}\) - \(4.7124 \times 10^{-7} \times 0.00109 = 5.14 \times 10^{-10} \, \text{kg/s}\) --- ## **Step 5: Time to disappear** \[ t = \frac{\text{initial mass}}{\text{rate of loss}} \] \[ t = \frac{1.16 \times 10^{-3}}{5.14 \times 10^{-10}} \approx 2.26 \times 10^6 \, \text{s} \] Convert seconds to days: \[ \text{Days} = \frac{2.26 \times 10^6}{86400} \approx 26.2~\text{days} \] --- ## **Final Answer** \[ \boxed{ t \approx 2.26 \times 10^6~\text{s} \approx 26.2~\text{days} } \] **It will take approximately 26 days for a 1.25 cm diameter naphthalene sphere to completely disappear at \(10^\circ C\) in a stagnant, infinite air medium.** --- ### **References / Notes** - Diffusivity and vapor pressure may vary slightly with sources, but the above are typical textbook values. - If you need references or want to account for natural convection (which would speed up the process), please specify. - The approximation assumes the sphere shrinks uniformly (quasi-steady state). For high accuracy, use the shrinking sphere model, but the result will not differ greatly for small spheres and slow rates.