Estimate the diffusivity of ethyl alcohol (C$_2$H$_5$OH) in dilute aqueous solution at $20^\circ\mathrm{C}$. For water as solvent, $\Phi_B = 2.26$; molecular weight of water, $M_B = 18.02$; viscosity of water, $\mu_B = 0.001005\ \mathrm{kg/(m\,s)}$; atomic volumes of C, H and O are 14.8, 3.7 and 7.4 $\mathrm{cm^3/mol}$, respectively.

A water droplet with an initial diameter of 10 mm is suspended in a large volume of air. The air is at a temperature of $26.65^\circ\mathrm{C}$ and a total pressure of 1 atm. Moisture present in the air exerts a partial pressure of 0.01036 atm. The wet-bulb temperature of the water droplet is $15.55^\circ\mathrm{C}$, at which temperature the vapour pressure of water is 0.01743 atm. The diffusivity of water vapour at its average temperature of $21.1^\circ\mathrm{C}$ has been estimated to be $0.2607\ \mathrm{cm^2/s}$. Calculate the time required for complete evaporation of the water droplet. The time required for evaporation from droplets is estimated from equation A1.1: $\theta = \frac{\rho_A \times r_s^2 \times R \times T}{2 \times M_A \times D_{AB} \times P \times \ln\left[\dfrac{P - P'_{A,b}}{P - P'_{A,s}}\right]}$ where $\theta$ is the time (s) required for the evaporation of a droplet with radius $r_s$ (cm); water molecular weight $M_A = 18\ \mathrm{g\,mol^{-1}}$, density $\rho_A = 1\ \mathrm{g/cm^3}$; $R$ the gas constant $8.206 \times 10^{-5}\ \mathrm{m^3\,atm/(K\,mol)}$; $P'_{A,b}$ and $P'_{A,s}$ are the vapour pressures at the bulk gas phase and the surface of the droplet respectively. Comment on the conditions that can lead to a faster evaporation time. The effect of curvature on the vapour pressure of water and convective effect may be neglected.

Hydrogen gas is diffusing through an unglazed neoprene rubber with a wall thickness of 50 mm at $25^\circ\mathrm{C}$ and pressure of 1 atm. The solubility of hydrogen in the rubber has been estimated to be $0.053\ \mathrm{cm^3}$ (at STP, $V_{mol} = 22.4$ litres at STP) of H$_2$ per $\mathrm{cm^3}$ of neoprene. The diffusivity of hydrogen through the rubber wall is $1.8 \times 10^{-6}\ \mathrm{cm^2/s}$. Calculate the mass diffusion rate ($\mathrm{g/(m^2\,s)}$) of hydrogen per square meter of the wall.