A researcher is using a 5 -cm-diameter Stefan tube to measure the mass diffusivity of chloroform in air at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). Initially, the liquid chloroform surface was \(7.00 \mathrm{~cm}\) from the top of the tube; and after 10 hours have elapsed, the liquid chloroform surface was \(7.44 \mathrm{~cm}\) from the top of the tube, which corresponds to \(222 \mathrm{~g}\) of chloroform being diffused. At \(25^{\circ} \mathrm{C}\), the chloroform vapor pressure is \(0.263 \mathrm{~atm}\), and the concentration of chloroform is zero at the top of the tube. If the molar mass of chloroform is \(119.39 \mathrm{~kg} / \mathrm{kmol}\), determine the mass diffusivity of chloroform in air.

Initial distance, \(L_1 = 7.00 \,\text{cm}\) Final distance, \(L_2 = 7.44 \,\text{cm}\) Time elapsed, \(t = 10\, \text{hours}\) Amount of chloroform diffused, \(m = 222\, \text{g}\) Vapor pressure of chloroform, \(P_{\text{CHCl}_3} = 0.263\, \text{atm}\) Molar mass of chloroform, \(M_{\text{CHCl}_3} = 119.39\, \frac{\text{kg}}{\text{kmol}}\) #Step 2: Determine the average concentration#

We can have the initial concentration, \(C_1 = 0\), since it is given that the concentration of chloroform is zero at the top of the tube. We can find the final concentration, \(C_2\), from the vapor pressure of chloroform and the ideal gas law, as follows: $$ C_2 = \frac{P_{\text{CHCl}_3}}{RT} \cdot M_{\text{CHCl}_3} $$ where \(R = 8314.5 \, \frac{\text{J}}{\text{kmol}\,\text{K}}\) is the universal gas constant, and \(T = 25 + 273.15 = 298.15\, \text{K}\) is the absolute temperature. Substituting the values, we get: $$ C_2 = \frac{0.263 \,\text{atm}}{8314.5 \, \frac{\text{J}}{\text{kmol}\,\text{K}} \cdot 298.15 \,\text{K}} \cdot 119.39\, \frac{\text{kg}}{\text{kmol}} \cdot \frac{101325\, \text{Pa}}{1\, \text{atm}} $$ $$ C_2 = 1.256 \, \frac{\text{g}}{\text{m}^3} $$ Now, determine the average concentration, \(C_{avg} = \frac{C_1 + C_2}{2}\): $$ C_{avg} = \frac{0 + 1.256}{2} = 0.628 \, \frac{\text{g}}{\text{m}^3} $$ #Step 3: Calculate the flux#

We can use Fick's first law of diffusion to write the equation for flux: $$ J = \frac{\Delta m}{A\Delta t} $$ where \(J\) is the flux, \(\Delta m = 222 \, \text{g}\) is the change in mass, \(A= \pi (\frac{0.05}{2})^2 \,\text{m}^2\) is the area of the Stefan tube, and \(\Delta t = 10 \cdot 3600 = 36000 \, \text{s}\) is the elapsed time in seconds. Calculate the flux, \(J\): $$ J = \frac{222\, \text{g}}{\pi (\frac{0.05}{2})^2 \,\text{m}^2 \cdot 36000\, \text{s}} = 0.2993\, \frac{\text{g}}{\text{m}^2 \text{s}} $$ #Step 4: Determine the mass diffusivity#

Finally, we can use Fick's first law of diffusion to calculate the mass diffusivity, \(D_m\): $$ J = -D_m \frac{\Delta C}{\Delta L} $$ $$ D_m = -\frac{J}{\frac{\Delta C}{\Delta L}} $$ where \(\Delta C = C_2 - C_1 = 1.256 - 0 = 1.256 \, \frac{\text{g}}{\text{m}^3}\) and \(\Delta L = L_2 - L_1 = 7.44 - 7.00 = 0.44\, \text{cm}\). Calculate the mass diffusivity, \(D_m\): $$ D_m = -\frac{0.2993\, \frac{\text{g}}{\text{m}^2 \text{s}}}{\frac{1.256 \, \frac{\text{g}}{\text{m}^3}}{0.0044 \,\text{m}}} = 1.07 \times 10^{-5}\, \frac{\text{m}^2}{\text{s}} $$ So, the mass diffusivity of chloroform in air is \(1.07 \times 10^{-5}\, \frac{\text{m}^2}{\text{s}}\).