Benzene-free air at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) enters a 5 -cm-diameter tube at an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The inner surface of the \(6-m\)-long tube is coated with a thin film of pure benzene at \(25^{\circ} \mathrm{C}\). The vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C}\) is \(13 \mathrm{kPa}\), and the solubility of air in benezene is assumed to be negligible. Calculate \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the molar concentration of benzene in the outlet air, and \((c)\) the evaporation rate of benzene in \(\mathrm{kg} / \mathrm{h}\).

Calculate the volumetric flow rate of air by using the given average velocity and tube dimensions. Volumetric flow rate (\(Q\)) can be calculated using the formula: \(Q = A * v\), where \(A\) is the cross-sectional area of the tube and \(v\) is the average velocity. The area can be calculated as: \(A = \pi * (D / 2)^2\), where \(D\) is the diameter of the tube. The diameter is \(5\,\text{cm}=0.05\,\text{m}\).

To find the molar flow rate, we will use the ideal gas equation: \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. The molar flow rate can be calculated as: \(n_m = \dfrac{Q*P_{air}}{R*T}\), where \(P_{air}\) is the partial pressure of air in the stream, \(R\) is the ideal gas constant \(8.314\,\text{J/mol K}\), and \(T\) is the temperature in Kelvin (which is \(25^{\circ}\text{C}+273.15=\,298.15\,\text{K}\)).

We know that the mass transfer coefficient, \(K\), is related to the mass transfer flux by the following equation: \(K = \dfrac{J}{x}\), where \(J\) is the mass transfer flux and \(x\) is the distance traveled. Using Fick's law, we can calculate the flux, \(J\), as: \(J \approx \frac{\Delta C}{\Delta x}\), where \(\Delta C\) is the difference in concentrations and \(\Delta x\) is the distance over which diffusion occurs. Since the molar flow rate of air is much higher than the benzene flux (due to its low vapor pressure), we can assume that the air molar flow rate remains unchanged during the process. Since we already know the benzene vapor pressure at \(25^{\circ}\text{C}\), we can assume that the molar concentration of benzene in the air at the exit is equal to the vapor pressure (\(13\,\text{kPa}\)). The distance over which diffusion occurs can be assumed to be the length of the tube, which is \(6\,\mathrm{m}\). Hence, the average mass transfer coefficient can be calculated as: \(K = \dfrac{J}{x}\).

Since we know the vapor pressure of benzene at \(25^{\circ}\text{C}\) (\(13\,\text{kPa}\)), we can assume that the molar concentration of benzene in the air at the exit is equal to the vapor pressure. It is important to note that the unit of this molar concentration of benzene in the air is \(kPa\).

The evaporation rate of benzene can be calculated using the concentration at the outlet, found in step 4, and the molar flow rate of air found in step 2. The molar rate of evaporation can be calculated as: \(n_{C_6H_6} = \left(\dfrac{P_{benzene}}{P_{air}}\right)* n_{air}\), where \(P_{benzene}\) is the vapor pressure of benzene in the air, and \(n_{air}\) is the molar flow rate of air. Finally, convert the molar rate of evaporation to mass rate using the molar mass of benzene (\(78.11\,\text{g/mol}\)): \(Evap\_Rate = n_{C_6H_6} * M_{C_6H_6}\). Convert the evaporation rate to \(\mathrm{kg} / \mathrm{h}\). Follow the steps given above to find the values of the average mass transfer coefficient, molar concentration of benzene in the outlet air, and evaporation rate of benzene in \(\mathrm{kg} / \mathrm{h}\).