Question

The mass diffusivity of ethanol $\left(\rho=789 \mathrm{~kg} / \mathrm{m}^{3}\right.\( and \)M=46 \mathrm{~kg} / \mathrm{kmol}$ ) through air was determined in a Stefan tube. The tube has a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). Initially, the ethanol surface was $10 \mathrm{~cm}$ from the top of the tube; after 10 hours elapsed, the ethanol surface was \(25 \mathrm{~cm}\) from the top of the tube, which corresponds to \(0.0445 \mathrm{~cm}^{3}\) of ethanol being evaporated. The ethanol vapor pressure is \(0.0684 \mathrm{~atm}\), and the concentration of ethanol is zero at the top of the tube. If the entire process was operated at $24^{\circ} \mathrm{C}$ and 1 atm, determine the mass diffusivity of ethanol in air.

Short answer

Answer: The mass diffusivity of ethanol in air in the given system is approximately 1.414 * 10^{-5} m^2/s.

Step-by-step solution

Calculate the molar concentration of ethanol at the bottom of the tube

To calculate the molar concentration of ethanol at the bottom of the tube, we can use the vapor pressure of ethanol and the gas constant R. The molar concentration, C_bot will be given by C_bot = P/(RT), where P is the vapor pressure, R is the gas constant, and T is the temperature in Kelvin. Let's first convert the temperature to Kelvin: T_K = 24°C + 273.15 = 297.15 K Now, we can calculate the molar concentration of ethanol at the bottom of the tube: C_bot = \frac{0.0684 atm}{(0.0821 \frac{L atm}{mol K})(297.15 K)} =\frac{0.0684}{24.398915} = 0.002803 \frac{mol}{L}

Calculate the mass concentration of ethanol at the bottom of the tube

To calculate the mass concentration (ρ_bot) of ethanol at the bottom of the tube, we can use the molar concentration (C_bot) and the molar mass (M) of ethanol. The mass concentration at the bottom of the tube can be calculated using the following equation: ρ_bot = C_bot * M ρ_bot = 0.002803 \frac{mol}{L} * 46 \frac{kg}{kmol} = 0.128938 \frac{kg}{m^3}

Calculate the mass flow rate of ethanol

We are given that 0.0445 cm³ of ethanol evaporated after 10 hours. To find the mass flow rate, we will first find the mass of evaporated ethanol by multiplying the volume with the density of ethanol. Then, divide the mass by the total time to calculate the mass flow rate (m_dot). Mass of evaporated ethanol = Volume * Density = 0.0445 cm³ * 789 kg/m³ = 0.035092 kg m_dot = \frac{0.035092 kg}{10*3600 s} = 9.7477 * 10^{-7}\frac{kg}{s}

Calculate the mass diffusivity of ethanol using Stefan's Law

Now that we have all the necessary values, we can use the Stefan's Law to calculate the mass diffusivity of ethanol in air (D). Stefan's Law states that m_dot = DA(dρ/dz), where A is the cross-sectional area of the tube, and (dρ/dz) is the concentration gradient along the tube. We can calculate the concentration gradient as follows: (dρ/dz) = \frac{ρ_bot - ρ_top}{z_f - z_i} = \frac{0.128938 - 0}{0.25 - 0.10} = \frac{0.128938}{0.15} = 0.859587 \frac{kg}{m^3.cm} Now, we can find the mass diffusivity of ethanol in air by substituting the values in Stefan's Law equation: D = \frac{m_dot}{A (dρ/dz)} = \frac{9.7477 * 10^{-7}\frac{kg}{s}}{(0.8 cm^2)(0.859587\frac{kg}{m^3.cm})} = \frac{9.7477 * 10^{-10}}{0.00068847} m^2/s D ≈ 1.414 * 10^{-5} m^2/s The mass diffusivity of ethanol in air is approximately 1.414 * 10^{-5} m^2/s.