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; and after 10 hours have 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 \(0.0684\) 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 \(1 \mathrm{~atm}\), determine the mass diffusivity of ethanol in air.

Short answer

Question: Determine the mass diffusivity of ethanol in air based on the given measurements from a Stefan tube experiment. Answer: First, you need to calculate the molar flow rate of ethanol using the given volume of evaporated ethanol and elapsed time. Then, use the ideal gas law to find the concentration of ethanol vapor at the surface of the liquid. Next, calculate the concentration of ethanol vapor at the top of the tube, where it is zero. Apply Fick's law of diffusion to find the mass diffusivity of ethanol by relating the mass flux and concentration gradient. Lastly, calculate the mass diffusivity of ethanol in air by plugging the values obtained from the previous steps into the equation.

Step-by-step solution

Calculate the molar flow rate of ethanol.

The first step is to determine the molar flow rate of ethanol evaporation. This is the amount of ethanol evaporated per unit time. Given volume of evaporated ethanol: \(0.0445 \mathrm{~cm}^{3}\) Time elapsed: \(10 \mathrm{~h}\) Molar mass of ethanol: \(M = 46 \mathrm{~kg} / \mathrm{kmol}\) The molar flow rate of ethanol, N, can be calculated as: $$ N = \frac{(0.0445 \mathrm{~cm}^3)(\frac{10^3}{46}\frac{\mathrm{kg}}{\mathrm{kmol}})}{(10 \mathrm{~h})(3600 \mathrm{s}^{-1}\mathrm{h}^{-1})} $$

Use the ideal gas law to find the concentration of ethanol vapor at the surface of the liquid.

Now that we have the molar flow rate of ethanol, we can use the ideal gas law to find the concentration of ethanol vapor at the surface of the liquid ethanol (C_s). The ethanol vapor pressure is given as \(0.0684 \mathrm{~atm}\) and the process is operated at \(24^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) $$ C_s = \frac{P}{RT} = \frac{0.0684 \, \mathrm{atm}}{(0.0821 \, \mathrm{(L \, atm)}/(\mathrm{K\, mol})) (273+24)\mathrm{K}} $$

Calculate the concentration of ethanol vapor at the top of the tube.

The concentration of ethanol is zero at the top of the tube (C_t).

Apply Fick's law of diffusion.

Now we can apply Fick's law of diffusion to find the mass diffusivity (D) of ethanol. The Fick's law of diffusion states that the mass flux (J) is proportional to the concentration gradient (ΔC) and is given by $$ J = -D\frac{dC}{dz} $$ The mass flux, J, is also given by the molar flow rate, N, divided by the cross-sectional area A, as follows: $$ J = \frac{N}{A} $$ So, we can equate the two expressions for mass flux, and solve for D as $$ D = -\frac{NA}{\frac{dC}{dz}} $$ The concentration gradient, \(\frac{dC}{dz}\), can be treated as a linear change in concentration over the distance from the ethanol surface to the top of the tube (Δz) as: $$ \frac{dC}{dz} = \frac{C_s-C_t}{\Delta z} $$ Putting all the values, we get: $$ D = -\frac{NA}{\frac{C_s-C_t}{\Delta z}} $$ where A = 0.8 cm^2, Δz = 25 cm - 10 cm = 15 cm

Calculate the mass diffusivity of ethanol in air.

Plug in the values obtained from previous steps into the equation to find the mass diffusivity of ethanol. $$ D = -\frac{NA}{\frac{C_s-C_t}{\Delta z}} $$

Key concepts

Fick's Law of Diffusion

Understanding the behavior of molecules as they move across concentration gradients is pivotal in fields ranging from chemical engineering to biophysics. Here, Fick's law of diffusion provides a foundational model, describing the rate at which a substance moves due to diffusion. According to Fick's first law, the diffusion flux is proportional to the negative of the gradient of the concentration of the substance. Mathematically, it can be represented as:
\[ J = -D \frac{dC}{dz} \]
The negative sign indicates that diffusion occurs in the direction of decreasing concentration. In simple terms, substances will naturally move from an area of high concentration to an area of low concentration. The proportionality constant, D, is known as the diffusivity or the diffusion coefficient.
In practical terms, to find the mass diffusivity of a substance such as ethanol in air, one would measure the molar flow rate at which ethanol evaporates, determine the concentration at various points (like the surface of the liquid and the top of the tube), and then apply Fick's law to relate these values to the diffusion flux. The result helps us to understand how quickly ethanol molecules are spreading out into the surrounding air, under specific conditions of temperature and pressure.

Ideal Gas Law

Moving on to a universal principle that describes the behavior of gases under various conditions, the ideal gas law is a cornerstone of thermodynamics and physical chemistry. It expresses the relationship among pressure (P), volume (V), temperature (T), and the number of moles of a gas (n). The ideal gas law equation is:
\[ PV = nRT \]
Here, R is the ideal gas constant. For a given situation, if we know any three of these variables, we can calculate the fourth.
In the context of our exercise, the ideal gas law plays a crucial role in determining the concentration of ethanol vapor at the liquid surface in the Stefan tube. Using the given vapor pressure, temperature, and the gas constant (R), one can compute the concentration, which is vital for applying Fick's law to find the mass diffusivity. By assuming the gas behaves ideally, we're able to simplify a complex mixture of ethanol and air into a more manageable calculation. This assumption is reasonable under the experimental conditions provided in the textbook exercise.

Molar Flow Rate

Lastly, we delve into the concept of molar flow rate. It is an expression of the number of moles of a substance that pass through a given cross-sectional area per time unit. This rate is crucial in the realm of chemical engineering and process control because it helps quantify the movement of substances in a system.
The molar flow rate, typically denoted as N, is calculated using the volume of evaporated substance, its molar mass, and the time over which the evaporation takes place. In the given exercise, the molar flow rate of ethanol is determined by the volume that has evaporated over the total time, measured in hours, and then converting this to a rate in moles per second. The equation looks like this:
\[ N = \frac{V \times (\frac{{Molar Density}}{{Molar Mass}})}{time \times 3600 \frac{s}{h}} \]
As an integral part of Fick's law, the molar flow rate aids in computing the mass flux, which allows us to eventually solve for the mass diffusivity of ethanol. Through combining this flow rate with the cross-sectional area of the Stefan tube and the concentration gradient, we understand how the diffusive movement is related to the physical setup of the experiment.