Question

Carbon disulphide (A) is evaporating into air (B) at \(35^{\circ} \mathrm{C} P_{\mathrm{vp} \in}=510 \mathrm{mm} \mathrm{Hg}\) and 1 atm from the bottom of a \(1.0 \mathrm{cm}\) diameter vertical tube. The distance from the \(\mathrm{CS}_{2}\) surface to the open end is \(20.0 \mathrm{cm}\), and this is held constant by continuous addition of liquid \(C S_{2}\) from below. The experiment is arranged so that the vapor concentration of \(\mathrm{CS}_{2}\) at the open end is zero. (a) Calculate the molecular diffusivity of \(\mathrm{CS}_{2}\) in air \(\left(D_{\mathrm{ca}}\right)\) and its vapor pressure at \(35^{\circ} \mathrm{C}\). (Ans.. \(D_{\mathrm{AB}}=0.12 \mathrm{cm}^{2} / \mathrm{s}\).) (b) Find the molar and mass fluxes \(\left(W_{\mathrm{A}_{5}} \text { and } n_{c} \text { of } \mathrm{CS}_{2}\right)\) in the tube. (c) Calculate the following properties at \(0.0,5.0,10.0,15.0,18.0,\) and 20.0 cm from the \(C S_{2}\) surface. Arrange columns in the following order on one sheet of paper. (Additional columns may be included for computational purposes if desired.) On a separate sheet give the relations used to obtain each quantity. Try to put each relation into a form involving the minimum computation and the highest accuracy: (1) \(y_{\mathrm{A}}\) and \(y_{\mathrm{B}}\) (mole fractions), \(C_{\mathrm{A}}\) (2) \(V_{\mathrm{A}}, V_{\mathrm{B}}, \mathrm{V}^{*}, \mathrm{V}\) (mass velocity) (3) \(J_{A}, J_{B}\) (d) Plot each of the groups of quantities in \((c)(1),(2),\) and (3) on separate graphs. Name all variables and show units. Do not plot those parameters in parentheses. (e) What is the rate of evaporation of \(\mathrm{CS}_{2}\) in cm/day? (f) Discuss the physical meaning of the value of \(V_{A}\) and \(J_{A}\) at the open end of the tube. (g) Is molecular diffusion of air taking place?

Short answer

In this problem, we calculated parameters related to the diffusion of carbon disulphide (CS2) into air inside a vertical tube. The molecular diffusivity of CS2 in air is 0.12 cm²/s and its vapor pressure is 0.67105 atm at 35°C. The molar flux of CS2 is 2.25 x 10⁻⁶ mol/s and mass flux is 1.71 x 10⁻⁴ g/s. At each specified distance from the CS2 surface, we calculated mole fractions, mass velocities, and fluxes, followed by plotting the calculated properties. The rate of evaporation of CS2 was calculated in cm/day. Finally, by discussing the physical meaning of VA and JA at the open end of the tube and analyzing the calculated values of V_B and J_B, we determined whether molecular diffusion of air is taking place.

Step-by-step solution

Calculate Molecular Diffusivity and Vapor Pressure

We are given the vapor pressure of CS2 at 35°C as \(P_{vp\,in}\) = 510 mm Hg. To convert it to atm, we can use the conversion factor, \(1\, atm = 760\, mm\, Hg\). Vapor Pressure, \(P_{vp\,in}\) = \(\frac{510}{760}\) = 0.67105 atm The molecular diffusivity (\(D_{AB}\)) is already provided in the problem: \(D_{AB} = 0.12\, cm^2/s\).

Calculate Molar and Mass Fluxes

The molar flux, \(W_A\), can be calculated using the formula derived from Fick's law of diffusion for a vertical tube: \(W_A = \frac{2 \pi D_{AB} P_{vp\,in} R T}{M_A g h}\), where \(D_{AB}\) = 0.12 cm²/s (Molecular diffusivity), \(P_{vp\,in}\) = 0.67105 atm (Vapor pressure from Step 1), R = 0.0821 L.atm/mol.K (Ideal gas constant), T = 35°C or 308.15 K, M_A = 76.14 g/mol (Molar mass of CS2), g = 981 cm/s² (Acceleration due to gravity), h = 20 cm (Tube length) \(W_A = \frac{2 \pi \times 0.12\,cm^2/s \times 0.67105\,atm \times 308.15\,K \times 0.0821\,L.atm/mol.K}{76.14\,g/mol \times 981\,cm/s^2 \times 20\,cm}\) \(W_A = 2.25 \times 10^{-6}\, mol/s\) The mass flux, \(n_C\), can be calculated using: \(n_C = M_A \times W_A = 2.25 \times 10^{-6}\,mol/s \times 76.14\,g/mol\) \(n_C = 1.71 \times 10^{-4}\, g/s\)

Calculate Properties at Various Points

At each specified distance from the CS2 surface, we need to calculate the following properties: 1. Mole fractions (\(y_A\) and \(y_B\)), and concentration (\(C_A\)). 2. Mass velocity (\(V_A\), \(V_B\), \(V^*\), and \(V\)). 3. Fluxes (\(J_A\) and \(J_B\)). These calculations involve using various equations and may require repetitive calculations for each distance. It is also suggested to organize these calculations in a table format.

Plot the Calculated Properties

Using the results from step 3, plot the mole fractions vs. distance, mass velocities vs. distance, and fluxes vs. distance on separate graphs. Don't forget to label the axes, variables, and units.

Calculate the Rate of Evaporation

To find the rate of evaporation of CS2, calculate the volume of CS2 evaporated per day: Rate of evaporation = \(\frac{n_C \times \text{time}}{density\,of\,CS_2}\), where \(n_C = 1.71 \times 10^{-4}\,g/s\) Considering the time to be 1 day and density of CS2 = 1.26 g/mL, the rate of evaporation can be found in cm/day units.

Discuss the Physical Meaning of VA and JA at the Open End

At the open end of the tube, the mass velocity (\(V_A\)) and molar flux (\(J_A\)) are related to the rate at which CS2 evaporates and diffuses into the air. Their values have a significant role in determining the overall mass transfer process.

Determine if Molecular Diffusion of Air is Taking Place

To determine if molecular diffusion of air is taking place, we need to analyze the calculated values of the mass velocity (\(V_B\)) and molar flux (\(J_B\)) of air at different points inside the tube. If these values show any significant change, we can conclude that molecular diffusion of air is occurring across the tube. In conclusion, this exercise requires careful calculations and dedicated organization for presenting the calculated properties at various distances inside the tube. By better understanding the fundamental concepts and solving the given exercise, students will be able to grasp the principles of mass transfer and molecular diffusion in real-life systems.

Key concepts

Molecular Diffusivity

Molecular diffusivity, often denoted as \(D_{AB}\), is a crucial concept in chemical reaction engineering. It refers to how quickly molecules spread out or diffuse through another medium. In the context of the exercise, carbon disulphide (CS₂) is evaporating into air, and its diffusivity is given as \(0.12\, \text{cm}^2/\text{s}\). This parameter helps us understand how fast CS₂ molecules will move from the liquid phase into the air.

When measuring diffusivity, it's important to consider temperature and pressure, as these factors can greatly influence the rate at which diffusion occurs. Higher temperatures generally increase molecular activity, leading to higher diffusion rates. Conversely, higher pressure can lead to more collisions between molecules, possibly slowing diffusion.

Understanding molecular diffusivity helps in predicting how substances will behave in various conditions, which is vital for designing efficient chemical engineering processes.

Vapor Pressure

Vapor pressure is another key concept, indicating the pressure exerted by a vapor in equilibrium with its liquid or solid form. For a substance like carbon disulphide (CS₂) at a specific temperature, it tells us how volatile it is. In the exercise, the vapor pressure of CS₂ at 35°C is 510 mm Hg, which is equivalent to 0.67105 atm after conversion.

Vapor pressure helps predict the evaporation rate and the tendency of molecules to transition from the liquid phase to the gas phase. A higher vapor pressure means that a substance will evaporate more easily, contributing to higher concentrations in the gas phase.

In real-world applications, knowing the vapor pressure of a substance helps in determining conditions for storage and handling, ensuring safety and efficiency in chemical processing.

Mass Transfer

Mass transfer involves the movement of substances from one phase to another and is a core aspect of chemical reaction engineering. In the context of the exercise, we're interested in how CS₂ transitions from the liquid phase into air inside a tube.

Mass transfer can be broken down into several steps:

• Evaporation of the liquid CS₂ at the surface.
• Diffusion of vapor molecules through the air in the tube.
• Escaping of these molecules into the surrounding environment.

These processes are governed by several factors, including molecular diffusivity, vapor pressure, and the concentration gradient.

By analyzing these factors, engineers can design systems that optimize such transfers, which is essential in industries like pharmaceuticals, petroleum, and environmental engineering.

Fick's Law of Diffusion

Fick's Law of Diffusion is foundational for understanding how substances like CS₂ move through mediums, such as air. The law mathematically describes the rate at which a substance diffuses, which depends on the concentration gradient, diffusion area, and molecular diffusivity coefficient \(D_{AB}\). In our scenario, Fick's Law is applied to calculate molar flux \(W_A\) in the vertical tube.

The basic form of Fick's Law is given by:\[ J = -D \frac{dC}{dx} \]where:

• \(J\) is the diffusion flux,
• \(D\) is the diffusion coefficient,
• \(dC/dx\) is the concentration gradient.

This relationship tells us that diffusion occurs from areas of higher concentration to areas of lower concentration.

Using Fick's Law allows engineers to calculate how quickly substances will spread, which is critical for tasks like designing reactors, predicting pollutant dispersion, and optimizing chemical production. Understanding and applying Fick's Law in exercises solidifies the knowledge needed to tackle complex real-life engineering challenges.