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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.

Short Answer

Expert verified
Answer: The mass diffusivity of chloroform in air is \(1.07 \times 10^{-5}\, \frac{\text{m}^2}{\text{s}}\).

Step by step solution

01

List given information

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#
02

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#
03

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#
04

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 \(1.07 \times 10^{-5}\, \frac{\text{m}^2}{\text{s}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fick's Law of Diffusion
Fick's Law of Diffusion is a foundational principle in the study of how substances move from areas of high concentration to areas of low concentration. This process is known as diffusion. In many physical, chemical, and biological systems, diffusion plays a critical role.

There are two key components to Fick's law:
  • Fick's First Law: It deals with the steady state of diffusion, establishing that the flux, or the mass of the substance that diffuses per unit area per unit time, is proportional to the negative gradient of concentrations.
  • Fick's Second Law: It represents the unsteady state or time-dependent changes in diffusion, elaborating on how diffusion changes across time.
In the context of the Stefan tube experiment from the original problem, Fick's First Law is used. Here, we calculate the diffusion flux of chloroform in air. This is given by the formula \[ J = -D_m \frac{\Delta C}{\Delta L} \]where \( J \) is the flux, \( D_m \) is the mass diffusivity, and \( \Delta C / \Delta L \) is the concentration gradient. The negative sign indicates that diffusion happens in the direction of decreasing concentration. This law allows us to determine how much chloroform moves through a specific area of air over a unit time.
Stefan Tube
A Stefan Tube is a piece of laboratory equipment used to measure the rate of diffusion of a vapor through a stationary gas. It's a type of apparatus that helps in studying the characteristics of mass transport by diffusion in gases.

This apparatus typically consists of a vertical cylindrical tube filled with a liquid solvent at the bottom. As the solvent evaporates, its vapor diffuses through the gas, typically air, filling the rest of the tube. Researchers use the Stefan Tube to measure various properties such as:
  • The rate of diffusion of the solvent vapor across the gas.
  • The change in height of the liquid surface over a set period.
  • The concentration of the vapor at different heights within the tube.
The ideal setup ensures that the vapor concentration at the top remains zero, creating a well-defined concentration gradient. In the given exercise, the Stefan tube measures the mass diffusivity of chloroform by observing changes in the liquid height and vapor diffusion over time, under controlled temperature and pressure.
Chloroform-Air Diffusion
Chloroform-air diffusion involves the movement of chloroform vapor away from a chloroform liquid surface into the surrounding air. This process is an excellent subject to study diffusion because chloroform is volatile, meaning it easily transitions between liquid and vapor phases, and it diffuses effectively through air.

In the given problem, measuring how much chloroform evaporates and diffuses into air over a given time allows scientists to calculate important transport properties such as mass diffusivity. Mass diffusivity indicates how fast one substance will diffuse through another. For chloroform-air systems, it's determined under specific conditions:
  • Temperature: Diffusion rates increase with temperature as the kinetic energy of molecules increases, enhancing movement.
  • Pressure: Influences vapor pressure and can affect the diffusion rate, although in this experiment, the total atmospheric pressure is constant.
  • Concentration Gradient: The difference in chloroform concentration between the surface and the surrounding air drives diffusion.
By quantifying how much chloroform evaporates over 10 hours, as calculated from the height change in the Stefan Tube, we find data that highlights its diffusion characteristics.
Vapor Pressure Calculation
Vapor pressure is the pressure exerted by the vapor present above a liquid in a closed system at equilibrium. It is a crucial concept for understanding and predicting how liquids behave when exposed to different temperatures and pressures.

In the problem, the vapor pressure of chloroform at 25°C is given as \(0.263\, \text{atm}\). This pressure indicates how much chloroform vapor is in the air directly above the liquid, under the given temperature. A higher vapor pressure means that the liquid evaporates more readily, producing more vapor.

Using vapor pressure in calculations:
  • Finding concentration: It can help us derive the concentration of chloroform in gaseous form using ideal gas law principles, \( C = \frac{P}{RT} \times M \).
  • Driving diffusion: The pressure difference between the surface and open air generates the concentration gradient that drives diffusion.
  • Temperature dependency: It implies that changes in temperature alter the vapor pressure and potentially the diffusion rate.
Understanding these principles allows for precise measurement and prediction of evaporative diffusion in experimental setups like the Stefan tube.

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Most popular questions from this chapter

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

Consider a tank that contains moist air at \(3 \mathrm{~atm}\) and whose walls are permeable to water vapor. The surrounding air at \(1 \mathrm{~atm}\) pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain.

The diffusion of water vapor through plaster boards and its condensation in the wall insulation in cold weather are of concern since they reduce the effectiveness of insulation. Consider a house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(97 \mathrm{kPa}\). The inside of the walls is finished with \(9.5\)-mm-thick gypsum wallboard. Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 8-\mathrm{m}\) section of a wall during a 24-h period. The permeance of the \(9.5\)-mm-thick gypsum wallboard to water vapor is \(2.86 \times 10^{-9} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\).

The solubility of hydrogen gas in steel in terms of its mass fraction is given as \(w_{\mathrm{H}_{2}}=2.09 \times 10^{-4} \exp (-3950 / T) P_{\mathrm{H}_{2}}^{0.5}\) where \(P_{\mathrm{H}_{2}}\) is the partial pressure of hydrogen in bars and \(T\) is the temperature in \(\mathrm{K}\). If natural gas is transported in a 1-cm-thick, 3-m-internal-diameter steel pipe at \(500 \mathrm{kPa}\) pressure and the mole fraction of hydrogen in the natural gas is 8 percent, determine the highest rate of hydrogen loss through a 100 -m-long section of the pipe at steady conditions at a temperature of \(293 \mathrm{~K}\) if the pipe is exposed to air. Take the diffusivity of hydrogen in steel to be \(2.9 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).

Explain how vapor pressure of the ambient air is determined when the temperature, total pressure, and relative humidity of the air are given.

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