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14-45 Consider a rubber membrane separating carbon dioxide gas that is maintained on one side at \(2 \mathrm{~atm}\) and on the opposite at \(1 \mathrm{~atm}\). If the temperature is constant at \(25^{\circ} \mathrm{C}\), determine (a) the molar densities of carbon dioxide in the rubber membrane on both sides and \((b)\) the molar densities of carbon dioxide outside the rubber membrane on both sides.

Short Answer

Expert verified
Answer: The molar densities of CO2 inside and outside the rubber membrane are: Side 1: \(\rho_{1} = 0.0812 \frac{mol}{L}\) Side 2: \(\rho_{2} = 0.0406 \frac{mol}{L}\)

Step by step solution

01

Convert temperature to Kelvin

To convert the given temperature in Celsius to Kelvin, add 273.15 to the Celsius value: \(T_{K} = 25 + 273.15 = 298.15 K\)
02

Find the molar density, ρ, inside the rubber membrane

To find the molar density, we need to rearrange the Ideal Gas Law equation as follows: ρ = n/V Substituting the Ideal Gas Law into this equation, we get: ρ = P / RT For side 1: \(\rho_{1} = \frac{P_{1}}{RT} = \frac{2 atm}{0.0821 \frac{L atm}{mol K} \times 298.15 K} = 0.0812 \frac{mol}{L}\) For side 2: \(\rho_{2} = \frac{P_{2}}{RT} = \frac{1 atm}{0.0821 \frac{L atm}{mol K} \times 298.15 K} = 0.0406 \frac{mol}{L}\) Therefore, the molar densities of CO2 inside the rubber membrane are: Side 1: \(\rho_{1} = 0.0812 \frac{mol}{L}\) Side 2: \(\rho_{2} = 0.0406 \frac{mol}{L}\) #b. Find the molar densities of CO2 outside the rubber membrane# Since the rubber membrane does not affect the molar densities of CO2 outside, the molar densities outside remain the same as inside the membrane on the respective sides. Side 1: \(\rho_{1} = 0.0812 \frac{mol}{L}\) Side 2: \(\rho_{2} = 0.0406 \frac{mol}{L}\)

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Key Concepts

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

Understanding the Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics that describes the behavior of an ideal gas. This law is generally expressed as the equation \( PV = nRT \), where \(P\) represents the pressure of the gas, \(V\) is the volume it occupies, \(n\) is the amount of substance in moles, \(R\) is the universal gas constant, and \(T\) is the absolute temperature in Kelvin.

In real-world applications, the Ideal Gas Law can help us predict how a gas will behave under various conditions. For example, if we know the pressure, temperature, and volume, we can calculate the amount of gas in moles. Conversely, knowing the amount of gas and the temperature can allow us to find the volume it will occupy at a certain pressure.

In the provided exercise, the molar density (moles of gas per unit volume) inside a rubber membrane is determined by rearranging the Ideal Gas Law to \( \rho = \frac{P}{RT} \), where \(\rho\) represents molar density. This equation illustrates that molar density is directly proportional to pressure \(P\) and inversely proportional to absolute temperature \(T\).

By plugging into this equation the known values of pressure and temperature, the student can calculate the molar density of the gas on either side of the membrane. This is an essential step for understanding gas behavior in enclosed systems and is a practical tool for engineers and scientists in fields such as chemical engineering and environmental science.
Exploring Membrane Separation
Membrane separation is a technology used to separate substances based on their physical or chemical properties. This process utilizes a membrane, which is a selective barrier; it allows some particles to pass through while blocking others. The selectivity of the membrane can depend on various factors such as size, charge, or chemical affinity.

In the context of gases, membrane separation is often applied in industries for gas purification or the separation of different gas mixtures. For instance, a rubber membrane might be used to separate carbon dioxide from other gases based on its permeability to CO2.

The effectiveness of membrane separation can be influenced by the pressure and temperature across the membrane, as well as the membrane's material properties. The example in the exercise shows a rubber membrane separating carbon dioxide at two different pressures. It is implied that the membrane allows for a certain degree of CO2 transfer, which may be related to industrial applications such as carbon capture or the production of enriched gas mixtures.

Understanding how membrane separation works and the factors that influence it is crucial for those designing and operating such systems, ensuring efficient and effective separation according to the desired outcomes.
Relation Between Gas Pressure and Temperature
The relationship between gas pressure and temperature is described by Gay-Lussac’s Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume is held constant. This is expressed by the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) when comparing two states of the same amount of gas.

This relationship is an essential component of the Ideal Gas Law. When the temperature increases, the kinetic energy of the gas particles increases, causing more frequent and forceful collisions with the walls of their container, thereby increasing pressure. Conversely, cooling the gas would decrease its pressure.

The exercise presented concerns temperature held constant, which means the pressure difference across the rubber membrane is solely due to the application of different pressures, not temperature change. By maintaining a constant temperature, one can isolate the variable of pressure and explore its direct effect on molar density, as shown in the exercise. This principle is vital in various engineering and scientific scenarios, such as when managing the pressure conditions of chemical reactions or controlling environmental conditions in cultivation industries.

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Most popular questions from this chapter

In transient mass diffusion analysis, can we treat the diffusion of a solid into another solid of finite thickness (such as the diffusion of carbon into an ordinary steel component) as a diffusion process in a semi-infinite medium? Explain.

You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

A 2-m-deep 20-m \(\times 20-\mathrm{m}\) heated swimming pool is maintained at a constant temperature of \(30^{\circ} \mathrm{C}\) at a location where the atmospheric pressure is \(1 \mathrm{~atm}\). If the ambient air is at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity and the effective sky temperature is \(0^{\circ} \mathrm{C}\), determine the rate of heat loss from the top surface of the pool by ( \(a\) ) radiation, \((b)\) natural convection, and (c) evaporation. ( \(d\) ) Assuming the heat losses to the ground to be negligible, determine the size of the heater.

The pressure in a pipeline that transports helium gas at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) is maintained at \(14.5\) psia by venting helium to the atmosphere through a \(0.25\)-in-internal-diameter tube that extends \(30 \mathrm{ft}\) into the air. Assuming both the helium and the atmospheric air to be at \(80^{\circ} \mathrm{F}\), determine \((a)\) the mass flow rate of helium lost to the atmosphere through the tube, (b) the mass flow rate of air that infiltrates into the pipeline, and \((c)\) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation.

Air flows through a wet pipe at \(298 \mathrm{~K}\) and 1 atm, and the diffusion coefficient of water vapor in air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). If the heat transfer coefficient is determined to be \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the mass transfer coefficient is (a) \(0.0326 \mathrm{~m} / \mathrm{s}\) (b) \(0.0387 \mathrm{~m} / \mathrm{s}\) (c) \(0.0517 \mathrm{~m} / \mathrm{s}\) (d) \(0.0583 \mathrm{~m} / \mathrm{s}\) (e) \(0.0707 \mathrm{~m} / \mathrm{s}\)

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