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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, 蟻, inside the rubber membrane

To find the molar density, we need to rearrange the Ideal Gas Law equation as follows: 蟻 = n/V Substituting the Ideal Gas Law into this equation, we get: 蟻 = 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鈥檚 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

Saturated water vapor at \(25^{\circ} \mathrm{C}\left(P_{\text {sat }}=3.17 \mathrm{kPa}\right)\) flows in a pipe that passes through air at \(25^{\circ} \mathrm{C}\) with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a \(7-\mathrm{mm}\) internal-diameter tube that extends \(10 \mathrm{~m}\) into the air. The diffusion coefficient of vapor through air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The amount of water vapor lost to the atmosphere through this individual tube by diffusion is (a) \(1.02 \times 10^{-6} \mathrm{~kg}\) (b) \(1.37 \times 10^{-6} \mathrm{~kg}\) (c) \(2.28 \times 10^{-6} \mathrm{~kg}\) (d) \(4.13 \times 10^{-6} \mathrm{~kg}\) (e) \(6.07 \times 10^{-6} \mathrm{~kg}\)

In natural convection mass transfer, the Grashof number is evaluated using density difference instead of temperature difference. Can the Grashof number evaluated this way be used in heat transfer calculations also?

A rubber object is in contact with nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(298 \mathrm{~K}\) and \(250 \mathrm{kPa}\). The solubility of nitrogen gas in rubber is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3}\).bar. The mass density of nitrogen at the interface is (a) \(0.049 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(0.064 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(0.077 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(0.092 \mathrm{~kg} / \mathrm{m}^{3}\) (e) \(0.109 \mathrm{~kg} / \mathrm{m}^{3}\)

In a manufacturing facility, \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\) wet brass plates coming out of a water bath are to be dried by passing them through a section where dry air at 1 atm and \(25^{\circ} \mathrm{C}\) is blown parallel to their surfaces at \(4 \mathrm{~m} / \mathrm{s}\). If the plates are at \(15^{\circ} \mathrm{C}\) and there are no dry spots, determine the rate of evaporation from both sides of a plate.

Express the mass flow rate of water vapor through a wall of thickness \(L\) in terms of the partial pressure of water vapor on both sides of the wall and the permeability of the wall to the water vapor.

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