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Chapter 14: Problem 29

Using Henry's constant data for a gas dissolved in a liquid, explain how you would determine the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature.

Short Answer

Expert verified
Question: Based on the step-by-step solution provided, determine the mole fraction of a gas dissolved in a liquid at the interface if the partial pressure of the gas above the solution is 500 atm, and the Henry's constant (KH) is 25,000 atm at the specified temperature. Answer: To determine the mole fraction (x) of the gas dissolved in the liquid, use the rearranged formula of Henry's law: x = P/KH. Substitute the given values: x = \frac{500 \,\text{atm}}{25,000 \,\text{atm}} x = 0.02 Hence, the mole fraction of the gas dissolved in the liquid at the interface at the specified temperature is 0.02.

Step by step solution

01

Identify the given data and variables

First, we need to identify the Henry's constant data given for the gas dissolved in a liquid, along with the temperature and the partial pressure of the gas above the solution at the specified temperature.
02

Write the formula of Henry's law

Henry's law is expressed as follows: P = KH * x where P is the partial pressure of the gas above the solution, KH is the Henry's constant, and x is the mole fraction of the gas dissolved in the liquid. Note that the Henry's constant is temperature-dependent, so make sure to use the data given for the specified temperature.
03

Rearrange the formula to solve for x

We need to find the mole fraction of the gas dissolved in the liquid, which is represented by x. Rearrange the equation to isolate x on one side: x = \frac{P}{KH}
04

Substitute the given values and calculate x

Substitute the values of the partial pressure (P) and Henry's constant (KH) into the equation: x = \frac{P}{KH} Make sure the units of pressure and Henry's constant are the same before calculating the mole fraction.
05

Interpret the result

The result x represents the mole fraction of the gas dissolved in the liquid at the interface at the specified temperature. This value gives a measure of the concentration of the gas in the liquid and helps to understand the behavior of the system under various conditions.

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Key Concepts

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

Mole Fraction
The mole fraction is a way to express concentration. It tells us how much of a specific component is present compared to the total amount of all components. \[ x = \frac{n_i}{n_{total}} \] This formula calculates the mole fraction where \( n_i \) is the moles of the component of interest, and \( n_{total} \) is the total moles of all components. Unlike percentages, mole fractions don't have units. They're especially valuable in chemistry because they reflect the real-life proportions of molecules. To find out how much gas is dissolved in a liquid, Henry's Law makes use of the mole fraction of the gas involved.
Partial Pressure
Partial pressure is crucial in understanding how gases behave, especially in mixtures. Each gas in a mixture exerts pressure as if it were the only gas present. This specific pressure is known as the 'partial pressure' of that gas. In calculations, it is usually represented by \( P \). For instance, if you have oxygen and nitrogen in a container, each gas will have its own partial pressure. The total pressure of the container is the sum of these partial pressures. \[ P_{total} = P_1 + P_2 + ... + P_n \] In the context of Henry's Law, you use the partial pressure of the gas above the solution to determine how much gas dissolves in the liquid.
Henry's Constant
Henry's constant (\( K_H \)) is a special coefficient that connects a gas's solubility to its partial pressure. It's unique for each gas-liquid pair and defined at specific temperatures. Henry's constant tells us how "willing" a gas is to dissolve in a liquid under a certain pressure. If Henry's constant is large, the gas is less soluble, meaning it requires a lot of pressure to dissolve a small amount of the gas. In the formula \( P = K_H \cdot x \), \( K_H \) acts as a critical factor translating the gas's properties into measurable quantities like the mole fraction.
Gas-Liquid Interface
The gas-liquid interface is the boundary between the gas above and the liquid below. It is a dynamic boundary where gas molecules can dissolve into the liquid or escape back into the gas phase. The concentration of gas at this interface directly affects how much gas dissolves in the liquid. For our calculations, we use the mole fraction of the gas at the interface to determine the dissolved amount. Interaction depends both on the properties of the gas and the liquid, including intermolecular forces and temperature.
Temperature Dependence
Temperature plays a huge role in gas solubility. Generally, as temperature increases, gases become less soluble in liquids. This is important because Henry's constant changes with temperature. When using Henry's Law, always ensure the constant used reflects the correct temperature conditions. If the temperature changes, it can make gases dissolve more or less than expected, affecting the calculated mole fraction. Maintaining constant temperature conditions ensures that our application of Henry's Law remains accurate and reliable.

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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.

What are the adverse effects of excess moisture on the wood and metal components of a house and the paint on the walls?

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) \(6.07 \times 10^{-6} \mathrm{~kg}\)

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.

Methanol ( \(\rho=791 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\left.M=32 \mathrm{~kg} / \mathrm{kmol}\right)\) undergoes evaporation in a vertical tube with a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). At the top of the tube, the methanol concentration is zero, and its surface is \(30 \mathrm{~cm}\) from the top of the tube (Fig. P14-104). The methanol vapor pressure is \(17 \mathrm{kPa}\), with a mass diffusivity of \(D_{A B}=0.162 \mathrm{~cm}^{2} / \mathrm{s}\) in air. The evaporation process is operated at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). (a) Determine the evaporation rate of the methanol in \(\mathrm{kg} / \mathrm{h}\) and \((b)\) plot the mole fraction of methanol vapor as a function of the tube height, from the methanol surface \((x=0)\) to the top of the tube \((x=L)\).
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