Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wall made of natural rubber separates \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) gases at \(25^{\circ} \mathrm{C}\) and \(750 \mathrm{kPa}\). Determine the molar concentrations of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) in the wall.

Consider a 20 -cm-thick brick wall of a house. The indoor conditions are \(25^{\circ} \mathrm{C}\) and 50 percent relative humidity while the outside conditions are \(50^{\circ} \mathrm{C}\) and 50 percent relative humidity. Assuming that there is no condensation or freezing within the wall, determine the amount of moisture flowing through a unit surface area of the wall during a \(24-\mathrm{h}\) period.

Dry air whose molar analysis is \(78.1\) percent \(\mathrm{N}_{2}\), \(20.9\) percent \(\mathrm{O}_{2}\), and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) during the process, determine (a) the molar analysis of the saturated air and \((b)\) the density of air before and after the process. What do you conclude from your results?

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.

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Energy Physics

Read Explanation

Electrostatics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fields in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free