/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A He-Xe mixture containing \(0.7... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A He-Xe mixture containing \(0.75\) mole fraction of helium is used for cooling of electronics in an avionics application. At a temperature of \(300 \mathrm{~K}\) and atmospheric pressure, calculate the mass fraction of helium and the mass density, molar concentration, and molecular weight of the mixture. If the cooling system capacity is 10 liters, whut is the mass of the coolant?

Short Answer

Expert verified
The mass fraction of He is 0.084, molar concentration is 0.0406 mol/L, mass density is 1.33 g/L, and 13.3 g of coolant fills 10 L.

Step by step solution

01

Calculate the Molecular Weight of the Mixture

To find the molecular weight of the mixture, we use the mole fraction and the molecular weights of helium \(\text{He}\) and xenon \(\text{Xe}\). The molecular weights are \(4\ \text{g/mol}\) for \(\text{He}\) and \(131.29\ \text{g/mol}\) for \(\text{Xe}\). Using the formula \[\text{Molecular Weight} = (x_{\text{He}} \times M_{\text{He}}) + (x_{\text{Xe}} \times M_{\text{Xe}})\]where \(x_{\text{He}}=0.75\) and \(x_{\text{Xe}}=0.25\), we get:\[\text{Molecular Weight} = (0.75 \times 4) + (0.25 \times 131.29) = 32.8225 \ \text{g/mol}\]
02

Determine the Mass Fraction of Helium

The mass fraction of a component is given by the mass of that component divided by the total mass of the mixture. The mass of each component is the product of its mole fraction and its molecular weight.\[\text{Mass of He} = x_{\text{He}} \times M_{\text{He}} = 0.75 \times 4 = 3 \ \text{g/mol}\]\[\text{Mass of Xe} = x_{\text{Xe}} \times M_{\text{Xe}} = 0.25 \times 131.29 = 32.8225 \ \text{g/mol}\]Therefore, the mass fraction of helium is:\[\text{Mass Fraction of He} = \frac{3}{3 + 32.8225} = 0.084\]
03

Calculate Molar Concentration

Molar concentration \(C\) is defined as the number of moles per unit volume. At 1 atm and 300 K, using the ideal gas law \(PV = nRT\), for 1 mole we have:- \(R = 0.0821 \ \text{L atm/mol K}\)To find volume of 1 mole: \[V = \frac{RT}{P} = \frac{0.0821 \times 300}{1} = 24.63 \text{L/mol}\]Thus, molar concentration is:\[C = \frac{1}{24.63} = 0.0406 \text{mol/L}\]
04

Compute the Mass Density

Mass density \(\rho\) is mass per unit volume. It can be calculated by multiplying the molar concentration \(C\) by the molecular weight of the gas mixture. \[\text{Mass Density} = C \times \text{Molecular Weight} = 0.0406 \ \text{mol/L} \times 32.8225 \ \text{g/mol} = 1.33 \ \text{g/L}\]
05

Find the Mass of Coolant in Cooling System

The mass of the coolant can be calculated by multiplying the volume capacity of the system by the mass density of the mixture.\[\text{Mass of Coolant} = \text{Volume} \times \text{Mass Density} = 10 \ \text{L} \times 1.33 \ \text{g/L} = 13.3 \ \text{g}\]

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Molecular Weight
Understanding molecular weight is crucial in chemistry, especially when dealing with gas mixtures. The molecular weight of a mixture is a weighted average based on the mole fraction and the individual molecular weights of its components. Each gas contributes to the total molecular weight proportionally to its presence in the mixture. In our specific helium (He) and xenon (Xe) mixture, helium has a mole fraction of 0.75 while xenon has 0.25. The molecular weight of each gas is 4 g/mol for helium and 131.29 g/mol for xenon. By calculating \[\text{Molecular Weight} = (0.75 \times 4) + (0.25 \times 131.29),\]we find the molecular weight of the mixture to be 32.8225 g/mol. This value helps us understand how heavy or light the gas molecules are on average.
Mass Fraction
The mass fraction is an important concept in chemistry that tells us how much of the total mass of the mixture is made up by each component. To find the mass fraction of helium, we first calculate the mass of each component using their mole fractions and molecular weights. Helium contributes 3 g/mol, while xenon contributes 32.8225 g/mol. The mass fraction of helium is calculated using \[\text{Mass Fraction of He} = \frac{3}{3 + 32.8225},\]resulting in 0.084. This shows that helium makes up only a small portion of the mixture's mass. Mass fraction is useful when assessing the proportionate presence of different elements in a compound.
Molar Concentration
Molar concentration, also known as molarity, refers to the number of moles of a solute per liter of solution. For gases, it is determined using the ideal gas law, which is a fundamental concept in chemistry describing how gases behave under different conditions. At 300 K and 1 atm, the ideal gas law \(PV = nRT\) helps us find the volume occupied by one mole of gas. With R as 0.0821 L atm/mol K, we calculate \[V = \frac{RT}{P} = \frac{0.0821 \times 300}{1} = 24.63 \text{ L/mol}.\]Thus, the molar concentration of the mixture is \[C = \frac{1}{24.63} = 0.0406 \text{ mol/L}.\]Molar concentration is vital for calculating reactant quantities in chemical reactions and processes.
Mass Density
Mass density is defined as the mass per unit volume of a substance. It's an essential property that allows us to compare how much mass is contained in a given volume of a material, which is especially applicable in gases. For our helium-xenon mixture, the mass density is determined by multiplying the molar concentration by the molecular weight of the mixture: \[\text{Mass Density} = 0.0406 \times 32.8225 = 1.33 \text{ g/L}.\]This means each liter of the gas mixture weighs 1.33 grams. Mass density helps engineers design systems where specific weight and capacity considerations are crucial, such as in cooling systems for electronics. Understanding this concept ensures efficient material usage and optimal system performance.

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

Hydrogen at a pressure of 2 atm flows within a tube of diameter \(40 \mathrm{~mm}\) and wall thickness \(0.5 \mathrm{~mm}\). The outer surface is exposed to a gas stream for which the hydrogen partial pressure is \(0.1\) atm. The mass diffusivity and solubility of hydrogen in the tube material are \(1.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) and \(160 \mathrm{kmol} / \mathrm{m}^{3}\). atm. respectively. When the system is at \(500 \mathrm{~K}\), what is the rate of hydrogen transfer through the tube per unit length \((\mathrm{kg} / \mathrm{s}\) · \(\mathrm{m})\) ?

Consider air in a closed, cylindrical container with its axis vertical and with opposite ends maintained at different temperatures. Assume that the total pressure of the air is uniform throughout the container. (a) If the bottom surface is colder than the top surface, what is the nature of conditions within the container? For example, will there be vertical gradients of the species \(\left(\mathrm{O}_{2}\right.\) and \(\left.\mathrm{N}_{2}\right)\) concentrations? Is there any motion of the air? Does mass transfer occur? (b) What is the nature of conditions within the container if it is inverted (i.e., the warm surface is now at the bottom)?

Assuming air to be composed exclusively of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), with their partial pressures in the ratio 0.21:0.79, what are their mass fractions?

polycarbonate to reduce mamufacturing cosis. Assume that a firs-onder homogeneous chemical reaction takes place between the polymer and oxygen; the reaction rate is proportional to the oxygen molar concentration. (a) Write the governing equation. boundary conditions, and initial condition for the oxygen molar concentration after the DVD is removed from the oxyeen- proof pouch, for a DVD of thickness \(2 \mathrm{~L}\). Do not solve. (b) The DVD will gradually become more opaque over time as the reaction proceeds. The ability to read the DVD will depend on how well the laser light can penetrate through the thickness of the DVD. Therefore, it is important to know the volume-averaged molar concentration of product, \(\bar{C}_{\text {moe }}\) as a function of time. Write an expression for \(\bar{C}_{\text {pued }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.

A spherical droplet of liquuid \(A\) and radius \(r_{e}\) evaporates into a stagnant layer of gas B. Derive an expession for the evaporation rate of species \(\mathrm{A}\) in terms of the saturation pressure of species A, \(p_{\mathrm{A}}\left(r_{\mathrm{n}}\right)=p_{\mathrm{A} \text { at }}\) the partial pressure of species \(\mathrm{A}\) at an arbitrary radius \(r_{\text {. }}\) \(p_{A}(r)\), the total pressure \(p\), and other pertinent quantities. Assume the droplet and the mixture are at a uniform pressure \(p\) and temperature \(T\).

See all solutions

Recommended explanations on Physics Textbooks

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.