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Chapter 11: Problem 23

In Part IV you'll learn to calculate that 1 mole \((6.02 \times\) \(10^{23}\) atoms) of helium atoms in the gas phase has \(3700 \mathrm{J}\) of microscopic kinetic energy at room temperature. If we assume that all atoms move with the same speed, what is that speed? The mass of a helium atom is \(6.68 \times 10^{-27} \mathrm{kg}.\)

Short Answer

Expert verified
The speed of helium atoms at room temperature is approximately 1360.20 m/s.

Step by step solution

01

Understanding problem description and applicable formula

From the problem description, we know that the microscopic kinetic energy of 1 mole of helium atoms is 3700 J and the mass of one helium atom is \(6.68 \times 10^{-27}\) kg. The kinetic energy is given by: \(Ke = 0.5nMv^2\). Our task is to solve for the speed (v). We first need to convert the mass of a helium atom to the mass of a mole of helium atoms.
02

Converting to the mass of a mole of helium atoms

As we know that 1 mole equals to \(6.02 \times 10^{23}\) atoms. We can calculate mass of 1 mole helium atoms: \(Mass = 6.68 \times 10^{-27} \mathrm{kg/atom} \times 6.02 \times 10^{23} \mathrm{atoms/mole} = 4.02 \times 10^{-3} \mathrm{kg/mole}\).
03

Solving for speed

Rearrange the kinetic energy equation to solve for v: \(v = \sqrt{Kinetic Energy \div 0.5 \times number of moles \times Molar mass}\). Substitute given values: \(v = \sqrt{3700 J \div 0.5 \times 1 mole \times 4.02 \times 10^{-3} \mathrm{kg/mole}} = 1360.20 m/s\). The speed of helium atoms at room temperature is approximately 1360.20 m/s.

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Key Concepts

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

Molar Mass of Helium
The molar mass of a substance is the mass of one mole of that substance. In the case of helium, which is a noble gas and one of the lightest elements, its molar mass is particularly significant in calculations involving gases.

Helium has an atomic mass of approximately 4.00 grams per mole (g/mol). In scientific notation, this is often expressed as 4.00 x 10^-3 kilograms per mole (kg/mol), since the mass of a single atom of helium is incredibly small. The step-by-step solution demonstrates the process of converting the atomic mass to the molar mass, which is essential in determining the kinetic energy of a gas.

Understanding molar mass is crucial when solving problems in physical chemistry, especially those related to gas laws and thermodynamics. It represents the bridge between the microscopic world of atoms and the macroscopic world of grams and kilograms that we can measure and observe.
Avogadro's Number
Avogadro's number, which is approximately 6.02 x 10^23, is a fundamental constant in chemistry that represents the number of atoms, ions, or molecules in one mole of a substance. Named after the Italian scientist Amedeo Avogadro, this number allows us to connect the microscopic atomic scale with the macroscopic scale in a quantifiable way.

Avogadro's number is utilized in the solution to convert the mass of a single helium atom to the mass of one mole of helium atoms. This conversion is vital because it aligns the microscopic kinetic energy of the individual atoms with the molar kinetic energy that we can calculate and measure experimentally.

Students often encounter Avogadro's number in stoichiometry calculations and when dealing with gas laws. It provides a consistent method to translate between the amount of substance and its associated mass or number of particles, which are central to many areas of chemistry and physics.
Root-Mean-Square Speed of Gas Particles
The root-mean-square (RMS) speed of gas particles is a measure of the average speed of the particles in a gas. It is grounded in the kinetic theory of gases, which postulates that gas particles are in constant, random motion. The RMS speed is an important concept because it correlates to the temperature and kinetic energy of the gas.

As illustrated in the solution, to determine the RMS speed, you use the formula that relates kinetic energy to the mass and speed of the particles: Ke = 0.5nMv^2. Once you have the total kinetic energy and the molar mass, you can solve for the RMS speed (v).

This speed is not the actual speed of each particle but rather an average speed that represents the energy distribution among the particles. It's an important parameter in understanding the behavior of gases, particularly in predicting the outcomes of reactions involving gases, and also in applications of thermodynamics such as engine efficiency and gas diffusion rates.

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Most popular questions from this chapter

A baggage handler throws a \(15 \mathrm{kg}\) suitcase horizontally along the floor of an airplane luggage compartment with an initial speed of \(1.2 \mathrm{m} / \mathrm{s}\). The suitcase slides \(2.0 \mathrm{m}\) before stopping. Use work and energy to find the suitcase's coefficient of kinetic friction on the floor.

a. A \(50 \mathrm{g}\) ice cube can slide without friction up and down a \(30^{\circ}\) slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring \(10 \mathrm{cm} .\) The spring constant is \(25 \mathrm{N} / \mathrm{m}\). When the ice cube is released, what total distance will it travel up the slope before reversing direction? b. The ice cube is replaced by a 50 g plastic cube whose coefficient of kinetic friction is \(0.20 .\) How far will the plastic cube travel up the slope?

A 50 kg ice skater is gliding along the ice, heading due north at \(4.0 \mathrm{m} / \mathrm{s} .\) The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but \(\mu_{\mathbf{k}}=0 .\) Suddenly, a wind from the northeast exerts a force of \(4.0 \mathrm{N}\) on the skater. a. Use work and energy to find the skater's speed after gliding \(100 \mathrm{m}\) in this wind. b. What is the minimum value of \(\mu\), that allows her to continue moving straight north?

A \(10.2 \mathrm{kg}\) weather rocket generates a thrust of \(200 \mathrm{N}\). The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is \(500 \mathrm{N} / \mathrm{m},\) is anchored to the ground. a. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed? b. After the engine is ignited, what is the rocket's speed when the spring has stretched \(40 \mathrm{cm} ?\) For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

Sam, whose mass is \(75 \mathrm{kg}\), straps on his skis and starts down a 50-m-high, \(20^{\circ}\) frictionless slope. A strong headwind exerts a horizontal force of \(200 \mathrm{N}\) on him as he skies. Find Sam's speed at the bottom (a) using work and energy, (b) using Newton's laws.
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