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a) What is the root-mean-square speed for a collection of helium- 4 atoms at \(300 . \mathrm{K} ?\) b) What is the root-mean-square speed for a collection of helium- 3 atoms at 300 . K?

Short Answer

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a) Calculate the root-mean-square speed of a collection of helium-4 atoms at 300 K. b) Calculate the root-mean-square speed of a collection of helium-3 atoms at 300 K.

Step by step solution

01

a) Finding the root-mean-square speed of helium-4 atoms

First, we need to find the mass of the helium-4 atom. Since one atomic mass unit is approximately \(1.66 \times 10^{-27} \, \text{kg}\), the mass of helium-4 is \(m_1 = 4 \times 1.66 \times 10^{-27} \, \text{kg}\). Now, we can plug the values into the formula for root-mean-square speed: $$v_{rms_1} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \, \text{J/K}) \times 300\,\text{K}}{4 \times (1.66 \times 10^{-27} \, \text{kg})}}$$ Calculate \(v_{rms_1}\) to obtain the root-mean-square speed for a collection of helium-4 atoms at 300 K.
02

b) Finding the root-mean-square speed of helium-3 atoms

Similar to the helium-4 atoms, we first need to determine the mass of the helium-3 atom. The mass of helium-3 is \(m_2 = 3 \times 1.66 \times 10^{-27} \, \text{kg}\). Using the formula for root-mean-square speed, we can now calculate \(v_{rms_2}\) for helium-3 atoms: $$v_{rms_2} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \, \text{J/K}) \times 300\, \text{K}}{3 \times (1.66 \times 10^{-27} \, \text{kg})}}$$ Calculate \(v_{rms_2}\) to obtain the root-mean-square speed for a collection of helium-3 atoms at 300 K.

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Key Concepts

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

Kinetic Theory of Gases
The kinetic theory of gases is a model that helps us understand the behavior of gases at the molecular level. It rests on a few fundamental assumptions: gases are made up of a large number of small particles, these particles are in constant random motion, and they frequently collide with each other and the walls of their container. During these collisions, the particles exchange energy but, overall, the total energy remains constant in a closed system.

Key to this theory is the concept of temperature. In the kinetic theory of gases, temperature is a measure of the average kinetic energy of the particles in the gas. When we talk about the root-mean-square (rms) speed of gas molecules, we're referring to the square root of the average of the squares of the individual speeds of the molecules. It's a sort of 'average' speed that signifies the speed of a typical molecule in the gas and is rooted in the distribution of kinetic energy among the particles.

The calculation of the root-mean-square speed allows us to relate macroscopic properties, like temperature and pressure, to the microscopic motion of individual gas particles. It's a fundamental part of understanding how gases behave and is crucial for fields ranging from chemistry to astrophysics.
Atomic Mass Unit
The atomic mass unit (amu) is a standard unit of mass that quantifies the mass of atoms and molecules. It's defined as one twelfth of the mass of an unbound neutral atom of carbon-12 at rest and in its ground state. The amu provides a convenient way for scientists to compare the masses of different atoms on a scale that is much more manageable than using kilograms.

This unit is particularly useful when dealing with atomic and subatomic particles because their masses are so small that traditional units like grams or kilograms would result in inconveniently small numbers. For example, in the provided exercise, the atomic mass unit is used to calculate the mass of helium-4 and helium-3 atoms for the subsequent computation of root-mean-square speeds. By knowing that 1 amu ≈ 1.66 x 10^-27 kg, we can easily convert the atomic mass of an element from amu to kilograms, a necessary step when applying formulas from kinetic theory.
Speed of Helium Atoms
The speed of helium atoms in a gas, as with any gas atoms or molecules, is not uniform. Instead, there's a distribution of speeds governed by the gas's temperature and the mass of the atoms. For helium, which is a light, noble gas, the average speeds can be particularly high due to its low atomic mass.

The root-mean-square speed we calculated in the exercise is an expression of this speed distribution. It's important to note that helium-4 and helium-3 have slightly different masses, with helium-4 being heavier due to an extra neutron. This difference results in helium-3 atoms having a higher root-mean-square speed at the same temperature. It's intriguing to see how a single subatomic particle can influence the behavior of an atom on a macroscopic level, like the speed of gas particles in a container.

Understanding the principles behind the speed of gas particles is not only critical for academic studies but also practical applications, such as the design of gas-based systems and the analysis of the thermal properties of materials.

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Most popular questions from this chapter

An ideal gas may expand from an initial pressure, \(p_{\mathrm{i}},\) and volume, \(V_{\mathrm{i}},\) to a final volume, \(V_{\mathrm{f}}\), isothermally, adiabatically, or isobarically. For which type of process is the heat that is added to the gas the largest? (Assume that \(p_{i}, V_{i}\) and \(V_{f}\) are the same for each process.) a) isothermal process b) adiabatic process c) isobaric process d) All the processes have the same heat flow.

The electrons in a metal that produce electric currents behave approximately as molecules of an ideal gas. The mass of an electron is \(m_{\mathrm{e}} \doteq 9.109 \cdot 10^{-31} \mathrm{~kg} .\) If the temperature of the metal is \(300.0 \mathrm{~K},\) what is the root-mean-square speed of the electrons?

Liquid nitrogen, which is used in many physics research labs, can present a safety hazard if a large quantity evaporates in a confined space. The resulting nitrogen gas reduces the oxygen concentration, creating the risk of asphyxiation. Suppose \(1.00 \mathrm{~L}\) of liquid nitrogen \(\left(\rho=808 \mathrm{~kg} / \mathrm{m}^{3}\right)\) evaporates and comes into equilibrium with the air at \(21.0^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\). How much volume will it occupy?

A tire on a car is inflated to a gauge pressure of \(32 \mathrm{lb} / \mathrm{in}^{2}\) at a temperature of \(27^{\circ} \mathrm{C}\). After the car is driven for \(30 \mathrm{mi}\), the pressure has increased to \(34 \mathrm{lb} / \mathrm{in}^{2} .\) What is the temperature of the air inside the tire at this point? a) \(40^{\circ} \mathrm{C}\) b) \(23^{\circ} \mathrm{C}\) c) \(32^{\circ} \mathrm{C}\) d) \(54^{\circ} \mathrm{C}\)

A sample of gas at \(p=1000 . \mathrm{Pa}, V=1.00 \mathrm{~L},\) and \(T=300 . \mathrm{K}\) is confined in a cylinder. a) Find the new pressure if the volume is reduced to half of the original volume at the same temperature. b) If the temperature is raised to \(400 . \mathrm{K}\) in the process of part (a), what is the new pressure? c) If the gas is then heated to \(600 . \mathrm{K}\) from the initial value and the pressure of the gas becomes \(3000 . \mathrm{Pa},\) what is the new volume?

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