/*! 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 33 Consult Interactive Solution \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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Consult Interactive Solution \(14.33\) at to see how this problem can be solved. Very fine smoke particles are suspended in air. The translational rms speed of a smoke particle is \(2.8 \times 10^{-3} \mathrm{~m} / \mathrm{s}\), and the temperature is \(301 \mathrm{~K}\). Find the mass of a particle.

Short Answer

Expert verified
The mass of the smoke particle is approximately \(1.591 \times 10^{-15} \text{ kg}\).

Step by step solution

01

Understanding the formula for translational rms speed

The translational root-mean-square (rms) speed \(v_{rms}\) of a particle in terms of its mass \(m\) and temperature \(T\) is given by the equation of kinetic theory:\[v_{rms} = \sqrt{\frac{3kT}{m}}\]where \(k\) is the Boltzmann constant \(1.38 \times 10^{-23} \text{ J/K}\). Our goal is to find the mass \(m\) of the smoke particle.
02

Rearrange the formula to solve for mass

Rearrange the formula to solve for \(m\): \[v_{rms}^2 = \frac{3kT}{m} \implies m = \frac{3kT}{v_{rms}^2}\]This equation will allow us to find the mass \(m\) given the translational rms speed \(v_{rms}\) and the temperature \(T\).
03

Substitute known values into the equation

Substitute the known values into the equation:- Translational rms speed, \(v_{rms} = 2.8 \times 10^{-3} \, \text{m/s}\)- Temperature, \(T = 301 \, \text{K}\)- Boltzmann constant, \(k = 1.38 \times 10^{-23} \, \text{J/K}\)\[m = \frac{3 \times 1.38 \times 10^{-23} \times 301}{(2.8 \times 10^{-3})^2}\]
04

Calculate the mass

Calculate the value of \(m\) by performing the arithmetic:First, calculate the denominator:\[(2.8 \times 10^{-3})^2 = 7.84 \times 10^{-6}\]Now, calculate the numerator:\[3 \times 1.38 \times 10^{-23} \times 301 \approx 1.24782 \times 10^{-20}\]Finally, divide the numerator by the denominator:\[m = \frac{1.24782 \times 10^{-20}}{7.84 \times 10^{-6}} \approx 1.591 \times 10^{-15} \, \text{kg}\]

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Key Concepts

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

Root-Mean-Square Speed
The root-mean-square speed, often abbreviated as rms speed, is a measure of the speed of particles in a gas. It is related to the temperature and mass of the particles. In the context of kinetic theory, the rms speed provides insight into the average speed of particles based on their energy.

The formula for calculating the translational rms speed, \( v_{rms} \), is: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where \( k \) is the Boltzmann constant, \( T \) is the temperature in kelvin, and \( m \) is the mass of a particle.

Understanding this formula is crucial. It links the microscopic properties of particles (mass and speed) with a macroscopic property (temperature). The rms speed is particularly useful for describing gases, as it gives a single value representing the speed distribution of an ensemble of particles. By using rms speed, we can understand how changing temperature or the size of particles will affect their average motion.

Remember, the rms speed is related to the mean kinetic energy of particles. Essentially, the higher the temperature or the lighter the particle, the faster the rms speed.
Boltzmann Constant
The Boltzmann constant \( (k) \) plays a fundamental role in the kinetic theory of gases. Named after the Austrian physicist Ludwig Boltzmann, this constant is central to connecting the microscopic and macroscopic worlds of physics.

The value of the Boltzmann constant is \( 1.38 \times 10^{-23} \text{ J/K} \). This constant serves as a bridge between energy scales and temperature scales. It allows us to express the energy per particle in terms of temperature:
  • It equates the physical temperature of a system to the kinetic energy of its particles.
  • The constant is crucial for equations, such as \( v_{rms} = \sqrt{\frac{3kT}{m}} \), as it helps translate the thermal energy of particles into their speed.
The importance of the Boltzmann constant stretches beyond just kinetic theory. It also appears in statistical mechanics and even in entropy calculations. Nevertheless, in the context of gases, it helps us appreciate how temperature influences the behavior and speed of individual gas particles. Understanding the Boltzmann constant lets us dive deeper into the physical phenomena associated with thermal energy.
Translational Motion
Translational motion refers to the type of motion in which every point of a body moves parallel. In gases, particles move through translational motion, bouncing off each other and their surroundings. This motion is essential in understanding gas behavior, especially when using concepts from kinetic theory.

It is important to consider how translational motion translates into speed measures like rms speed. As particles move, they possess kinetic energy, which is directly related to their translational speeds. The kinetic theory equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \) reflects this energy in terms of speed.

Translational motion can explain various phenomena in gases:
  • It is why gas fills any container shape uniformly, as particles are in constant motion.
  • This random movement plays a vital role in diffusion, where particles spread from high concentration areas to low concentration areas.
  • The energy exchanges during particle collisions form the basis for temperature, as these interactions signify the average kinetic energy or speed.
Lastly, understanding translational motion is key to appreciating broader thermodynamic laws, which describe how gases behave, interact, and change under different conditions.

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Most popular questions from this chapter

Four closed tanks, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) each contain identical numbers of moles of an ideal gas. The table gives the absolute pressure and volume of the gas in each tank. Which tanks (if any) have the same temperature? Account for your answer.$$ \begin{array}{|l|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} \\ \hline \text { Absolute pressure (Pa) } & 25.0 & 30.0 & 20.0 & 2.0 \\ \hline \text { Volume (m }^{3} \text { ) } & 4.0 & 5.0 & 5.0 & 75 \\ \hline \end{array} $$ Problem Each of the tanks contains 0.10 mol of gas. Using this number and the data in the table, compute the temperature of each gas. Verify that your answers are consistent with your answer to the Concept Question.

It takes \(0.16 \mathrm{~g}\) of helium (He) to fill a balloon. How many grams of nitrogen \(\left(\mathrm{N}_{2}\right)\) would be required to fill the balloon to the same pressure, volume, and temperature?

Interactive Solution 14.24 at offers one approach to this problem. One assumption of the ideal gas law is that the atoms or molecules themselves occupy a negligible volume. Verify that this assumption is reasonable by considering gaseous xenon (Xe). Xenon has an atomic radius of \(2.0 \times 10^{-10} \mathrm{~m}\). For STP conditions, calculate the percentage of the total volume occupied by the atoms.

In a TV, electrons with a speed of \(8.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) strike the screen from behind, causing it to glow. The electrons come to a halt after striking the screen. Each electron has a mass of \(9.11 \times 10^{-31} \mathrm{~kg}\), and there are \(6.2 \times 10^{16}\) electrons per second hitting the screen over an area of \(\mathrm{m}^{2}\). What is the pressure that the electrons exert on the screen?

If the translational rms speed of the water vapor molecules \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) in air is \(648 \mathrm{~m} / \mathrm{s}\), what is the translational rms speed of the carbon dioxide molecules \(\left(\mathrm{CO}_{2}\right)\) in the same air? Both gases are at the same temperature.

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