/*! 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 37 Fifteen identical particles have... [FREE SOLUTION] | 91Ó°ÊÓ

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Fifteen identical particles have various speeds: one has a speed of \(2.00 \mathrm{m} / \mathrm{s} ;\) two have speeds of \(3.00 \mathrm{m} / \mathrm{s} ;\) three have speeds of \(5.00 \mathrm{m} / \mathrm{s} ;\) four have speeds of \(7.00 \mathrm{m} / \mathrm{s}\) three have speeds of \(9.00 \mathrm{m} / \mathrm{s} ;\) and two have speeds of \(12.0 \mathrm{m} / \mathrm{s} .\) Find (a) the average speed, (b) the rms speed, and (c) the most probable speed of these particles.

Short Answer

Expert verified
The average speed of the particles is 6.80 m/s. The rms speed of the particles is approximately 4.92 m/s. The most probable speed of the particles is 7.00 m/s.

Step by step solution

01

Compute the total speed

Sum up the overall speed of all particles. Using the given speeds and number of particles, speeds of all particles are: one particle at 2.00 m/s (1 x 2.00 = 2.00 m/s), two particles at 3.00 m/s (2 x 3 = 6.00 m/s), three particles at 5.00 m/s (3 x 5 = 15.00 m/s), four particles at 7.00 m/s (4 x 7 = 28.00 m/s), three particles at 9.00 m/s (3 x 9 = 27.00 m/s), and two particles at 12.00 m/s (2 x 12= 24.00 m/s). As a result, the total speed equals 2.00 + 6.00 + 15.00 + 28.00 + 27.00 + 24.00 = 102.00 m/s.
02

Compute the average speed

The average speed is computed by dividing the total speed by the total number of particles. Given that there are 15 particles in total, the average speed equals 102.00 m/s / 15 = 6.80 m/s.
03

Calculate the rms speed

The root mean square (rms) speed is computed by finding the square root of the average of the squares of the speeds. First compute the sum of the squares of all speed values. This is done by squaring the speeds and adding them up: 1 * (2^2) + 2 * (3^2) + 3 * (5^2) + 4 * (7^2) + 3 * (9^2) + 2 * (12^2) = 364. Then, divide this total by the number of particles (364 / 15 = 24.26). Finally, find the square root of this result: \(\sqrt{24.26}\) = 4.92 m/s.
04

Determine the most probable speed

The most probable speed is the one that the largest number of particles have. Here, the largest number of particles (four) have a speed of 7.00 m/s, so this is the most probable speed.

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Key Concepts

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

Average Speed
When we talk about the average speed of gas particles, we're referring to the mean value of their speeds. It’s a way of summarizing the overall motion of a large number of particles in a single, simple figure. This concept is important because it allows us to characterize the movement of particles in a gas without having to describe each one individually.

To find the average speed, as in the provided textbook example, you sum up the speeds of all particles and then divide by the total number of particles. In this case, the calculation led to an average speed of 6.80 meters per second (m/s). It provides a macroscopic view of the particles’ velocities, which is particularly useful when you’re dealing with a large number of particles, as in samples of gases.
Root Mean Square (rms) Speed
The root mean square (rms) speed is a particularly important concept in the kinetic theory of gases. It represents a sort of 'average' speed in that it gives us a single value to describe the motion of particles. However, rather than taking a simple mean, it takes into account the square of the speeds.

The reason we use rms speed, rather than average, is because it factors in the kinetic energy of the particles, which is proportional to the square of the velocity. To calculate the rms speed, you square each speed, average those squares, and then take the square root. In the exercise, rms speed was calculated to be approximately 4.92 m/s. This speed is often used in physics to describe the energy of the particles in a gas.
Most Probable Speed
The most probable speed is a term used to describe the speed that is most commonly found among particles in a gas. It is a mode of the distribution of speeds and can be thought of as the peak on a graph of the number of particles versus speed.

In the given exercise, because the largest number of particles (four) share a speed of 7.00 m/s, this is declared as the most probable speed. It's essential to understand that the most probable speed is not the same as average or rms speed; it's the value most frequently occurring within the set of data and is highly useful when predicting the behavior of a gas under given conditions.
Statistical Mechanics
The study and application of these various speeds fall under the branch of physics known as statistical mechanics. This field provides a framework for relating the microscopic properties of individual atoms and molecules to the macroscopic or bulk properties of materials that can be observed on the human scale, like temperature and pressure.

Link to Kinetic Theory

Statistical mechanics provides the foundation for the kinetic theory of gases, which describes a gas as a large number of submicroscopic particles (atoms or molecules), all of which are in constant, random motion. The theory explains how the microscopic properties of individual gas particles, such as speed and energy, translate into the macroscopic properties that we can measure, like temperature and volume.

Understanding these concepts allows us to make predictions about the behavior of gases, and by extension many other phases of matter, which are invaluable to the fields of thermodynamics, engineering, chemistry, and biophysics.

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Most popular questions from this chapter

(a) Show that \(1 \mathrm{Pa}=1 \mathrm{J} / \mathrm{m}^{3} .\) (b) Show that the density in space of the translational kinetic energy of an ideal gas is \(3 P / 2.\)

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A sample of monatomic ideal gas occupies \(5.00 \mathrm{L}\) at atmospheric pressure and \(300 \mathrm{K}\) (point \(A\) in Figure \(\mathrm{P} 21.67\) ). It is heated at constant volume to 3.00 atm (point \(B\) ). Then it is allowed to expand isothermally to 1.00 atm (point \(C\) ) and at last compressed isobarically to its original state. (a) Find the number of moles in the sample. (b) Find the temperature at points \(B\) and \(C\) and the volume at point \(C\). (c) Assuming that the molar specific heat does not depend on temperature, so that \(E_{\text {int }}=3 n R T / 2,\) find the internal energy at points \(A, B,\) and \(C\) (d) Tabulate \(P, V, T,\) and \(E_{\text {int }}\) for the states at points \(A, B,\) and \(C .\) (e) Now consider the processes \(A \rightarrow B, B \rightarrow C\), and \(C \rightarrow A\). Describe just how to carry out each process experimentally. (f) Find \(Q, W,\) and \(\Delta E_{\text {int }}\) for each of the processes. (g) For the whole cycle \(A \rightarrow B \rightarrow C \rightarrow A\) find \(Q, W,\) and \(\Delta E_{\text {int }}.\)

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