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Chapter 19: Problem 35

Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s}\), two at \(500 \mathrm{~m} / \mathrm{s}\), and four at \(600 \mathrm{~m} / \mathrm{s}\). Calculate their (a) average and (b) rms speeds. (c) Is \(v_{\text {rms }}>v_{\text {avg }}\) ?

Short Answer

Expert verified
(a) 420 m/s, (b) 458.26 m/s, (c) Yes, \(v_{\text{rms}} > v_{\text{avg}}\).

Step by step solution

01

Calculate Total Speeds Summation

We need to find the sum of all the particle speeds. Given four particles at 200 m/s, two particles at 500 m/s, and four particles at 600 m/s, the total speed summation is:\[4 \times 200 + 2 \times 500 + 4 \times 600 = 800 + 1000 + 2400 = 4200 \text{ m/s}\]
02

Calculate Average Speed

The average speed is the total speed divided by the number of particles. There are ten particles:\[v_{\text{avg}} = \frac{4200}{10} = 420 \text{ m/s}\]
03

Calculate the Sum of Speeds Squared

To find the rms speed, first calculate the sum of the squares of their speeds:\[(4 \times 200^2) + (2 \times 500^2) + (4 \times 600^2) = 4 \times 40000 + 2 \times 250000 + 4 \times 360000 = 160000 + 500000 + 1440000 = 2100000\]
04

Calculate RMS Speed

The rms speed is the square root of the average of the squared speeds sum:\[v_{\text{rms}} = \sqrt{\frac{2100000}{10}} = \sqrt{210000} \approx 458.26 \text{ m/s}\]
05

Compare Average and RMS Speeds

To decide if \(v_{\text{rms}} > v_{\text{avg}}\), compare the values:\[\]Given \(v_{\text{rms}} \approx 458.26 \text{ m/s}\) and \(v_{\text{avg}} = 420 \text{ m/s}\), it is clear that \(v_{\text{rms}} > v_{\text{avg}}\).

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Key Concepts

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

Particle Motion
Understanding particle motion involves comprehending how particles move through space with varying speeds. In this context, we are dealing with ten particles moving with assorted speeds. These speeds are categorized into three groups: four particles at 200 m/s, two particles at 500 m/s, and four particles at 600 m/s. Each particle's speed contributes to the overall dynamics of the system. When dealing with problems involving particle speeds, it is essential to understand how the distribution of individual particle speeds impacts the calculation of average and root mean square (rms) speeds.
Average Speed Calculation
The average speed calculation gives an overall idea of how fast the particles are moving collectively. To find the average speed (\(v_{\text{avg}}\)), sum up the individual speeds and divide by the total number of particles. In our problem, the sum of all particle speeds is calculated as:
  • 4 particles at 200 m/s contribute 800 m/s (4 x 200).
  • 2 particles at 500 m/s contribute 1000 m/s (2 x 500).
  • 4 particles at 600 m/s contribute 2400 m/s (4 x 600).
So, the total sum of speeds is 4200 m/s. Thus, the average speed is:\[v_{\text{avg}} = \frac{4200}{10} = 420 \text{ m/s}\]This value represents an central trend around which particle speeds are distributed.
RMS Speed Calculation
The root mean square speed calculation is slightly more involved than average speed. RMS provides insights into the average magnitude of the velocities, offering a glimpse into energy distributions. To calculate the rms speed (\(v_{\text{rms}}\)), perform the following steps:
  • First, calculate the square of each speed, then multiply by the number of particles at each speed.
  • Square these results and sum them up: \((4 \times 200^2) + (2 \times 500^2) + (4 \times 600^2)\) becomes 2100000.
  • Next, average this sum over the number of particles by dividing by 10, giving 210000.
  • Finally, take the square root of the result: \[v_{\text{rms}} = \sqrt{210000} \approx 458.26 \text{ m/s}\]
The rms speed reflects a greater sensitivity to higher speeds due to the squaring process.
Comparison of Speeds
Comparing the average speed and rms speed reveals significant insights into the distribution of particle speeds. In our exercise, the calculated average speed is 420 m/s, while the rms speed is approximately 458.26 m/s. This comparison shows that:
  • RMS speed (\(v_{\text{rms}}\)) is greater than the average speed (\(v_{\text{avg}}\)).
  • This relationship holds true because rms speed emphasizes larger speeds due to the squaring effect used in its calculation.
This contrast in speeds provides valuable information about the variability in speeds among the particles. High rms speed compared to average speed suggests a wider range of speeds in the group, potentially indicating the presence of faster-moving particles within the ensemble.

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Most popular questions from this chapter

In an interstellar gas cloud at \(50.0 \mathrm{~K}\), the pressure is \(1.00 \times 10^{-8} \mathrm{~Pa}\). Assuming that the molecular diameters of the gases in the cloud are all \(20.0 \mathrm{~nm}\), what is their mean free path?

An ideal gas undergoes isothermal compression from an initial volume of \(4.00 \mathrm{~m}^{3}\) to a final volume of \(3.00 \mathrm{~m}^{3}\). There is \(3.50 \mathrm{~mol}\) of the gas, and its temperature is \(10.0^{\circ} \mathrm{C}\). (a) How much work is done by the gas? (b) How much energy is transferred as heat between the gas and its environment?

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at \(1.0 \mathrm{~atm}\) pressure and \(0.00^{\circ} \mathrm{C} ?\) The molecular diameter is \(3.0 \times 10^{-8} \mathrm{~cm}\).

In a certain particle accelerator, protons travel around a circular path of diameter \(23.0 \mathrm{~m}\) in an evacuated chamber, whose residual gas is at \(295 \mathrm{~K}\) and \(1.00 \times 10^{-6}\) torr pressure. (a) Calculate the number of gas molecules per cubic centimeter at this pressure. (b) What is the mean free path of the gas molecules if the molecular diameter is \(2.00 \times 10^{-8} \mathrm{~cm} ?\)

In an industrial process the volume of \(25.0 \mathrm{~mol}\) of a monatomic ideal gas is reduced at a uniform rate from \(0.616 \mathrm{~m}^{3}\) to \(0.308 \mathrm{~m}^{3}\) in \(2.00 \mathrm{~h}\) while its temperature is increased at a uniform rate from \(27.0^{\circ} \mathrm{C}\) to \(450^{\circ} \mathrm{C}\). Throughout the process, the gas passes through thermodynamic equilibrium states. What are (a) the cumulative work done on the gas, (b) the cumulative energy absorbed by the gas as heat, and (c) the molar specific heat for the process? (Hint: To evaluate the integral for the work, you might use $$\int \frac{a+b x}{A+B x} d x=\frac{b x}{B}+\frac{a B-b A}{B^{2}} \ln (A+B x)$$ an indefinite integral.) Suppose the process is replaced with a twostep process that reaches the same final state. In step 1 , the gas volume is reduced at constant temperature, and in step 2 the temperature is increased at constant volume. For this process, what are (d) the cumulative work done on the gas, (e) the cumulative energy absorbed by the gas as heat, and (f) the molar specific heat for the process?
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