Answered: 33. ssm mmh Very fine smoke particles…

33. ssm mmh Very fine smoke particles are suspended in air. The trans- lational rms speed of a smoke particle is 2.8 × 10-3 m/s, and the temper- ature is 301 K. Find the mass of a particle.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter16: Temperature And The Kinetic Theory Of Gases

Section: Chapter Questions

Problem 9OQ: A gas is at 200 K. If we wish to double the rms speed of the molecules of the gas, to what value...

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Question

33. ssm mmh Very fine smoke particles are suspended in air. The trans-
lational rms speed of a smoke particle is 2.8 × 10-3 m/s, and the temper-
ature is 301 K. Find the mass of a particle.

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(a) How many atoms of helium gas fill a spherical balloon of diameter 30.0 cm at 20.0C and 1.00 atm? (b) What is the average kinetic energy of the helium atoms? (c) What is the rms speed of the helium atoms?
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Consider the Maxwell-Boltzmann distribution function plotted in Problem 28. For those parameters, determine the rms velocity and the most probable speed, as well as the values of f(v) for each of these values. Compare these values with the graph in Problem 28. 28. Plot the Maxwell-Boltzmann distribution function for a gas composed of nitrogen molecules (N2) at a temperature of 295 K. Identify the points on the curve that have a value of half the maximum value. Estimate these speeds, which represent the range of speeds most of the molecules are likely to have. The mass of a nitrogen molecule is 4.68 1026 kg. Equation 20.18 can be used to find the rms velocity given the temperature, Boltzmanns constant, and the mass of the atom or molecule. The mass of a nitrogen molecule is 4.68 1026 kg. vrms=3kBTm=3(1.381023J/K)4.681026kg=511m/s Using the results of Problem 28 and the rms velocity, we can calculate the value of f(v). f(vrms) = (3.11 108)(511)2 e(5.75106(511)2) = 0.00181 The most probable speed, for which this function has its maximum value, is given by Equation 20.20. vmp=2kBTm=2(1.381023J/K)(295K)4.681026kg=417m/s f(vmp) = (3.11108)(417)2 e(5.75106(417)2) = 0.00199 We plot these points on the speed distribution. The most probable speed is indeed at the peak of the distribution function. Since the function is not symmetric, the rms velocity is somewhat higher than the most probable speed. Figure P20.29ANS
A gas is at 200 K. If we wish to double the rms speed of the molecules of the gas, to what value must we raise its temperature? (a) 283 K (b) 400 K (c) 566 K (d) 800 K (e) 1 130 K
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