77 SSM Figure 19-28 shows a hy- pothetical speed distribution for particles of a certain gas: P(v) = Cv2 for 0 Vg. Find (a) an expression for C in terms of vo, (b) the average speed of the particles, and (c) their rms speed. %3D Vo Speed Figure 19-28 Problem 77. (4)d

77 SSM Figure 19-28 shows a hy- pothetical speed distribution for particles of a certain gas: P(v) = Cv2 for 0 Vg. Find (a) an expression for C in terms of vo, (b) the average speed of the particles, and (c) their rms speed. %3D Vo Speed Figure 19-28 Problem 77. (4)d

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 46P: From the MaxwellBoltzmann speed distribution, show that the most probable speed of a gas molecule is...

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The product of the pressure and volume of a sample of hydrogen gas at 0.00 is 80.0 J. (a) How many moles of hydrogen are present? (b) What is the average translational kinetic energy of the hydrogen molecules? (c) What is the value of the product of pressure and volume at 200 ?
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
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