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Find the molar mass and the number of degrees of freedom of molecules in a gas if cV=0.65J/gK and cP=0.91J/g K.

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter2: The Kinetic Theory Of Gases

Section: Chapter Questions

Problem 71P: Find (a) the most probable speed, (b) the average speed, and (c) the rms speed for nitrogen...

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Concept explainers

Diatomic Gas

Molecule, is a collection of two or more atoms together (which constitute the smallest observable unit) into a pure material. There are some ways to represent the structure of a molecule. Factor symbols represent atoms in Lewis structures, and dots represent electrons that accompany them. A pair of mutual electrons (covalent bond) may be represented by a single dash. The ball-and-stick model depicts the spatial structure of the atoms better.

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Find the molar mass and the number of degrees of freedom of molecules in a gas if cV=0.65J/gK and cP=0.91J/g K.

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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