Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 19, Problem 45PQ
To determine
The number of molecules in the device.
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Physics for Scientists and Engineers: Foundations and Connections
Ch. 19.1 - The Fahrenheit scale remains useful in part due to...Ch. 19.2 - Prob. 19.2CECh. 19.3 - Prob. 19.3CECh. 19.3 - Prob. 19.4CECh. 19.4 - Prob. 19.5CECh. 19.5 - Prob. 19.6CECh. 19.6 - Prob. 19.7CECh. 19 - Prob. 1PQCh. 19 - Prob. 2PQCh. 19 - Prob. 3PQ
Ch. 19 - Prob. 4PQCh. 19 - Prob. 5PQCh. 19 - Prob. 6PQCh. 19 - Prob. 7PQCh. 19 - Prob. 8PQCh. 19 - Object A is placed in thermal contact with a very...Ch. 19 - Prob. 10PQCh. 19 - Prob. 11PQCh. 19 - Prob. 12PQCh. 19 - Prob. 13PQCh. 19 - The tallest building in Chicago is the Willis...Ch. 19 - Prob. 15PQCh. 19 - Prob. 16PQCh. 19 - At 22.0C, the radius of a solid aluminum sphere is...Ch. 19 - Prob. 18PQCh. 19 - Prob. 19PQCh. 19 - Prob. 20PQCh. 19 - The distance between telephone poles is 30.50 m in...Ch. 19 - Prob. 22PQCh. 19 - Prob. 23PQCh. 19 - Prob. 24PQCh. 19 - Prob. 25PQCh. 19 - Prob. 26PQCh. 19 - Prob. 27PQCh. 19 - Prob. 28PQCh. 19 - Prob. 29PQCh. 19 - Prob. 30PQCh. 19 - Prob. 31PQCh. 19 - Prob. 32PQCh. 19 - Prob. 33PQCh. 19 - Prob. 34PQCh. 19 - Prob. 35PQCh. 19 - Prob. 36PQCh. 19 - Prob. 37PQCh. 19 - Prob. 38PQCh. 19 - Prob. 39PQCh. 19 - On a hot summer day, the density of air at...Ch. 19 - Prob. 41PQCh. 19 - Prob. 42PQCh. 19 - Prob. 43PQCh. 19 - Prob. 44PQCh. 19 - Prob. 45PQCh. 19 - Prob. 46PQCh. 19 - Prob. 47PQCh. 19 - A triple-point cell such as the one shown in...Ch. 19 - An ideal gas is trapped inside a tube of uniform...Ch. 19 - Prob. 50PQCh. 19 - Prob. 51PQCh. 19 - Case Study When a constant-volume thermometer is...Ch. 19 - An air bubble starts rising from the bottom of a...Ch. 19 - Prob. 54PQCh. 19 - Prob. 55PQCh. 19 - Prob. 56PQCh. 19 - Prob. 57PQCh. 19 - Prob. 58PQCh. 19 - Prob. 59PQCh. 19 - Prob. 60PQCh. 19 - Prob. 61PQCh. 19 - Prob. 62PQCh. 19 - Prob. 63PQCh. 19 - Prob. 64PQCh. 19 - Prob. 65PQCh. 19 - Prob. 66PQCh. 19 - Prob. 67PQCh. 19 - Prob. 68PQCh. 19 - Prob. 69PQCh. 19 - Prob. 70PQCh. 19 - Prob. 71PQCh. 19 - A steel plate has a circular hole drilled in its...Ch. 19 - Prob. 73PQCh. 19 - A gas is in a container of volume V0 at pressure...Ch. 19 - Prob. 75PQCh. 19 - Prob. 76PQCh. 19 - Prob. 77PQCh. 19 - Prob. 78PQCh. 19 - Prob. 79PQCh. 19 - Prob. 80PQCh. 19 - Two glass bulbs of volumes 500 cm3 and 200 cm3 are...
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- 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 Karrow_forwardA 0.500-m3 container holding 3.00 mol of ozone (O3) is kept at a temperature of 250 K. Assume the molecules have radius r = 2.50 1010 m. What are the a. mean free path and b. mean free time between collisions for an ozone molecule in the container?arrow_forwardCylinder A contains oxygen (O2) gas, and cylinder B contains nitrogen (N2) gas. If the molecules in the two cylinders have the same rms speeds, which of the following statements is false? (a) The two gases haw different temperatures. (b) The temperature of cylinder B is less than the temperature of cylinder A. (c) The temperature of cylinder B is greater than the temperature of cylinder A. (d) The average kinetic energy of the nitrogen molecules is less than the average kinetic energy of the oxygen molecules.arrow_forward
- On a hot summer day, the density of air at atmospheric pressure at 35.0C is 1.1455 kg/m3. a. What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure? b. Avogadros number of air molecules has a mass of 2.85 102 kg. What is the mass of 1.00 m3 of air? c. Does the value calculated in part (b) agree with the stated density of air at this temperature?arrow_forwardHow many moles are there in (a) 0.0500 g of N2 gas (M = 28.0 g/mol)? (b) 10.0 g of CO2 gas (M = 44.0 g/mol)? (c) How many molecules are present in each case?arrow_forwardOne cylinder contains helium gas and another contains krypton gas at the same temperature. Mark each of these statements true, false, or impossible to determine from the given information. (a) The rms speeds of atoms in the two gases are the same. (b) The average kinetic energies of atoms in the two gases are the same. (c) The internal energies of 1 mole of gas in each cylinder are the same. (d) The pressures in the two cylinders ale the same.arrow_forward
- A sample of a monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fig. P21.65). It is warmed 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. Find (b) the temperature at point B, (c) the temperature at point C, and (d) the volume at point C. (e) Now consider the processes A B, B C, and C A. Describe how to carry out each process experimentally, (f) Find Q, W, and Eint for each of the processes, (g) For the whole cycle A B C A, find Q, W, and Eint.arrow_forwardConsider 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 P20.29ANSarrow_forward
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