Physics for Scientists and Engineers: Foundations and Connections
15th Edition
ISBN: 9781305289963
Author: Debora M. Katz
Publisher: Cengage Custom Learning
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Chapter 20, Problem 12PQ
To determine
The rms speed of Hydrogen molecules at
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Physics for Scientists and Engineers: Foundations and Connections
Ch. 20.2 - In Example 20.1, we found that the rms value of a...Ch. 20.3 - If the temperature of a gas is doubled, what...Ch. 20.3 - Prob. 20.3CECh. 20.5 - Prob. 20.4CECh. 20.7 - Prob. 20.5CECh. 20.8 - Prob. 20.6CECh. 20 - Prob. 1PQCh. 20 - Prob. 2PQCh. 20 - Prob. 3PQCh. 20 - Prob. 4PQ
Ch. 20 - Prob. 5PQCh. 20 - Prob. 6PQCh. 20 - Prob. 7PQCh. 20 - Prob. 8PQCh. 20 - Particles in an ideal gas of molecular oxygen (O2)...Ch. 20 - Prob. 10PQCh. 20 - Prob. 11PQCh. 20 - Prob. 12PQCh. 20 - Prob. 13PQCh. 20 - Prob. 14PQCh. 20 - The mass of a single hydrogen molecule is...Ch. 20 - Prob. 16PQCh. 20 - The noble gases neon (atomic mass 20.1797 u) and...Ch. 20 - Prob. 18PQCh. 20 - Prob. 19PQCh. 20 - Prob. 20PQCh. 20 - Prob. 22PQCh. 20 - Prob. 23PQCh. 20 - Prob. 24PQCh. 20 - Prob. 25PQCh. 20 - Prob. 26PQCh. 20 - Prob. 27PQCh. 20 - Prob. 28PQCh. 20 - Consider the Maxwell-Boltzmann distribution...Ch. 20 - Prob. 30PQCh. 20 - Prob. 31PQCh. 20 - Prob. 32PQCh. 20 - Prob. 33PQCh. 20 - Prob. 34PQCh. 20 - Prob. 35PQCh. 20 - Prob. 36PQCh. 20 - Prob. 37PQCh. 20 - Prob. 38PQCh. 20 - Prob. 39PQCh. 20 - Prob. 40PQCh. 20 - Prob. 41PQCh. 20 - Prob. 42PQCh. 20 - Prob. 43PQCh. 20 - Prob. 44PQCh. 20 - Figure P20.45 shows a phase diagram of carbon...Ch. 20 - Prob. 46PQCh. 20 - Prob. 47PQCh. 20 - Consider water at 0C and initially at some...Ch. 20 - Prob. 49PQCh. 20 - Prob. 50PQCh. 20 - Prob. 51PQCh. 20 - Prob. 52PQCh. 20 - Prob. 53PQCh. 20 - Prob. 54PQCh. 20 - Prob. 55PQCh. 20 - Prob. 56PQCh. 20 - Consider again the box and particles with the...Ch. 20 - Prob. 58PQCh. 20 - The average kinetic energy of an argon atom in a...Ch. 20 - For the exam scores given in Table P20.60, find...Ch. 20 - Prob. 61PQCh. 20 - Prob. 62PQCh. 20 - Prob. 63PQCh. 20 - Prob. 64PQCh. 20 - Prob. 65PQCh. 20 - Prob. 66PQCh. 20 - Determine the rms speed of an atom in a helium...Ch. 20 - Consider a gas filling two connected chambers that...Ch. 20 - Prob. 69PQCh. 20 - Prob. 70PQCh. 20 - A 0.500-m3 container holding 3.00 mol of ozone...Ch. 20 - Prob. 72PQCh. 20 - Prob. 73PQ
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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_forwardUsing the approximation v1v1+v f(v)dvf(v1)v for small v , estimate the fraction of nitrogen molecules at a temperature of 3.00102 K that have speeds between 290 m/s and 291 m/s.arrow_forwardWhat is the average velocity of the air molecules in the room where you are right now?arrow_forward
- Find (a) the most probable speed, (b) the average speed, and (c) the rms speed for nitrogen molecules at 295 K.arrow_forwardHelium gas is in thermal equilibrium with liquid helium at 4.20 K. Even though it is on the point of condensation, model the gas as ideal and determine the most probable speed of a helium atom (mass = 6.64 1027 kg) in it.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
- Cylinder 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_forwardAt what temperature is the average speed of carbon dioxide molecules ( M=44.0 g/mol) 510 m/s?arrow_forwardWhat is the gauge pressure inside a tank of 4.86104 mol of compressed nitrogen with a volume of 6.56 m3 if the rms speed is 514 m/s?arrow_forward
- Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of 5.00 m/s, four have speeds of 7.00 m/s, three have speeds of 9.00 m/s, and two have speeds of 12.0 m/s. Find (a) the average speed, (b) the rms speed, and (c) the most probable speed of these particles.arrow_forwardFind the ratio f(vp)/f(vrms) for hydrogen gas ( M=2.02 g/mol) at a temperature of 77.0 K.arrow_forwardReview. This problem is a continuation of Problem 16.29 in Chapter 16. A hot-air balloon consists of an envelope of constant volume 400 m3. Not including the air inside, the balloon and cargo have mass 200 kg. The air outside and originally inside is a diatomic ideal gas at 10.0C and 101 kPa, with density 1.25 kg/m3. A propane burner at the center of the spherical envelope injects energy into the air inside. The air inside stays at constant pressure. Hot air, at just the temperature required to make the balloon lift off, starts to fill the envelope at its closed top, rapidly enough so that negligible energy flows by heat to the cool air below it or out through the wall of the balloon. Air at 10C leaves through an opening at the bottom of the envelope until the whole balloon is filled with hot air at uniform temperature. Then the burner is shut off and the balloon rises from the ground. (a) Evaluate the quantity of energy the burner must transfer to the air in the balloon. (b) The heat value of propanethe internal energy released by burning each kilogramis 50.3 MJ/kg. What mass of propane must be burned?arrow_forward
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