Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 20, Problem 44PQ
To determine
The explanation for which the Van der Waals equation reduced to
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Give the temperature T of 1 mole of ideal gas as a function of the pressure P, volume V, and the gas constant R and give the internal energy U of a rigid diatomic ideal gas as a function of its temperature T and the gas constant R.
A diatomic ideal gas at pressure p and volume V is expanding to three times its initial volume under constant pressure. W=2pV.
a) If the initial temperature equals T, express the final temperature Tf in terms of the original temperature T.
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The only form of energy possessed by molecules of a monatomic ideal gas is translational kinetic energy. Using the results from the discussion of kinetic theory in Section 10.5, show that the internal energy of a monatomic ideal gas at pressure P and occupying volume V may be written as U = 3/2PV
Chapter 20 Solutions
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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- 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 P20.29ANSarrow_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_forwardFrom the MaxwellBoltzmann speed distribution, show that the most probable speed of a gas molecule is given by Equation 16.23. Note: The most probable speed corresponds to the point at which the slope of the speed distribution curve dNv/dv is zero.arrow_forward
- Suppose 4.00 mol of an ideal diatomic gas, with molecular rotation but not oscillation, experienced a temperature increase of 60.0 K under constant-pressure conditions. What are (a) the energy transferred as heat Q, (b) the change Eint in internal energy of the gas, (c) the work W done by the gas, and (d) the change K in the total translational kinetic energy of the gas?arrow_forwardHow many molecules are present in a sample of an ideal gas that occupies a volume of 1.70 cm3, is at a temperature of 20°C, and is at atmospheric pressure? How many molecules of the gas are present if the volume and temperature are the same as in part (a), but the pressure is now 2.80 ✕ 10−11 Pa (an extremely good vacuum)?arrow_forwardThe density of helium gas at 0°C is ρ0 = 0.179 kg/m3. The temperature is then raised to T = 170°C, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density ρf of the gas. kg/m3arrow_forward
- Expand 1.00 mol of an monatomic gas initially at 5.00 kPaand 600 K from initial volume Vi = 1.00 m3 to final volume Vf =2.00 m3. At any instant during the expansion, the pressure p and volumeV of the gas are related by p = 5.00 exp[(Vi - V)/a], with p inkilopascals, Vi and V in cubic meters, and a = 1.00 m3.What are thefinal (a) pressure and (b) temperature of the gas? (c) How muchwork is done by the gas during the expansion? (d) What is S for theexpansion? (Hint: Use two simple reversible processes to find S.)arrow_forwardA sealed cubical container 11.0 cm on a side contains a gas with five times Avogadro's number of neon atoms at a temperature of 25.0°C. (a) Find the internal energy (in J) of the gas. J (b) The total translational kinetic energy (in J) of the gas. J (c)Calculate the average kinetic energy (in J) per atom.J (d)Use P = 2 3 N V 1 2 mv2 to calculate the gas pressure (in Pa). (e)Calculate the gas pressure (in Pa) using the ideal gas law (PV = nRT). Paarrow_forwardA 3.65-mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1210 m3 to 0.750 m3. Initially the pressure was 1.00 atm. Determine: (a) the initial and final temperatures; (b) the change in internal energy; (c) the heat lost by the gas; (d) the work done on the gas. (Assume no molecular vibration.)arrow_forward
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