Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 5 g/mol ( ) Assuming that the volume of the solid contains many small cubes, with one atom at the center of each, then there are ( 6.0 × 10 28 )1/3 atoms/m = 3.9 × 10 9 atoms/m along each edge of a 1.00-m3 cube. Therefore, the spacing between the centers of neighboring atoms is d= 1 = 2.6 × 10 –10 m 3.9 × 10 /m ______________________________________________________________________________ 9 61. REASONING Since the xenon atom does not interact with any other atoms or molecules on its way up, we can apply the principle of conservation of mechanical energy (see Section 6.5) and set the final kinetic plus potential energy equal to the initial kinetic plus potential energy. Thus, during the rise, the atom’s initial kinetic energy is converted entirely into gravitational potential energy, because the atom comes to a momentary halt at the top of its 2 trajectory. The initial kinetic energy 1 mv0 is equal to the average translational kinetic 2 2 energy. Therefore, 1 mv0 = KE = 3 kT , according to Equation 14.6, where k is Boltzmann’s 2 2 constant and T is the Kelvin temperature. The gravitational potential energy is mgh, according to Equation 6.5. SOLUTION Equation 6.9b gives the principle of conservation of mechanical energy: 1 mv 2 + mgh f 24 14f 244 3 Final mechanical energy = 1 mv 2 + mgh 0 24 140244 3 Initial mechanical energy 2 In this expression, we know that 1 mv0 = KE = 3 kT and that 1 mvf2 = 0 J (since the atom 2 2 2 comes to a halt at the top of its trajectory). Furthermore, we can take the height at the earth’s surface to be h0 = 0 m. Taking this information into account, we can write the energy-conservation equation as follows: Chapter 14 Problems mghf = 3 kT 2 or hf = 761 3kT 2mg Using M to denote the molecular mass (in kilograms per mole) and recognizing that M m= , where NA is Avogadro’s number and is the number of xenon atoms per mole, we NA have 3kN AT 3kT 3kT hf = = = 2mg 2 Mg M 2 g NA Recognizing that kNA = R and that M = 131.29 g/mol = 131.29 × 10−3 kg/mol, we find hf = 3 8.31 J/ ( mol ⋅ K ) ( 291 K ) 3kN AT 3RT = = = 2820 m 2 Mg 2 Mg 2 131.29 × 10−3 kg/mol 9.80 m/s 2 ( )( ) 62. REASONING AND SOLUTION Since we are treating the air as a diatomic ideal gas (PV = nRT), it follows that U = 5 nRT = 5 PV = 2 2 5 2 (7.7×106 Pa ) (5.6 ×105 m3 ) = 1.1×1013 J The number of joules of energy consumed per day by one house is J ⋅ h 3600 s 8 30.0 kW ⋅ h = 30.0 ×103 = 1.08 ×10 J s 1 h The number of homes that could be served for one day by 1.1 × 1013 J of energy is 1 (1.1×1013 J ) 1.08home8 J = ×10 1.0 ×105 homes ______________________________________________________________________________ 63. REASONING When perspiration absorbs heat from the body, the perspiration vaporizes. The amount Q of heat required to vaporize a mass mperspiration of perspiration is given by Equation 12.5 as Q = mperspirationLv, where Lv is the latent heat of vaporization for water at body temperature. The average energy E given to a single water molecule is equal to the heat Q divided by the number N of water molecules. 762 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION Since E = Q/N and Q = mperspirationLv, we have E= Q mperspiration Lv = N N But the mass of perspiration is equal to the mass mH 2O molecule of a single water molecule times the number N of water molecules. The mass of a single water molecule is equal to its molecular mass (18.0 u), converted into kilograms. The average energy given to a single water molecule is E= mperspiration Lv N = mH O molecule N Lv 2 N 1.66 × 10−27 kg 6 −20 E = mH O molecule Lv = (18.0 u ) 2.42 × 10 J/kg = 7.23 × 10 J 2 1u ______________________________________________________________________________ ( ) 64. REASONING AND SOLUTION At the instant just before the balloon lifts off, the buoyant force from the outside air has a magnitude that equals the magnitude of the total weight. According to Archimedes’ principle, the buoyant force is the weight of the displaced outside air (density ρ0 = 1.29 × 103 kg/m3). The mass of the displaced outside air is ρ0V, where V = 650 m3. The corresponding weight is the mass times the magnitude g of the acceleration due to gravity. Thus, we have ρ V )g (124 43 0 Buoyant force = mtotal g { Total weight of balloon (1) The total mass of the balloon is mtotal = mload + mair, where mload = 320 kg and mair is the mass of the hot air within the balloon. The mass of the hot air can be calculated from the ideal gas law by using it to obtain the number of moles n of air and multiplying n by the mass per mole of air, M = 29 × 10–3 kg/mol: PV mair = n M = RT M Thus, the total mass of the balloon is mtotal = mload + PVM/(RT) and Equation (1) becomes PV RT ρ 0V = mload + M Chapter 14 Problems 763 Solving for T gives T= PVM ( ρ0V – mload ) R (1.01 × 10 Pa )( 650 m )( 29 × 10 kg/mol ) = = (1.29 × 10 kg/m )( 650 m ) – 320 kg 8.31 J/ ( mol ⋅ K ) 5 3 3 3 3 –3 440 K ______________________________________________________________________________ CHAPTER 15 THERM...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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