Physics Solution Manual for 1100 and 2101

The coefficient of performance is the heat qc removed

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Unformatted text preview: resh air contains approximately 21% oxygen, the total number of oxygen molecules in a normal breath is (0.21) nN A or (0.21)(1.94 ×10−2 mol)(6.022 ×1023 mol −1 )= 2.5 ×1021 Chapter 14 Problems 757 ______________________________________________________________________________ 55. SSM REASONING According to Equation 11.1, the mass density ρ of a substance is defined as its mass m divided by its volume V: ρ = m / V . The mass of nitrogen is equal to the number n of moles of nitrogen times its mass per mole: m = n (Mass per mole). The number of moles can be obtained from the ideal gas law (see Equation 14.1) as n = (PV)/(RT). The mass per mole (in g/mol) of nitrogen has the same numerical value as its molecular mass (which we know). SOLUTION Substituting m = n (Mass per mole) into ρ = m / V , we obtain ρ= m n (Mass per mole) = V V (1) Substituting n = (PV)/(RT) from the ideal gas law into Equation 1 gives the following result: PV n (Mass per mole) RT ρ= = V (Mass per mole) P (Mass per mole) = RT V 5 The pressure is 2.0 atmospheres, or P = 2 (1.013 × 10 Pa). The molecular mass of nitrogen is given as 28 u, which means that its mass per mole is 28 g/mol. Expressed in terms of kilograms per mol, the mass per mole is g 1 kg Mass per mole = 28 mol 103 g The density of the nitrogen gas is g 1 kg 2 (1.013 ×105 Pa ) 28 mol 103 g P(Mass per mole) ρ= = = 2.2 kg/m3 [8.31 J/ ( mol ⋅ K )] ( 310 K ) RT _____________________________________________________________________________________________ 56. REASONING The diffusion of the glycine is driven by the concentration difference ∆C = C2 – C1 between the ends of the tube, where C2 is the higher concentration and C1 = C2 − ∆C is the lower concentration. The concentration difference is related to the mass m DA ∆C rate of diffusion by Fick’s law: (Equation 14.8), where D is the diffusion = t L constant of glycine in water, and A and L are, respectively, the cross-sectional area and length of the tube. 758 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION Solving m DA ∆C (Equation 14.8) for ∆C, we obtain = t L m L ∆C = t DA (1) Substituting Equation (1) into C1 = C2 − ∆C yields m L C1 = C2 − ∆C = C2 − t DA (2) We note that the length of the tube is L = 2.0 cm = 0.020 m. Substituting this and the other given values into Equation (2), we find that the lower concentration is C1 = 8.3 × 10−3 kg/m3 − 57. ( 0.020 m ) ( 4.2 × 10−14 kg/s ) (1.06 ×10−9 m 2 /s )(1.5 × 10−4 m 2 ) = 3.0 × 10−3 kg/m3 SSM REASONING AND SOLUTION Using the expressions for v 2 and ( v ) given in the statement of the problem, we obtain: 2 1 3 a. 2 2 2 v 2 = (v1 + v2 + v3 ) = b. (v) 2 1 (3.0 3 m/s)2 + (7.0 m/s) 2 + (9.0 m/s)2 = 46.3 m 2 /s 2 2 2 = 1 (v1 + v2 + v3 ) = 1 (3.0 m/s + 7.0 m/s + 9.0 m/s) = 40.1 m 2 /s 2 3 3 v 2 and ( v ) are not equal, because they are two different physical quantities. ______________________________________________________________________________ 2 58. REASONING The translational rms-speed vrms is related to the Kelvin temperature T by 1 2 2 3 mvrms = 2 kT (Equation 14.6), where m is the mass of the oxygen molecule and k is Boltzmann’s constant. Solving this equation for the rms-speed gives vrms = 3kT / m . This relation will be used to find the ratio of the speeds. SOLUTION The rms-speeds in the ionosphere and near the earth’s surface are ( vrms )ionosphere = 3k Tionosphere m and ( vrms )earth's surface = 3kTearth's surface m Chapter 14 Problems 759 Dividing the first equation by the second gives ( vrms )ionosphere ( vrms )earth's surface 3kTionosphere = m 3kTearth's surface m = Tionosphere Tearth's surface = 3 = 1.73 ____________________________________________________________________________________________ 59. SSM REASONING The mass m of nitrogen that must be removed from the tank is equal to the number of moles withdrawn times the mass per mole. The number of moles withdrawn is the initial number ni minus the final number of moles nf, so we can write m= −n ( ni 24) ( Mass per mole ) 14 f3 (1) Number of moles withdrawn The final number of moles is related to the initial number by the ideal gas law. SOLUTION From the ideal gas law (Equation 14.1), we have nf = Pf V RT and ni = PV i RT Note that the volume V and temperature T do not change. Dividing the first by the second equation gives Pf V nf P P RT = =f or nf = ni f PV ni Pi i Pi RT Substituting this expression for nf into Equation 1 gives P m = ( ni − nf ) ( Mass per mole ) = ni − ni f ( Mass per mole ) Pi The molecular mass of nitrogen (N2) is 2 (14.0067 u) = 28.0134 u. Therefore, the mass per mole is 28.0134 g/mol. The mass of nitrogen that must be removed is 760 THE IDEAL GAS LAW AND KINETIC THEORY P m = ni − ni f Pi ( Mass per mole ) g 25 atm = 0.85 mol − ( 0.85 mol ) 28.0134 = 8.1 g mol 38 atm ______________________________________________________________________________ 3 60. REASONING AND SOLUTION Since the density of aluminum is 2700 kg/m , the number of atoms of aluminum per cubic meter is, using the given data, 2700 × 103 g/m3 23 −1 28 3 N /V = 6.022 × 10 mol = 6.0 × 10 atoms/m 26.981...
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