M18 YOUN7066 13 ISM c18 kho tài liệu bách khoa

THERMAL PROPERTIES OF MATTER 18.1 18 (a) IDENTIFY: We are asked about a single state of the system SET UP: Use Eq (18.2) to calculate the number of moles and then apply the ideal-gas equation 4.86 × 10−4 kg m EXECUTE: n = tot = = 0.122 mol M 4.00 × 10−3 kg/mol (b) pV = nRT implies p = nRT /V T must be in kelvins, so T = (18 + 273) K = 291 K p= (0.122 mol)(8.3145 J/mol ⋅ K)(291 K) 20.0 × 10−3 m3 = 1.47 × 104 Pa p = (1.47 × 104 Pa)(1.00 atm/1.013 × 105 Pa) = 0.145 atm 18.2 EVALUATE: The tank contains about 1/10 mole of He at around standard temperature, so a pressure around 1/10 atmosphere is reasonable IDENTIFY: pV = nRT SET UP: T1 = 41.0°C = 314 K R = 0.08206 L ⋅ atm/mol ⋅ K EXECUTE: n and R are constant so pV pV pV = nR is constant 1 = 2 T T1 T2 ⎛ p ⎞⎛ V ⎞ T2 = T1 ⎜ ⎟⎜ ⎟ = (314 K)(2)(2) = 1.256 × 103 K = 983°C ⎝ p1 ⎠⎝ V1 ⎠ pV (0.180 atm)(2.60 L) = = 0.01816 mol (b) n = RT (0.08206 L ⋅ atm/mol ⋅ K)(314 K) mtot = nM = (0.01816 mol)(4.00 g/mol) = 0.0727 g 18.3 18.4 EVALUATE: T is directly proportional to p and to V, so when p and V are each doubled the Kelvin temperature increases by a factor of IDENTIFY: pV = nRT SET UP: T is constant EXECUTE: nRT is constant so p1V1 = p2V2 ⎛ 0.110 m3 ⎞ ⎛V ⎞ p2 = p1 ⎜ ⎟ = (0.355 atm) ⎜ = 0.100 atm ⎜ 0.390 m3 ⎟⎟ ⎝ V2 ⎠ ⎝ ⎠ EVALUATE: For T constant, p decreases as V increases IDENTIFY: pV = nRT SET UP: T1 = 20.0°C = 293 K EXECUTE: (a) n, R and V are constant p p p nR = = constant = T V T1 T2 ⎛p ⎞ ⎛ 1.00 atm ⎞ T2 = T1 ⎜ ⎟ = ( 293 K ) ⎜ ⎟ = 97.7 K = −175°C p ⎝ 3.00 atm ⎠ ⎝ 1⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-1 18-2 Chapter 18 (b) p2 = 1.00 atm, V2 = 3.00 L p3 = 3.00 atm n, R and T are constant so pV = nRT = constant p2V2 = p3V3 18.5 ⎛p ⎞ ⎛ 1.00 atm ⎞ V3 = V2 ⎜ ⎟ = ( 3.00 L ) ⎜ ⎟ = 1.00 L p ⎝ 3.00 atm ⎠ ⎝ 3⎠ EVALUATE: The final volume is one-third the initial volume The initial and final pressures are the same, but the final temperature is one-third the initial temperature IDENTIFY: We know the pressure and temperature and want to find the density of the gas The ideal gas law applies pM SET UP: M CO2 = (12 + 2[16]) g/mol = 44 g/mol M N = 28 g/mol ρ = RT R = 8.315 J/mol ⋅ K T must be in kelvins Express M in kg/mol and p in Pa atm = 1.013 × 105 Pa EXECUTE: (a) Mars: ρ= (650 Pa)(44 × 10−3 kg/mol) = 0.0136 kg/m3 (8.315 J/mol ⋅ K)(253 K) (92 atm)(1.013 × 105 Pa/atm)(44 × 10−3 kg/mol) = 67.6 kg/m3 (8.315 J/mol ⋅ K)(730 K) Titan: T = −178 + 273 = 95 K Venus: ρ = ρ= (1.5 atm)(1.013 × 105 Pa/atm)(28 × 10−3 kg/mol) = 5.39 kg/m3 (8.315 J/mol ⋅ K)(95 K) EVALUATE: (b) Table 12.1 gives the density of air at 20°C and p = atm to be 1.20 kg/m3 The density 18.6 of the atmosphere of Mars is much less, the density for Venus is much greater and the density for Titan is somewhat greater IDENTIFY: pV = nRT and the mass of the gas is mtot = nM SET UP: The temperature is T = 22.0°C = 295.15 K The average molar mass of air is M = 28.8 × 10−3 kg/mol For helium M = 4.00 × 10−3 kg/mol EXECUTE: (a) mtot = nM = (b) mtot = nM = (1.00 atm)(0.900 L)(28.8 × 10−3 kg/mol) pV = 1.07 × 10−3 kg M= (0.08206 L ⋅ atm/mol ⋅ K)(295.15 K) RT (1.00 atm)(0.900 L)(4.00 × 10−3 kg/mol) pV = 1.49 × 10−4 kg M= (0.08206 L ⋅ atm/mol ⋅ K)(295.15 K) RT N pV = says that in each case the balloon contains the same number of molecules N A RT The mass is greater for air since the mass of one molecule is greater than for helium IDENTIFY: We are asked to compare two states Use the ideal gas law to obtain T2 in terms of T1 and EVALUATE: n = 18.7 ratios of pressures and volumes of the gas in the two states SET UP: pV = nRT and n, R constant implies pV/T = nR = constant and p1V1 /T1 = p2V2 /T2 EXECUTE: T1 = (27 + 273) K = 300 K p1 = 1.01 × 105 Pa p2 = 2.72 × 106 Pa + 1.01 × 105 Pa = 2.82 × 106 Pa (in the ideal gas equation the pressures must be absolute, not gauge, pressures) ⎛ 2.82 × 106 Pa ⎞⎛ 46.2 cm3 ⎞ ⎛ p ⎞⎛ V ⎞ T2 = T1 ⎜ ⎟⎜ ⎟ = 300 K ⎜ = 776 K ⎜ 1.01 × 105 Pa ⎟⎜ ⎟⎜ 499 cm3 ⎟⎟ ⎝ p1 ⎠⎝ V1 ⎠ ⎝ ⎠⎝ ⎠ T2 = (776 − 273)°C = 503°C EVALUATE: The units cancel in the V2 /V1 volume ratio, so it was not necessary to convert the volumes in cm3 to m3 It was essential, however, to use T in kelvins © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18.8 IDENTIFY: 18-3 pV = nRT and m = nM SET UP: We must use absolute pressure in pV = nRT p1 = 4.01 × 105 Pa, p2 = 2.81 × 105 Pa T1 = 310 K, T2 = 295 K p1V1 (4.01 × 105 Pa)(0.075 m3 ) = = 11.7 mol RT1 (8.315 J/mol ⋅ K)(310 K) m = nM = (11.7 mol)(32.0 g/mol) = 374 g EXECUTE: (a) n1 = p2V2 (2.81 × 105 Pa)(0.075 m3 ) = = 8.59 mol m = 275 g RT2 (8.315 J/mol ⋅ K)(295 K) The mass that has leaked out is 374 g − 275 g = 99 g (b) n2 = 18.9 EVALUATE: In the ideal gas law we must use absolute pressure, expressed in Pa, and T must be in kelvins IDENTIFY: pV = nRT SET UP: T1 = 300 K, T2 = 430 K EXECUTE: (a) n, R are constant so pV pV pV = nR = constant 1 = 2 T T1 T2 ⎛ 0.750 m3 ⎞ ⎛ 430 K ⎞ ⎛ V ⎞⎛ T ⎞ p2 = p1 ⎜ ⎟⎜ ⎟ = (7.50 × 103 Pa) ⎜ = 1.68 × 104 Pa ⎜ 0.480 m3 ⎟⎟ ⎜⎝ 300 K ⎟⎠ V T ⎝ ⎠⎝ ⎠ ⎝ ⎠ EVALUATE: Since the temperature increased while the volume decreased, the pressure must have increased In pV = nRT , T must be in kelvins, even if we use a ratio of temperatures 18.10 IDENTIFY: Use the ideal-gas equation to calculate the number of moles, n The mass mtotal of the gas is mtotal = nM SET UP: The volume of the cylinder is V = π r 2l , where r = 0.450 m and l = 1.50 m T = 22.0°C = 293.15 K atm = 1.013 × 105 Pa M = 32.0 × 10−3 kg/mol R = 8.314 J/mol ⋅ K EXECUTE: (a) pV = nRT gives n= pV (21.0 atm)(1.013 × 105 Pa/atm)π (0.450 m) (1.50 m) = = 827 mol RT (8.314 J/mol ⋅ K)(295.15 K) (b) mtotal = (827 mol)(32.0 × 10−3 kg/mol) = 26.5 kg EVALUATE: In the ideal-gas law, T must be in kelvins Since we used R in units of J/mol ⋅ K we had to 18.11 express p in units of Pa and V in units of m3 IDENTIFY: We are asked to compare two states Use the ideal-gas law to obtain V1 in terms of V2 and the ratio of the temperatures in the two states SET UP: pV = nRT and n, R, p are constant so V/T = nR/p = constant and V1/T1 = V2 /T2 EXECUTE: T1 = (19 + 273) K = 292 K (T must be in kelvins) V2 = V1(T2 /T1) = (0.600 L)(77.3 K/ 292 K) = 0.159 L 18.12 EVALUATE: p is constant so the ideal-gas equation says that a decrease in T means a decrease in V IDENTIFY: Apply pV = nRT and the van der Waals equation (Eq 18.7) to calculate p SET UP: 400 cm3 = 400 × 10−6 m3 R = 8.314 J/mol ⋅ K EXECUTE: (a) The ideal gas law gives p = nRT/V = 7.28 × 106 Pa while Eq (18.7) gives 5.87 × 106 Pa (b) The van der Waals equation, which accounts for the attraction between molecules, gives a pressure that is 20% lower 18.13 (c) The ideal gas law gives p = 7.28 × 105 Pa Eq (18.7) gives p = 7.13 × 105 Pa, for a 2.1% difference EVALUATE: (d) As n/V decreases, the formulas and the numerical values for the two equations approach each other IDENTIFY: We know the volume of the gas at STP on the earth and want to find the volume it would occupy on Venus where the pressure and temperature are much greater © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-4 Chapter 18 SET UP: STP is T = 273 K and p = atm Set up a ratio using pV = nRT with nR constant TV = 1003 + 273 = 1276 K EXECUTE: pV = nRT gives 18.14 pV p V p V = nR = constant, so E E = V V T TE TV ⎛ p ⎞⎛ T ⎞ ⎛ atm ⎞⎛ 1276 K ⎞ VV = VE ⎜ E ⎟⎜ V ⎟ = V ⎜ ⎟⎜ ⎟ = 0.0508V p T ⎝ 92 atm ⎠⎝ 273 K ⎠ ⎝ V ⎠⎝ E ⎠ EVALUATE: Even though the temperature on Venus is higher than it is on Earth, the pressure there is much greater than on Earth, so the volume of the gas on Venus is only about 5% what it is on Earth IDENTIFY: pV = nRT SET UP: T1 = 277 K T2 = 296 K Assume the number of moles of gas in the bubble remains constant EXECUTE: (a) n, R are constant so 18.15 pV pV pV = nR = constant 1 = 2 and T T1 T2 V2 ⎛ p1 ⎞⎛ T2 ⎞ ⎛ 3.50 atm ⎞⎛ 296 K ⎞ = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = 3.74 V1 ⎝ p2 ⎠⎝ T1 ⎠ ⎝ 1.00 atm ⎠⎝ 277 K ⎠ (b) This increase in volume of air in the lungs would be dangerous EVALUATE: The large decrease in pressure results in a large increase in volume IDENTIFY: We are asked to compare two states First use pV = nRT to calculate p1 Then use it to obtain T2 in terms of T1 and the ratio of pressures in the two states (a) SET UP: EXECUTE: SET UP: pV = nRT Find the initial pressure p1 p1 = nRT1 (11.0 mol)(8.3145 J/mol ⋅ K)(23.0 + 273.15)K = = 8.737 × 106 Pa V 3.10 × 10−3 m3 p2 = 100 atm(1.013 × 105 Pa/1 atm) = 1.013 × 107 Pa p/T = nR/V = constant, so p1/T1 = p2 /T2 18.16 ⎛ 1.013 × 107 Pa ⎞ ⎛p ⎞ EXECUTE: T2 = T1 ⎜ ⎟ = (296.15 K) ⎜ ⎟⎟ = 343.4 K = 70.2°C ⎜ ⎝ p1 ⎠ ⎝ 8.737 × 10 Pa ⎠ (b) EVALUATE: The coefficient of volume expansion for a gas is much larger than for a solid, so the expansion of the tank is negligible IDENTIFY: F = pA and pV = nRT SET UP: For a cube, V/A = L EXECUTE: (a) The force of any side of the cube is F = pA = (nRT/V ) A = (nRT ) /L, since the ratio of area to volume is A/V = 1/L For T = 20.0°C = 293.15 K, F= nRT (3 mol)(8.3145 J/mol ⋅ K)(293.15 K ) = = 3.66 × 104 N L 0.200 m (b) For T = 100.00°C = 373.15 K, F= 18.17 nRT (3 mol)(8.3145 J/mol ⋅ K)(373.15 K) = = 4.65 × 104 N L 0.200 m EVALUATE: When the temperature increases while the volume is kept constant, the pressure increases and therefore the force increases The force increases by the factor T2 /T1 IDENTIFY: Example 18.4 assumes a temperature of 0°C at all altitudes and neglects the variation of g with elevation With these approximations, p = p0e− Mgy/RT SET UP: ln(e − x ) = − x For air, M = 28.8 × 10−3 kg/mol EXECUTE: We want y for p = 0.90 p0 so 0.90 = e− Mgy/RT and y = − RT ln(0.90) = 850 m Mg © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-5 EVALUATE: This is a commonly occurring elevation, so our calculation shows that 10% variations in atmospheric pressure occur at many locations 18.18 IDENTIFY: From Example 18.4, the pressure at elevation y above sea level is p = p0e− Mgy/RT SET UP: The average molar mass of air is M = 28.8 × 10−3 kg/mol EXECUTE: At an altitude of 100 m, Mgy1 (28.8 × 10−3 kg/mol)(9.80 m/s )(100 m) = = 0.01243, and the RT (8.3145 J/mol ⋅ K)(273.15 K) percent decrease in pressure is − p/p0 = − e−0.01243 = 0.0124 = 1.24, At an altitude of 1000 m, Mgy2 /RT = 0.1243 and the percent decrease in pressure is − e−0.1243 = 0.117 = 11.7, EVALUATE: These answers differ by a factor of (11.7%)/(1.24%) = 9.44, which is less than 10 because the 18.19 variation of pressure with altitude is exponential rather than linear IDENTIFY: We know the volume, pressure and temperature of the gas and want to find its mass and density SET UP: V = 3.00 × 10−3 m3 T = 295 K p = 2.03 × 10−8 Pa The ideal gas law, pV = nRT , applies EXECUTE: (a) pV = nRT gives n= pV (2.03 × 10−8 Pa)(3.00 × 10−3 m3 ) = = 2.48 × 10−14 mol The mass of this amount of gas is RT (8.315 J/mol ⋅ K)(295 K) m = nM = (2.48 × 10−14 mol)(28.0 × 10−3 kg/mol) = 6.95 × 10−16 kg m 6.95 × 10−16 kg = = 2.32 × 10−13 kg/m3 V 3.00 × 10−3 m3 EVALUATE: The density at this level of vacuum is 13 orders of magnitude less than the density of air at STP, which is 1.20 kg/m3 IDENTIFY: p = p0e− Mgy/RT from Example 18.4 gives the variation of air pressure with altitude The (b) ρ = 18.20 pM , so ρ is proportional to the pressure p Let ρ0 be the density at the RT surface, where the pressure is p0 density ρ of the air is ρ = SET UP: From Example 18.4, EXECUTE: p = p0e − (1.244 × 10 Mg (28.8 × 10−3 kg/mol)(9.80 m/s ) = = 1.244 × 10−4 m −1 RT (8.314 J/mol ⋅ K)(273 K) −4 m −1 )(1.00 × 103 m) = 0.883 p0 ρ p = M ρ ρ = constant, so = and p p0 RT ⎛ p ⎞ ⎟ = 0.883ρ0 ⎝ p0 ⎠ ρ = ρ0 ⎜ 18.21 The density at an altitude of 1.00 km is 88.3% of its value at the surface EVALUATE: If the temperature is assumed to be constant, then the decrease in pressure with increase in altitude corresponds to a decrease in density IDENTIFY: Use Eq (18.5) and solve for p SET UP: ρ = pM/RT and p = RT ρ /M T = ( −56.5 + 273.15) K = 216.6 K For air M = 28.8 × 10−3 kg/mol (Example 18.3) 18.22 (8.3145 J/mol ⋅ K)(216.6 K)(0.364 kg/m3 ) = 2.28 × 104 Pa 28.8 × 10−3 kg/mol EVALUATE: The pressure is about one-fifth the pressure at sea-level IDENTIFY: The molar mass is M = N A m, where m is the mass of one molecule EXECUTE: p= SET UP: N A = 6.02 × 1023 molecules/mol EXECUTE: M = N A m = (6.02 × 1023 molecules/mol)(1.41 × 10−21 kg/molecule) = 849 kg/mol © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-6 Chapter 18 EVALUATE: For a carbon atom, M = 12 × 10−3 kg/mol If this molecule is mostly carbon, so the average mass of its atoms is the mass of carbon, the molecule would contain 18.23 849 kg/mol 12 × 10−3 kg/mol = 71,000 atoms IDENTIFY: The mass mtot is related to the number of moles n by mtot = nM Mass is related to volume by ρ = m/V SET UP: For gold, M = 196.97 g/mol and ρ = 19.3 × 103 kg/m3 The volume of a sphere of radius r is V = 43 π r EXECUTE: (a) mtot = nM = (3.00 mol)(196.97 g/mol) = 590.9 g The value of this mass of gold is (590.9 g)($14.75 /g) = $8720 (b) V = m ρ = 0.5909 kg 19.3 × 103 kg/m3 1/3 1/3 ⎛ 3[3.06 × 10−5 m3 ] ⎞ =⎜ ⎟⎟ = 0.0194 m = 1.94 cm The diameter is 2r = 3.88 cm ⎜ 4π ⎝ ⎠ EVALUATE: The mass and volume are directly proportional to the number of moles IDENTIFY: Use pV = nRT to calculate the number of moles and then the number of molecules would be ⎛ 3V ⎞ r =⎜ ⎟ ⎝ 4π ⎠ 18.24 = 3.06 × 10−5 m3 V = 43 π r gives N = nN A SET UP: atm = 1.013 × 105 Pa 1.00 cm3 = 1.00 × 10−6 m3 N A = 6.022 × 1023 molecules/mol EXECUTE: (a) n = pV (9.00 × 10−14 atm)(1.013 × 105 Pa/atm)(1.00 × 10−6 m3 ) = = 3.655 × 10−18 mol RT (8.314 J/mol ⋅ K)(300.0 K) N = nN A = (3.655 × 10−18 mol)(6.022 × 1023 molecules/mol) = 2.20 × 106 molecules (b) N = 18.25 N N N VN A pVN A = = constant and = so RT p1 p2 p RT ⎛p ⎞ 1.00 atm ⎛ ⎞ 19 N = N1 ⎜ ⎟ = (2.20 × 106 molecules) ⎜ ⎟ = 2.44 × 10 molecules 14 − atm ⎠ ⎝ 9.00 × 10 ⎝ p1 ⎠ EVALUATE: The number of molecules in a given volume is directly proportional to the pressure Even at the very low pressure in part (a) the number of molecules in 1.00 cm3 is very large IDENTIFY: We are asked about a single state of the system SET UP: Use the ideal-gas law Write n in terms of the number of molecules N (a) EXECUTE: pV = nRT , n = N/N A so pV = ( N/N A ) RT ⎛ N ⎞⎛ R ⎞ p = ⎜ ⎟⎜ ⎟T ⎝ V ⎠ ⎝ NA ⎠ 8.3145 J/mol ⋅ K ⎛ 80 molecules ⎞⎛ ⎞ −12 p=⎜ Pa ⎟⎜ ⎟ (7500 K) = 8.28 × 10 ⎝ × 10−6 m3 ⎠⎝ 6.022 × 1023 molecules/mol ⎠ p = 8.2 × 10−17 atm This is much lower than the laboratory pressure of × 10−14 atm in Exercise 18.24 18.26 (b) EVALUATE: The Lagoon Nebula is a very rarefied low pressure gas The gas would exert very little force on an object passing through it IDENTIFY: pV = nRT = NkT SET UP: At STP, T = 273 K, p = 1.01 × 105 Pa N = × 109 molecules EXECUTE: V = NkT (6 × 109 molecules)(1.381 × 10−23 J/molecule ⋅ K)(273 K) = = 2.24 × 10−16 m3 p 1.01 × 105 Pa L3 = V so L = V 1/3 = 6.1 × 10−6 m EVALUATE: This is a small cube © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18.27 IDENTIFY: n = 18-7 m N = M NA SET UP: N A = 6.022 × 1023 molecules/mol For water, M = 18 × 10−3 kg/mol EXECUTE: n = m 1.00 kg = = 55.6 mol M 18 × 10−3 kg/mol N = nN A = (55.6 mol)(6.022 × 1023 molecules/mol) = 3.35 × 1025 molecules 18.28 EVALUATE: Note that we converted M to kg/mol N IDENTIFY: Use pV = nRT and n = with N = to calculate the volume V occupied by molecule NA The length l of the side of the cube with volume V is given by V = l SET UP: T = 27°C = 300 K p = 1.00 atm = 1.013 × 105 Pa R = 8.314 J/mol ⋅ K N A = 6.022 × 1023 molecules/mol The diameter of a typical molecule is about 10−10 m 0.3 nm = 0.3 × 10−9 m N EXECUTE: (a) pV = nRT and n = gives NA (1.00)(8.314 J/mol ⋅ K)(300 K) NRT = = 4.09 × 10−26 m3 l = V 1/ = 3.45 × 10−9 m N A p (6.022 × 1023 molecules/mol)(1.013 × 105 Pa) (b) The distance in part (a) is about 10 times the diameter of a typical molecule (c) The spacing is about 10 times the spacing of atoms in solids EVALUATE: There is space between molecules in a gas whereas in a solid the atoms are closely packed together (a) IDENTIFY and SET UP: Use the density and the mass of 5.00 mol to calculate the volume ρ = m/V implies V = m/ρ, where m = mtot , the mass of 5.00 mol of water V= 18.29 EXECUTE: mtot = nM = (5.00 mol)(18.0 × 10−3 kg/mol) = 0.0900 kg Then V = m ρ = 0.0900 kg 1000 kg/m3 = 9.00 × 10−5 m3 (b) One mole contains N A = 6.022 × 1023 molecules, so the volume occupied by one molecule is 9.00 × 10−5 m3/mol (5.00 mol)(6.022 × 1023 molecules/mol) = 2.989 × 10−29 m3/molecule V = a3, where a is the length of each side of the cube occupied by a molecule a3 = 2.989 × 10−29 m3 , so a = 3.1 × 10−10 m 18.30 (c) EVALUATE: Atoms and molecules are on the order of 10−10 m in diameter, in agreement with the above estimates 3RT IDENTIFY: K av = 32 kT vrms = M SET UP: M Ne = 20.180 g/mol, M Kr = 83.80 g/mol and M Rn = 222 g/mol EXECUTE: (a) K av = 32 kT depends only on the temperature so it is the same for each species of atom in the mixture v v M Kr 83.80 g/mol M Rn 222 g/mol (b) rms,Ne = = = 2.04 rms,Ne = = = 3.32 vrms,Kr M Ne 20.18 g/mol vrms,Rn M Ne 20.18 g/mol vrms,Kr vrms,Rn = 222 g/mol M Rn = = 1.63 83.80 g/mol M Kr EVALUATE: The average kinetic energies are the same The gas atoms with smaller mass have larger vrms © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-8 Chapter 18 18.31 IDENTIFY and SET UP: vrms = 3RT M EXECUTE: (a) vrms is different for the two different isotopes, so the 235 isotope diffuses more rapidly (b) vrms,235 vrms,238 = 0.352 kg/mol M 238 = = 1.004 M 235 0.349 kg/mol EVALUATE: The vrms values each depend on T but their ratio is independent of T 18.32 IDENTIFY and SET UP: With the multiplicity of each score denoted by ni , the average score is 1/2 18.33 ⎡⎛ ⎞ ⎛ ⎞ 2⎤ ⎜ ⎟ ∑ ni xi and the rms score is ⎢⎜⎝ 150 ⎟⎠ ∑ ni xi ⎥ 150 ⎝ ⎠ ⎣ ⎦ EXECUTE: (a) 54.6 (b) 61.1 EVALUATE: The rms score is higher than the average score since the rms calculation gives more weight to the higher scores N m IDENTIFY: pV = nRT = RT = tot RT NA M SET UP: We know that VA = VB and that TA > TB EXECUTE: (a) p = nRT/V ; we don’t know n for each box, so either pressure could be higher ⎛ N ⎞ pVN A (b) pV = ⎜ , where N A is Avogadro’s number We don’t know how the pressures ⎟ RT so N = RT ⎝ NA ⎠ compare, so either N could be larger (c) pV = (mtot /M ) RT We don’t know the mass of the gas in each box, so they could contain the same gas or different gases (d) 12 m(v )av = 32 kT TA > TB and the average kinetic energy per molecule depends only on T, so the statement must be true (e) vrms = 3kT/m We don’t know anything about the masses of the atoms of the gas in each box, so either set of molecules could have a larger vrms 18.34 EVALUATE: Only statement (d) must be true We need more information in order to determine whether the other statements are true or false IDENTIFY: We can relate the temperature to the rms speed and the temperature to the pressure using the ideal gas law The target variable is the pressure 3RT SET UP: vrms = and pV = nRT, where n = m/M M EXECUTE: Use vrms to calculate T: vrms = T= Mvrms (28.014 × 10−3 kg/mol)(182 m/s) nRT = = 37.20 K The ideal gas law gives p = 3R 3(8.314 J/mol ⋅ K) V n= m 0.226 × 10−3 kg = = 8.067 × 10−3 mol Solving for p gives M 28.014 × 10−3 kg/mol (8.067 × 10−3 mol)(8.314 J/mol ⋅ K)(37.20 K) = 1.69 × 103 Pa 1.48 × 10−3 m3 EVALUATE: This pressure is around 1% of atmospheric pressure, which is not unreasonable since we have only around 1% of a mole of gas 3kT IDENTIFY: vrms = m p= 18.35 3RT so M © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-9 SET UP: The mass of a deuteron is m = mp + mn = 1.673 × 10−27 kg + 1.675 × 10−27 kg = 3.35 × 10−27 kg c = 3.00 × 108 m/s k = 1.381 × 10−23 J/molecule ⋅ K EXECUTE: (a) vrms = 3(1.381× 10−23 J/molecule ⋅ K)(300 × 106 K) 3.35 × 10−27 kg = 1.93 × 106 m/s vrms = 6.43 × 10−3 c ⎛ ⎞ 3.35 × 10−27 kg ⎛m⎞ 10 (b) T = ⎜ ⎟ (vrms )2 = ⎜ ⎟ (3.0 × 10 m/s) = 7.3 × 10 K −23 ⎜ k J/molecule ⋅ K) ⎠⎟ ⎝ ⎠ ⎝ 3(1.381 × 10 EVALUATE: Even at very high temperatures and for this light nucleus, vrms is a small fraction of the speed of light 18.36 IDENTIFY: vrms = n p 3RT , where T is in kelvins pV = nRT gives = V RT M SET UP: R = 8.314 J/mol ⋅ K M = 44.0 × 10−3 kg/mol EXECUTE: (a) For T = 0.0°C = 273.15 K, vrms = 3(8.314 J/mol ⋅ K)(273.15 K) 44.0 × 10−3 kg/mol = 393 m/s For T = −100.0°C = 173 K, vrms = 313 m/s The range of speeds is 393 m/s to 313 m/s n 650 Pa = = 0.286 mol/m3 For T = 173.15 K, V (8.314 J/mol ⋅ K)(273.15 K) (b) For T = 273.15 K, 18.37 n = 0.452 mol/m3 The range of densities is 0.286 mol/m3 to 0.452 mol/m3 V EVALUATE: When the temperature decreases the rms speed decreases and the density increases IDENTIFY and SET UP: Apply the analysis of Section 18.3 EXECUTE: (a) 12 m(v )av = 23 kT = 32 (1.38 × 10−23 J/molecule ⋅ K)(300 K) = 6.21 × 10−21 J (b) We need the mass m of one molecule: M 32.0 × 10−3 kg/mol m= = = 5.314 × 10−26 kg/molecule N A 6.022 × 1023 molecules/mol Then m(v ) av (v )av = = 6.21 × 10−21 J (from part (a)) gives 2(6.21 × 10−21 J) 2(6.21 × 10−21 J) = = 2.34 × 105 m /s m 5.314 × 10−26 kg (c) vrms = (v ) rms = 2.34 × 104 m /s = 484 m/s (d) p = mvrms = (5.314 × 10−26 kg)(484 m/s) = 2.57 × 10−23 kg ⋅ m/s (e) Time between collisions with one wall is t = 0.20 m 0.20 m = = 4.13 × 10−4 s vrms 484 m/s G In a collision v changes direction, so Δp = 2mvrms = 2(2.57 × 10−23 kg ⋅ m/s) = 5.14 × 10−23 kg ⋅ m/s F= Δp 5.14 × 10−23 kg ⋅ m/s dp = = 1.24 × 10−19 N so Fav = Δt dt 4.13 × 10−4 s (f) pressure = F/A = 1.24 × 10−19 N/(0.10 m) = 1.24 × 10−17 Pa (due to one molecule) (g) pressure = atm = 1.013 × 105 Pa Number of molecules needed is 1.013 × 105 Pa/(1.24 × 10−17 Pa/molecule) = 8.17 × 1021 molecules (h) pV = NkT (Eq 18.18), so N = (i) From the factor of pV (1.013 × 105 Pa)(0.10 m)3 = = 2.45 × 1022 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(300 K) in (vx2 )av = 13 (v )av © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-10 18.38 Chapter 18 EVALUATE: This exercise shows that the pressure exerted by a gas arises from collisions of the molecules of the gas with the walls IDENTIFY: Apply Eq (18.22) and calculate λ SET UP: atm = 1.013 × 105 Pa, so p = 3.55 × 10−8 Pa r = 2.0 × 10−10 m and k = 1.38 × 10−23 J/K kT (1.38 × 10−23 J/K)(300 K) = 1.6 × 105 m 4π 2r p 4π 2(2.0 × 10−10 m) (3.55 × 10−8 Pa) EVALUATE: At this very low pressure the mean free path is very large If v = 484 m/s, as in Example 18.8, EXECUTE: λ = then tmean = 18.39 λ v = = 330 s Collisions are infrequent IDENTIFY and SET UP: Use equal vrms to relate T and M for the two gases vrms = 3RT/M (Eq 18.19), so vrms /3R = T/M , where T must be in kelvins Same vrms so same T/M for the two gases and TN /M N = TH /M H ⎛ MN ⎞ ⎛ 28.014 g/mol ⎞ ⎟ = ((20 + 273)K) ⎜ EXECUTE: TN = TH ⎜ ⎟ = 4.071 × 10 K ⎜ MH ⎟ 016 g/mol ⎝ ⎠ ⎠ ⎝ TN = (4071 − 273)°C = 3800°C EVALUATE: A N molecule has more mass so N gas must be at a higher temperature to have the same vrms 18.40 IDENTIFY: vrms = 3kT m SET UP: k = 1.381 × 10−23 J/molecule ⋅ K EXECUTE: (a) vrms = 3(1.381 × 10−23 J/molecule ⋅ K)(300 K) 3.00 × 10−16 kg = 6.44 × 10−3 m/s = 6.44 mm/s EVALUATE: (b) No The rms speed depends on the average kinetic energy of the particles At this T, H2 molecules would have larger vrms than the typical air molecules but would have the same average kinetic 18.41 energy and the average kinetic energy of the smoke particles would be the same IDENTIFY: Use Eq (18.24), applied to a finite temperature change SET UP: CV = 5R/2 for a diatomic ideal gas and CV = 3R/2 for a monatomic ideal gas EXECUTE: (a) Q = nCV ΔT = n ( 52 R ) ΔT Q = (2.5 mol)( 52 )(8.3145 J/mol ⋅ K)(50.0 K) = 2600 J (b) Q = nCV ΔT = n( 32 R ) ΔT Q = (2.5 mol)( 32 )(8.3145 J/mol ⋅ K)(50.0 K) = 1560 J 18.42 EVALUATE: More heat is required for the diatomic gas; not all the heat that goes into the gas appears as translational kinetic energy, some goes into energy of the internal motion of the molecules (rotations) IDENTIFY: The heat Q added is related to the temperature increase ΔT by Q = nCV ΔT SET UP: For ideal H (a diatomic gas), CV ,H = 5/2 R, and for ideal Ne (a monatomic gas), CV ,Ne = 3/2 R Q = constant, so CV ,H ΔTH = CV ,Ne ΔTNe n ⎛ CV ,H ⎞ ⎛ 5/2 R ⎞ (2.50 C°) = 4.17 C° = 4.17 Κ =⎜ ΔT = ⎜ CV ,Ne ⎟⎟ H ⎜⎝ 3/2 R ⎟⎠ ⎝ ⎠ EXECUTE: CV ΔT = ΔTNe EVALUATE: The same amount of heat causes a smaller temperature increase for H since some of the 18.43 energy input goes into the internal degrees of freedom IDENTIFY: C = Mc, where C is the molar heat capacity and c is the specific heat capacity m pV = nRT = RT M © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-14 Chapter 18 Solving for ρ hot gives ρ hot = ρair − ρair = 18.56 290 kg 290 kg pM = 1.23 kg/m3 − = 0.65 kg/m3 ρhot = RThot V 500.0 m3 pM ρ hotThot = ρairTair so RTair ⎛ 1.23 kg/m3 ⎞ ⎛ρ ⎞ = 545 K = 272°C Thot = Tair ⎜ air ⎟ = (288 K) ⎜ ⎜ 0.65 kg/m3 ⎟⎟ ⎝ ρ hot ⎠ ⎝ ⎠ EVALUATE: This temperature is well above normal air temperatures, so the air in the balloon would need considerable heating IDENTIFY: ΔV = βV0 ΔT − V0 k Δp SET UP: For steel, β = 3.6 × 10−5 K −1 and k = 6.25 × 10−12 Pa −1 EXECUTE: βV0 ΔT = (3.6 × 10−5 K −1 )(11.0 L)(21 C°) = 0.0083 L 18.57 −kVo Δp = −(6.25 × 10−12 /Pa)(11 L)(2.1 × 107 Pa) = −0.0014 L The total change in volume is ΔV = 0.0083 L − 0.0014 L = 0.0069 L (b) Yes; ΔV is much less than the original volume of 11.0 L EVALUATE: Even for a large pressure increase and a modest temperature increase, the magnitude of the volume change due to the temperature increase is much larger than that due to the pressure increase IDENTIFY: We are asked to compare two states Use the ideal-gas law to obtain m2 in terms of m1 and the ratio of pressures in the two states Apply Eq (18.4) to the initial state to calculate m1 SET UP: pV = nRT can be written pV = (m/M ) RT T, V, M, R are all constant, so p/m = RT/MV = constant So p1/m1 = p2 /m2 , where m is the mass of the gas in the tank EXECUTE: p1 = 1.30 × 106 Pa + 1.01× 105 Pa = 1.40 × 106 Pa p2 = 2.50 × 105 Pa + 1.01 × 105 Pa = 3.51 × 105 Pa m1 = p1VM/RT ; V = hA = hπ r = (1.00 m)π (0.060 m) = 0.01131 m3 m1 = (1.40 × 106 Pa)(0.01131 m3 )(44.1× 10−3 kg/mol) = 0.2845 kg (8.3145 J/mol ⋅ K)((22.0 + 273.15)K) ⎛ 3.51 × 105 Pa ⎞ ⎛p ⎞ Then m2 = m1 ⎜ ⎟ = (0.2845 kg) ⎜ ⎟⎟ = 0.0713 kg ⎜ ⎝ p1 ⎠ ⎝ 1.40 × 10 Pa ⎠ m2 is the mass that remains in the tank The mass that has been used is m1 − m2 = 0.2845 kg − 0.0713 kg = 0.213 kg 18.58 EVALUATE: Note that we have to use absolute pressures The absolute pressure decreases by a factor of four and the mass of gas in the tank decreases by a factor of four IDENTIFY: Apply pV = nRT to the air inside the diving bell The pressure p at depth y below the surface of the water is p = patm + ρ gy SET UP: p = 1.013 × 105 Pa T = 300.15 K at the surface and T ′ = 280.15 K at the depth of 13.0 m EXECUTE: (a) The height h′ of the air column in the diving bell at this depth will be proportional to the volume, and hence inversely proportional to the pressure and proportional to the Kelvin temperature: p T′ patm T′ =h h′ = h p′ T patm + ρ gy T (1.013 × 105 Pa) ⎛ 280.15 K ⎞ ⎜ ⎟ = 0.26 m (1.013 × 10 Pa) + (1030 kg/m )(9.80 m/s )(73.0 m) ⎝ 300.15 K ⎠ The height of the water inside the diving bell is h − h′ = 2.04 m h′ = (2.30 m) (b) The necessary gauge pressure is the term ρ gy from the above calculation, pgauge = 7.37 × 105 Pa EVALUATE: The gauge pressure required in part (b) is about atm © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18.59 IDENTIFY: pV = NkT gives 18-15 N p = V kT SET UP: atm = 1.013 × 105 Pa TK = TC + 273.15 k = 1.381 × 10−23 J/molecule ⋅ K EXECUTE: (a) TC = TK − 273.15 = 94 K − 273.15 = −179°C (b) N p (1.5 atm)(1.013 × 105 Pa/atm) = = = 1.2 × 1026 molecules/m3 V kT (1.381 × 10−23 J/molecule ⋅ K)(94 K) (c) For the earth, p = 1.0 atm = 1.013 × 105 Pa and T = 22°C = 295 K 18.60 N (1.0 atm)(1.013 × 105 Pa/atm) = = 2.5 × 1025 molecules/m3 The atmosphere of Titan is about V (1.381 × 10−23 J/molecule ⋅ K)(295 K) five times denser than earth’s atmosphere EVALUATE: Though it is smaller than Earth and has weaker gravity at its surface, Titan can maintain a dense atmosphere because of the very low temperature of that atmosphere IDENTIFY: For constant temperature, the variation of pressure with altitude is calculated in Example 18.4 3RT to be p = p0e− Mgy/RT vrms = M SET UP: g Earth = 9.80 m/s T = 460°C = 733 K M = 44.0 g/mol = 44.0 × 10−3 kg/mol EXECUTE: (a) Mgy (44.0 × 10−3 kg/mol)(0.894)(9.80 m/s2 )(1.00 × 103 m) = = 0.06326 RT (8.314 J/mol ⋅ K)(733 K) p = p0e − Mgy/RT = (92 atm)e −0.06326 = 86 atm The pressure is 86 earth-atmospheres, or 0.94 Venus- atmospheres 3RT 3(8.314 J/mol ⋅ K)(733 K) (b) vrms = = = 645 m/s vrms has this value both at the surface and at an M 44.0 × 10−3 kg/mol 18.61 altitude of 1.00 km EVALUATE: vrms depends only on T and the molar mass of the gas For Venus compared to earth, the surface temperature, in kelvins, is nearly a factor of three larger and the molecular mass of the gas in the atmosphere is only about 50% larger, so vrms for the Venus atmosphere is larger than it is for the earth’s atmosphere IDENTIFY: pV = nRT SET UP: In pV = nRT we must use the absolute pressure T1 = 278 K p1 = 2.72 atm T2 = 318 K EXECUTE: n, R constant, so 18.62 pV pV pV = nR = constant 1 = 2 and T T1 T2 ⎛ 0.0150 m3 ⎞ ⎛ 318 K ⎞ ⎛ V ⎞⎛ T ⎞ = 2.94 atm The final gauge pressure is p2 = p1 ⎜ ⎟⎜ ⎟ = (2.72 atm) ⎜ ⎜ 0.0159 m3 ⎟⎟ ⎜⎝ 278 K ⎟⎠ ⎝ V2 ⎠⎝ T1 ⎠ ⎝ ⎠ 2.94 atm − 1.02 atm = 1.92 atm EVALUATE: Since a ratio is used, pressure can be expressed in atm But absolute pressures must be used The ratio of gauge pressures is not equal to the ratio of absolute pressures IDENTIFY: In part (a), apply pV = nRT to the ethane in the flask The volume is constant once the stopcock is in place In part (b) apply pV = mtot RT to the ethane at its final temperature and pressure M SET UP: 1.50 L = 1.50 × 10−3 m3 M = 30.1 × 10−3 kg/mol Neglect the thermal expansion of the flask EXECUTE: (a) p2 = p1 (T2 /T1 ) = (1.013 × 105 Pa)(300 K/490 K) = 6.20 × 104 Pa ⎛ (6.20 × 104 Pa)(1.50 × 10−3 m3 ) ⎞ ⎛pV⎞ (b) mtot = ⎜ ⎟ M = ⎜ (30.1 × 10−3 Kg/mol) = 1.12 g ⎜ (8.3145 J/mol ⋅ K)(300 K) ⎟⎟ ⎝ RT2 ⎠ ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-16 Chapter 18 EVALUATE: We could also calculate mtot with p = 1.013 × 105 Pa and T = 490 K, and we would obtain 18.63 the same result Originally, before the system was warmed, the mass of ethane in the flask was ⎛ 1.013 × 105 Pa ⎞ m = (1.12 g) ⎜ = 1.83 g ⎜ 6.20 × 104 Pa ⎟⎟ ⎝ ⎠ (a) IDENTIFY: Consider the gas in one cylinder Calculate the volume to which this volume of gas expands when the pressure is decreased from (1.20 × 106 Pa + 1.01 × 105 Pa) = 1.30 × 106 Pa to 1.01× 105 Pa Apply the ideal-gas law to the two states of the system to obtain an expression for V2 in terms of V1 and the ratio of the pressures in the two states SET UP: pV = nRT n, R, T constant implies pV = nRT = constant, so p1V1 = p2V2 ⎛ 1.30 × 106 Pa ⎞ EXECUTE: V2 = V1 ( p1 /p2 ) = (1.90 m3 ) ⎜ = 24.46 m3 ⎜ 1.01 × 105 Pa ⎟⎟ ⎝ ⎠ The number of cylinders required to fill a 750 m3 balloon is 750 m3 / 24.46 m3 = 30.7 cylinders EVALUATE: The ratio of the volume of the balloon to the volume of a cylinder is about 400 Fewer cylinders than this are required because of the large factor by which the gas is compressed in the cylinders (b) IDENTIFY: The upward force on the balloon is given by Archimedes’s principle (Chapter 12): B = weight of air displaced by balloon = ρairVg Apply Newton’s second law to the balloon and solve for the weight of the load that can be supported Use the ideal-gas equation to find the mass of the gas in the balloon SET UP: The free-body diagram for the balloon is given in Figure 18.63 mgas is the mass of the gas that is inside the balloon; mL is the mass of the load that is supported by the balloon EXECUTE: ∑ Fy = ma y B − mL g − mgas g = Figure 18.63 ρairVg − mL g − mgas g = mL = ρairV − mgas Calculate mgas , the mass of hydrogen that occupies 750 m3 at 15°C and p = 1.01 × 105 Pa pV = nRT = ( mgas /M ) RT gives mgas = pVM/RT = (1.01 × 105 Pa)(750 m3 )(2.02 × 10−3 kg/mol) = 63.9 kg (8.3145 J/mol ⋅ K)(288 K) Then mL = (1.23 kg/m3 )(750 m3 ) − 63.9 kg = 859 kg, and the weight that can be supported is wL = mL g = (859 kg)(9.80 m/s ) = 8420 N (c) mL = ρairV − mgas mgas = pVM/RT = (63.9 kg)((4.00 g/mol) /(2.02 g/mol)) = 126.5 kg (using the results of part (b)) Then mL = (1.23 kg/m3 )(750 m3 ) − 126.5 kg = 796 kg wL = mL g = (796 kg)(9.80 m/s ) = 7800 N EVALUATE: A greater weight can be supported when hydrogen is used because its density is less © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18.64 18-17 IDENTIFY: The upward force exerted by the gas on the piston must equal the piston’s weight Use pV = nRT to calculate the volume of the gas, and from this the height of the column of gas in the cylinder SET UP: F = pA = pπ r , with r = 0.100 m and p = 0.500 atm = 5.065 × 104 Pa For the cylinder, V = π r h pπ r (5.065 × 104 Pa)π (0.100 m) = = 162 kg g 9.80 m/s (b) V = πr2h and V = nRT/p Combining these equations gives h = nRT/πr2p, which gives (1.80 mol)(8.314 J/mol ⋅ K)(293.15 K) h= = 276 m π (0.100 m)2 (5.065 × 104 Pa) EVALUATE: The calculation assumes a vacuum ( p = 0) in the tank above the piston IDENTIFY: Apply Bernoulli’s equation to relate the efflux speed of water out the hose to the height of water in the tank and the pressure of the air above the water in the tank Use the ideal-gas equation to relate the volume of the air in the tank to the pressure of the air (a) SET UP: Points and are shown in Figure 18.65 (a) pπ r = mg and m = EXECUTE: 18.65 p1 = 4.20 × 105 Pa p2 = pair = 1.00 × 105 Pa large tank implies v1 ≈ Figure 18.65 EXECUTE: ρ v2 2 p1 + ρ gy1 + 12 ρ v12 = p2 + ρ gy2 + 12 ρ v22 = p1 − p2 + ρ g ( y1 − y2 ) v2 = (2/ρ)( p1 − p2 ) + g ( y1 − y2 ) v2 = 26.2 m/s (b) h = 3.00 m The volume of the air in the tank increases so its pressure decreases pV = nRT = constant, so pV = p0V0 ( p0 is the pressure for h0 = 3.50 m and p is the pressure for h = 3.00 m) p(4.00 m − h) A = p0 (4.00 m − h0 ) A ⎛ 4.00 m − h0 ⎞ ⎛ 4.00 m − 3.50 m ⎞ 5 p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ = 2.10 × 10 Pa ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − 3.00 m ⎠ Repeat the calculation of part (a), but now p1 = 2.10 × 105 Pa and y1 = 3.00 m v2 = (2/ρ)( p1 − p2 ) + g ( y1 − y2 ) v2 = 16.1 m/s h = 2.00 m ⎛ 4.00 m − h0 ⎞ ⎛ 4.00 m − 3.50 m ⎞ 5 p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ = 1.05 × 10 Pa ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − 2.00 m ⎠ v2 = (2/ρ)( p1 − p2 ) + g ( y1 − y2 ) v2 = 5.44 m/s (c) v2 = means (2 / ρ )( p1 − p2 ) + g ( y1 − y2 ) = p1 − p2 = − ρ g ( y1 − y2 ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-18 Chapter 18 y1 − y2 = h − 1.00 m ⎛ 0.50 m ⎞ ⎛ 0.50 m ⎞ p = p0 ⎜ ⎟ = (4.20 × 10 Pa) ⎜ ⎟ This is p1, so ⎝ 4.00 m − h ⎠ ⎝ 4.00 m − h ⎠ ⎛ 0.50 m ⎞ (4.20 × 105 Pa) ⎜ ⎟ − 1.00 × 10 Pa = (9.80 m/s )(1000 kg/m )(1.00 m − h) − 00 m h ⎝ ⎠ (210 / (4.00 − h)) − 100 = 9.80 − 9.80h, with h in meters 210 = (4.00 − h)(109.8 − 9.80h) 9.80h − 149h + 229.2 = and h − 15.20h + 23.39 = ( ) quadratic formula: h = 12 15.20 ± (15.20) − 4(23.39) = (7.60 ± 5.86) m h must be less than 4.00 m, so the only acceptable value is h = 7.60 m − 5.86 m = 1.74 m EVALUATE: The flow stops when p + ρ g ( y1 − y2 ) equals air pressure For h = 1.74 m, p = 9.3 × 104 Pa and ρ g ( y1 − y2 ) = 0.7 × 104 Pa, so p + ρ g ( y1 − y2 ) = 1.0 × 105 Pa, which is air pressure 18.66 IDENTIFY: Use the ideal gas law to find the number of moles of air taken in with each breath and from this calculate the number of oxygen molecules taken in Then find the pressure at an elevation of 2000 m and repeat the calculation SET UP: The number of molecules in a mole is N A = 6.022 × 1023 molecules/mol R = 0.08206 L ⋅ atm/mol ⋅ K Example 18.4 shows that the pressure variation with altitude y, when constant temperature is assumed, is p = p0e − Mgy/RT For air, M = 28.8 × 10−3 kg/mol EXECUTE: (a) pV = nRT gives n = pV (1.00 atm)(0.50 L) = = 0.0208 mol RT (0.08206 L ⋅ atm/mol ⋅ K)(293.15 K) N = (0.210) nN A = (0.210)(0.0208 mol)(6.022 × 1023 molecules/mol) = 2.63 × 1021 molecules (b) Mgy (28.8 × 10−3 kg/mol)(9.80 m/s )(2000 m) = = 0.2316 RT (8.314 J/mol ⋅ K)(293.15 K) p = p0e − Mgy/RT = (1.00 atm)e −0.2316 = 0.793 atm N is proportional to n, which is in turn proportional to p, so ⎛ 0.793 atm ⎞ 21 21 N =⎜ ⎟ (2.63 × 10 molecules) = 2.09 × 10 molecules ⎝ 1.00 atm ⎠ (c) Less O is taken in with each breath at the higher altitude, so the person must take more breaths per 18.67 minute EVALUATE: A given volume of gas contains fewer molecules when the pressure is lowered and the temperature is kept constant IDENTIFY and SET UP: Apply Eq.(18.2) to find n and then use Avogadro’s number to find the number of molecules EXECUTE: Calculate the number of water molecules N m 50 kg Number of moles: n = tot = = 2.778 × 103 mol M 18.0 × 10−3 kg/mol N = nN A = (2.778 × 103 mol)(6.022 × 1023 molecules/mol) = 1.7 × 1027 molecules Each water molecule has three atoms, so the number of atoms is 3(1.7 × 1027 ) = 5.1 × 1027 atoms EVALUATE: We could also use the masses in Example 18.5 to find the mass m of one H 2O molecule: m = 2.99 × 10−26 kg Then N = mtot /m = 1.7 × 1027 molecules, which checks 18.68 N RT Deviations will be noticeable when the volume V of a molecule is on the NA order of 1% of the volume of gas that contains one molecule IDENTIFY: pV = nRT = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-19 SET UP: The volume of a sphere of radius r is V = π r 3 RT EXECUTE: The volume of gas per molecule is , and the volume of a molecule is about NA p V0 = π (2.0 × 10−10 m)3 = 3.4 × 10−29 m3 Denoting the ratio of these volumes as f, p= f 18.69 RT (8.3145 J/mol ⋅ K)(300 K) = f = (1.2 × 108 Pa) f N AV0 (6.022 × 1023 molecules/mol)(3.4 × 10−29 m3 ) “Noticeable deviations” is a subjective term, but f on the order of 1.0% gives a pressure of 106 Pa EVALUATE: The forces between molecules also cause deviations from ideal-gas behavior IDENTIFY: Eq (18.16) says that the average translational kinetic energy of each molecule is equal to 32 kT vrms = 3kT m SET UP: k = 1.381 × 10−23 J/molecule ⋅ K EXECUTE: (a) m (v ) av depends only on T and both gases have the same T, so both molecules have the same average translational kinetic energy vrms is proportional to m −1/2 , so the lighter molecules, A, have the greater vrms (b) The temperature of gas B would need to be raised T T T vrms (c) = = constant, so A = B m A mB m 3k ⎛ 5.34 × 10−26 kg ⎞ ⎛m ⎞ TB = ⎜ B ⎟ TA = ⎜ (283.15 K) = 4.53 × 103 K = 4250°C ⎜ 3.34 × 10−27 kg ⎟⎟ m ⎝ A⎠ ⎝ ⎠ (d) TB > TA so the B molecules have greater translational kinetic energy per molecule 3kT the temperature T must be in kelvins m IDENTIFY: The equations derived in the subsection Collisions Between Molecules in Section 18.3 can be applied to the bees The average distance a bee travels between collisions is the mean free path, λ The average dN time between collisions is the mean free time, tmean The number of collisions per second is = dt tmean EVALUATE: In 18.70 m(v ) av = 32 kT and vrms = SET UP: V = (1.25 m)3 = 1.95 m3 r = 0.750 × 10−2 m v = 1.10 m/s N = 2500 EXECUTE: (a) λ = V 4π 2r N (b) λ = vtmean , so tmean = λ = = 1.95 m3 4π 2(0.750 × 10−2 m) (2500) = 0.780 m = 78.0 cm 0.780 m = 0.709 s 1.10 m/s v dN 1 (c) = = = 1.41 collisions/s dt tmean 0.709 s 18.71 EVALUATE: The calculation is valid only if the motion of each bee is random IDENTIFY: The mass of one molecule is the molar mass, M, divided by the number of molecules in a mole, N A The average translational kinetic energy of a single molecule is 12 m(v )av = 23 kT Use pV = NkT to calculate N, the number of molecules SET UP: k = 1.381 × 10−23 J/molecule ⋅ K M = 28.0 × 10−3 kg/mol T = 295.15 K The volume of the balloon is V = 43 π (0.250 m)3 = 0.0654 m3 p = 1.25 atm = 1.27 × 105 Pa © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-20 Chapter 18 EXECUTE: (a) m = (b) m(v ) av M 28.0 × 10−3 kg/mol = = 4.65 × 10−26 kg N A 6.022 × 1023 molecules/mol = 23 kT = 23 (1.381 × 10−23 J/molecule ⋅ K)(295.15 K) = 6.11 × 10−21 J pV (1.27 × 105 Pa)(0.0654 m3 ) = = 2.04 × 1024 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(295.15 K) (d) The total average translational kinetic energy is (c) N = N ( m (v ) av ) = (2.04 ×10 24 molecules)(6.11 × 10−21 J/molecule) = 1.25 × 104 J EVALUATE: The number of moles is n = N 2.04 × 1024 molecules = = 3.39 mol N A 6.022 × 1023 molecules/mol K tr = 32 nRT = 32 (3.39 mol)(8.314 J/mol ⋅ K)(295.15 K) = 1.25 × 104 J, which agrees with our results in part (d) 18.72 IDENTIFY: U = mgy The mass of one molecule is m = M/N A K av = 32 kT SET UP: Let y = at the surface of the earth and h = 400 m N A = 6.022 × 1023 molecules/mol and k = 1.38 × 10−23 J/K 15.0°C = 288 K ⎛ ⎞ M 28.0 × 10−3 kg/mol −22 gh = ⎜ J ⎟⎟ (9.80 m/s )(400 m) = 1.82 × 10 23 ⎜ NA 022 10 molecules/mol × ⎝ ⎠ ⎛ 1.82 × 10−22 J ⎞ (b) Setting U = kT , T = ⎜ ⎟ = 8.80 K ⎜⎝ 1.38 × 10−23 J/K ⎟⎠ EVALUATE: (c) The average kinetic energy at 15.0°C is much larger than the increase in gravitational potential energy, so it is energetically possible for a molecule to rise to this height But Example 18.8 shows that the mean free path will be very much less than this and a molecule will undergo many collisions as it rises These numerous collisions transfer kinetic energy between molecules and make it highly unlikely that a given molecule can have very much of its translational kinetic energy converted to gravitational potential energy IDENTIFY and SET UP: At equilibrium F (r ) = The work done to increase the separation from r2 to ∞ EXECUTE: (a) U = mgh = 18.73 is U (∞) − U (r2 ) (a) EXECUTE: U (r ) = U 0[( R0 /r )12 − 2( R0 /r )6 ] Eq (14.26): F (r ) = 12(U /R0 )[( R0 /r )13 − ( R0 /r )7 ] The graphs are given in Figure 18.73 Figure 18.73 (b) equilibrium requires F = 0; occurs at point r2 r2 is where U is a minimum (stable equilibrium) (c) U = implies [( R0 /r )12 2( R0 /r )6 ] = (r1/R0 )6 = 1/2 and r1 = R0 /(2)1/6 F = implies [( R0 /r )13 − ( R0 /r )7 ] = (r2 /R0 )6 = and r2 = R0 Then r1/r2 = ( R0 /21/6 )/R0 = 2−1/6 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-21 (d) Wother = ΔU At r → ∞, U = 0, so W = −U ( R0 ) = −U 0[( R0 /R0 )12 − 2( R0 /R0 )6 ] = +U EVALUATE: The answer to part (d), U , is the depth of the potential well shown in the graph of U (r ) 18.74 IDENTIFY: Use pV = nRT to calculate the number of moles, n Then K tr = 32 nRT The mass of the gas, mtot , is given by mtot = nM SET UP: 5.00 L = 5.00 × 10−3 m3 EXECUTE: (a) n = pV (1.01 × 105 Pa)(5.00 × 10−3 m3 ) = = 0.2025 moles RT (8.314 J/mol ⋅ K)(300 K) K tr = 32 (0.2025 mol)(8.314 J/mol ⋅ K)(300 K) = 758 J (b) mtot = nM = (0.2025 mol)(2.016 × 10−3 kg/mol) = 4.08 × 10−4 kg The kinetic energy due to the speed of the jet is K = 12 mv = 12 (4.08 × 10−4 kg)(300.0 m/s)2 = 18.4 J The total kinetic energy is K tot = K + K tr = 18.4 J + 758 J = 776 J The percentage increase is K 18.4 J × 100% = × 100% = 2.37% K tot 776 J (c) No The temperature is associated with the random translational motion, and that hasn’t changed EVALUATE: Eq (18.13) gives K tr = 32 pV = 32 (1.01 × 105 Pa)(5.00 × 10−3 m3 ) = 758 J, which agrees with 3RT = 1.93 × 103 m/s vrms is a lot larger than the speed of the jet, so the M percentage increase in the total kinetic energy, calculated in part (b), is small IDENTIFY and SET UP: Apply Eq (18.19) for vrms The equation preceeding Eq (18.12) relates vrms and our result in part (a) vrms = 18.75 (vx ) rms EXECUTE: (a) vrms = 3RT/M vrms = 3(8.3145 J/mol ⋅ K)(300 K) 28.0 × 10−3 kg/mol (b) (vx2 )av = 13 (v )av so ( = 517 m/s (vx2 )av = 1/ ) ( ) ( ) (v )av = 1/ vrms = 1/ (517 m/s) = 298 m/s EVALUATE: The speed of sound is approximately equal to (vx )rms since it is the motion along the 18.76 direction of propagation of the wave that transmits the wave 3kT IDENTIFY: vrms = m SET UP: M = 1.99 × 1030 kg, R = 6.96 × 108 m and G = 6.673 × 10−11 N ⋅ m /kg EXECUTE: (a) vrms = (b) vescape = 18.77 3kT 3(1.38 × 10−23 J/K)(5800 K) = = 1.20 × 104 m/s m (1.67 × 10−27 kg) 2GM 2(6.673 × 10−11 N ⋅ m /kg )(1.99 × 1030 kg) = = 6.18 × 105 m/s R (6.96 × 108 m) EVALUATE: (c) The escape speed is about 50 times the rms speed, and any of Figure 18.23 in the textbook, Eq (18.32) or Table (18.2) will indicate that there is a negligibly small fraction of molecules with the escape speed (a) IDENTIFY and SET UP: Apply conservation of energy K1 + U1 + Wother = K + U , where U = −Gmmp /r Let point be at the surface of the planet, where the projectile is launched, and let point be far from the earth Just barely escapes says v2 = EXECUTE: Only gravity does work says Wother = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-22 Chapter 18 U1 = −Gmmp /Rp ; r2 → ∞ so U = 0; v2 = so K = The conservation of energy equation becomes K1 − Gmmp /Rp = and K1 = Gmmp /Rp But g = Gmp /Rp2 so Gmp /Rp = Rp g and K1 = mgRp , as was to be shown EVALUATE: The greater gRp is, the more initial kinetic energy is required for escape (b) IDENTIFY and SET UP: Set K1 from part (a) equal to the average kinetic energy of a molecule as given by Eq (18.16) EXECUTE: T = m(v ) av = mgRp (from part (a)) But also, m(v ) av = 32 kT , so mgRp = 32 kT 2mgRp 3k nitrogen mN = (28.0 × 10−3 kg/mol)/(6.022 × 1023 molecules/mol) = 4.65 × 10−26 kg/molecule T= 2mgRp 3k = 2(4.65 × 10−26 kg/molecule)(9.80 m/s )(6.38 × 106 m) 3(1.381× 10 −23 J/molecule ⋅ K) = 1.40 × 105 K hydrogen mH = (2.02 × 10−3 kg/mol)/(6.022 × 1023 molecules/mol) = 3.354 × 10−27 kg/molecule T= 2mgRp 3k (c) T = = 2(3.354 × 10−27 kg/molecule)(9.80 m/s )(6.38 × 106 m) 3(1.381 × 10−23 J/molecule ⋅ K) = 1.01 × 104 K 2mgRp 3k nitrogen T= 18.78 2(4.65 × 10−26 kg/molecule)(1.63 m/s )(1.74 × 106 m) 3(1.381 × 10−23 J/molecule ⋅ K) = 6370 K hydrogen 2(3.354 × 10−27 kg/molecule)(1.63 m/s )(1.74 × 106 m) T= = 459 K 3(1.381 × 10−23 J/molecule ⋅ K) (d) EVALUATE: The “escape temperatures” are much less for the moon than for the earth For the moon a larger fraction of the molecules at a given temperature will have speeds in the Maxwell-Boltzmann distribution larger than the escape speed After a long time most of the molecules will have escaped from the moon 3RT IDENTIFY: vrms = M SET UP: M H = 2.02 × 10−3 kg/mol M O2 = 32.0 × 10−3 kg/mol For earth, M = 5.97 × 1024 kg and R = 6.38 × 106 m For Jupiter, M = 1.90 × 1027 kg and R = 6.91 × 107 m For a sphere, M = ρV = ρ π r 3 The escape speed is vescape = 2GM R EXECUTE: (a) Jupiter: vrms = 3(8.3145J/mol ⋅ K)(140 K) / (2.02 × 10−3 kg/mol) = 1.31 × 103 m/s vescape = 6.06 × 104 m/s vrms = 0.022vescape Earth: vrms = 3(8.3145 J/mol ⋅ K)(220 K) / (2.02 × 10−3 kg/mol) = 1.65 × 103 m/s vescape = 1.12 × 104 m/s vrms = 0.15vescape (b) Escape from Jupiter is not likely for any molecule, while escape from earth is much more probable © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-23 (c) vrms = 3(8.3145 J/mol ⋅ K)(200 K) / (32.0 × 10−3 kg/mol) = 395 m/s The radius of the asteroid is R = (3M/ 4πρ )1/ = 4.68 × 105 m, and the escape speed is vescape = 2GM/R = 542 m/s Over time the O molecules would essentially all escape and there can be no such atmosphere EVALUATE: As Figure 18.23 in the textbook shows, there are some molecules in the velocity distribution that have speeds greater than vrms But as the speed increases above vrms the number with speeds in that range decreases 18.79 3kT m NA The number of molecules in an object of mass m is N = nN A = m M SET UP: The volume of a sphere of radius r is V = π r 3 IDENTIFY: vrms = EXECUTE: (a) m = 3kT vrms = 3(1.381 × 10−23 J/K) ( 300 K ) (0.0010 m/s) = 1.24 × 10−14 kg (b) N = mN A /M = (1.24 × 10−14 kg)(6.022 × 1023 molecules/mol) / (18.0 × 10−3 kg/mol) N = 4.16 × 1011 molecules 1/3 ⎛ 3V ⎞ (c) The diameter is D = 2r = ⎜ ⎟ ⎝ 4π ⎠ 1/3 ⎛ 3m/ρ ⎞ = 2⎜ ⎟ ⎝ 4π ⎠ 1/3 ⎛ 3(1.24 × 10−14 kg) ⎞ = 2⎜ ⎜ 4π (920 kg/m3 ) ⎟⎟ ⎝ ⎠ = 2.95 × 10−6 m which is too small to see EVALUATE: vrms decreases as m increases 18.80 IDENTIFY: For a simple harmonic oscillator, x = A cos ωt and vx = −ω A sin ωt , with ω = k/m SET UP: The average value of cos(2ωt ) over one period is zero, so (sin ωt )av = (cos ωt )av = 12 EXECUTE: x = A cos ωt , vx = −ω A sin ωt , U av = 12 kA2 (cos ωt )av , K av = 12 mω A2 (sin ωt )av Using (sin ωt )av = (cos ωt )av = 12 and mω = k shows that K av = U av 18.81 EVALUATE: In general, at any given instant of time U ≠ K It is only the values averaged over one period that are equal IDENTIFY: The equipartition principle says that each atom has an average kinetic energy of 12 kT for each 18.82 degree of freedom There is an equal average potential energy SET UP: The atoms in a three-dimensional solid have three degrees of freedom and the atoms in a twodimensional solid have two degrees of freedom EXECUTE: (a) In the same manner that Eq (18.28) was obtained, the heat capacity of the twodimensional solid would be R = 16.6 J/mol ⋅ K (b) The heat capacity would behave qualitatively like those in Figure 18.21 in the textbook, and the heat capacity would decrease with decreasing temperature EVALUATE: At very low temperatures the equipartition theorem doesn’t apply Most of the atoms remain in their lowest energy states because the next higher energy level is not accessible IDENTIFY: The equipartition principle says that each molecule has average kinetic energy of 12 kT for each degree of freedom I = 2m( L/ 2) , where L is the distance between the two atoms in the molecule K rot = 12 I ω ωrms = (ω )av SET UP: The mass of one atom is m = M/N A = (16.0 × 10−3 kg/mol)/(6.022 × 1023 molecules/mol) = 2.66 × 10−26 kg EXECUTE: (a) The two degrees of freedom associated with the rotation for a diatomic molecule account for two-fifths of the total kinetic energy, so K rot = nRT = (1.00 mol)(8.3145 J/mol ⋅ K)(300 K) = 2.49 ì 103 J â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-24 Chapter 18 ⎛ ⎞ 16.0 × 10−3 kg/mol (6.05 × 10−11 m) = 1.94 × 10−46 kg ⋅ m (b) I = 2m( L/2) = ⎜ ⎜ 6.022 × 1023 molecules/mol ⎟⎟ ⎝ ⎠ (c) Since the result in part (b) is for one mole, the rotational kinetic energy for one atom is K rot /N A and ω rms = K rot /N A 2(2.49 × 103 J) = = 6.52 × 1012 rad/s This is 46 − I (1.94 × 10 kg ⋅ m )(6.022 × 1023 molecules/mol) much larger than the typical value for a piece of rotating machinery 2π rad EVALUATE: The average rotational period, T = , for molecules is very short ωrms 18.83 IDENTIFY: CV = N ( 12 R ) , where N is the number of degrees of freedom SET UP: There are three translational degrees of freedom EXECUTE: For CO , N = and the contribution to CV other than from vibration is R = 20.79 J/mol ⋅ K and CV − 52 R = 0.270 CV So 27% of CV is due to vibration For both SO2 and H2S, N = and the contribution to CV other than from vibration is R = 24.94 J/mol ⋅ K The respective fractions of CV from vibration are 21% and 3.9% EVALUATE: The vibrational contribution is much less for H 2S In H 2S the vibrational energy steps are 18.84 larger because the two hydrogen atoms have small mass and ω = k/m IDENTIFY: Evaluate the integral, as specified in the problem SET UP: Use the integral formula given in Problem 18.85, with α = m/ 2kT 3/2 3/ ∞ − mv /2 kT ⎞ π ⎛ m ⎞ ⎛ m ⎞ ⎛ f (v) dv = 4π ⎜ dv = 4π ⎜ =1 ⎟ ⎟ ∫0 v e ⎟ ⎜ kT kT 4( m /2 kT ) m /2 kT π π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ EVALUATE: (b) f (v ) dv is the probability that a particle has speed between v and v + dv; the probability that the particle has some speed is unity, so the sum (integral) of f (v )dv must be EXECUTE: (a) 18.85 ∞ ∫0 IDENTIFY and SET UP: Evaluate the integral in Eq (18.31) as specified in the problem EXECUTE: ∞ ∫0 v f (v) dv = 4π (m/2π kT )3/2 ∞ − av ∫0 v e The integral formula with n = gives Apply with a = m/ 2kT , ∞ ∫0 v EVALUATE: Eq (18.16) says 18.86 ∞ − mv / kT ∫0 v e dv dv = (3/8a ) π /a f (v) dv = 4π (m/2π kT )3/2 (3/8)(2kT/m) 2π kT/m = (3/2)(2kT/m) = 3kT/m m(v ) av = 3kT/ 2, so (v )av = 3kT/m, in agreement with our calculation IDENTIFY: Follow the procedure specified in the problem SET UP: If v = x, then dx = 2vdv EXECUTE: ∞ ∫0 ⎛ m ⎞ vf (v)dv = 4π ⎜ ⎟ ⎝ 2π kT ⎠ 3/2 ∞ − mv /2 kT ∫0 v e dv Making the suggested change of variable, v = x 2vdv = dx, v3 dv = (1/2) x dx, and the integral becomes ⎛ m ⎞ ∫ vf (v)dv = 2π ⎜⎝ 2π kT ⎟⎠ which is Eq (18.35) ∞ EVALUATE: The integral ∞ 3/2 ∞ ∫0 ∫ vf (v)dv ⎛ m ⎞ xe− mx/2 kT dx = 2π ⎜ ⎟ ⎝ 2π kT ⎠ 3/2 2 ⎛ 2kT ⎞ ⎜ ⎟ = π ⎝ m ⎠ 2kT 8kT = πm m is the definition of vav © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18.87 IDENTIFY: 18-25 f (v)dv is the probability that a particle has a speed between v and v + dv Eq (18.32) gives f (v) vmp is given by Eq (18.34) SET UP: For O2 , the mass of one molecule is m = M/N A = 5.32 × 10−26 kg EXECUTE: (a) f (v)dv is the fraction of the particles that have speed in the range from v to v + dv The number of particles with speeds between v and v + dv is therefore dN = Nf (v) dv and ΔN = N ∫ v + Δv v f (v) dv (b) Setting v = vmp = 2kT ⎛ m ⎞ in f (v ) gives f (vmp ) = 4π ⎜ ⎟ m ⎝ 2π kT ⎠ 3/2 ⎛ 2kT ⎞ −1 For oxygen ⎜ ⎟e = e π vmp ⎝ m ⎠ gas at 300 K, vmp = 3.95 × 102 m/s and f (v) Δv = 0.0421 (c) Increasing v by a factor of changes f by a factor of e −48 , and f (v) Δv = 2.94 × 10−21 (d) Multiplying the temperature by a factor of increases the most probable speed by a factor of 2, and −21 18.88 the answers are decreased by 2: 0.0297 and 2.08 × 10 (e) Similarly, when the temperature is one-half what it was in parts (b) and (c), the fractions increase by to 0.0595 and 4.15 × 10−21 EVALUATE: (f) At lower temperatures, the distribution is more sharply peaked about the maximum (the most probable speed), as is shown in Figure 18.23a in the textbook m IDENTIFY: Apply the definition of relative humidity given in the problem pV = nRT = tot RT M SET UP: M = 18.0 × 10−3 kg/mol EXECUTE: (a) The pressure due to water vapor is (0.60)(2.34 × 103 Pa) = 1.40 × 103 Pa MpV (18.0 × 10−3 kg/mol)(1.40 × 103 Pa)(1.00 m3 ) = = 10 g RT (8.3145 J/mol ⋅ K)(293.15 K) EVALUATE: The vapor pressure of water vapor at this temperature is much less than the total atmospheric pressure of 1.0 × 105 Pa IDENTIFY: The measurement gives the dew point Relative humidity is defined in Problem 18.88 partial pressure of water vapor at temperature T SET UP: relative humidity = vapor pressure of water at temperature T EXECUTE: The experiment shows that the dew point is 16.0°C, so the partial pressure of water vapor at (b) mtot = 18.89 30.0°C is equal to the vapor pressure at 16.0°C, which is 1.81 × 103 Pa Thus the relative humidity = 18.90 1.81 × 103 Pa = 0.426 = 42.6% 4.25 × 103 Pa EVALUATE: The lower the dew point is compared to the air temperature, the smaller the relative humidity IDENTIFY: Use the definition of relative humidity in Problem 18.88 and the vapor pressure table in Problem 18.89 SET UP: At 28.0°C the vapor pressure of water is 3.78 × 103 Pa EXECUTE: For a relative humidity of 35%, the partial pressure of water vapor is (0.35)(3.78 × 103 Pa) = 1.323 × 103 Pa This is close to the vapor pressure at 12°C, which would be at an altitude (30°C − 12°C)/(0.6 C° /100 m) = km above the ground For a relative humidity of 80%, the vapor pressure will be the same as the water pressure at around 24°C, corresponding to an altitude of about km EVALUATE: Clouds form at a lower height when the relative humidity at the surface is larger © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18-26 18.91 Chapter 18 IDENTIFY: Eq (18.21) gives the mean free path λ In Eq (18.20) use vrms = pV = nRT = NkT The escape speed is vescape = 3RT in place of υ M 2GM R SET UP: For atomic hydrogen, M = 1.008 × 10−3 kg/mol EXECUTE: (a) From Eq (18.21), λ = (4π 2r ( N /V )) −1 = (4π 2(5.0 × 10−11 m) (50 × 106 m −3 ))−1 = 4.5 × 1011 m (b) vrms = 3RT/M = 3(8.3145 J/mol ⋅ K)(20 K) / (1.008 × 10−3 kg/mol) = 703 m/s, and the time between collisions is then (4.5 × 1011 m) / (703 m/s) = 6.4 × 108 s, about 20 yr Collisions are not very important (c) p = ( N/V )kT = (50 / 1.0 × 10−6 m3 )(1.381 × 10−23 J/K)(20 K) = 1.4 × 10−14 Pa (d) vescape = 2GM 2G ( Nm/V )(4π R3/3) = = (8π / 3)G ( N/V )mR R R vescape = (8π /3)(6.673 × 10−11 N ⋅ m /kg )(50 × 106 m −3 )(1.67 × 10−27 kg)(10 × 9.46 × 1015 m)2 vescape = 650 m/s This is lower than vrms and the cloud would tend to evaporate (e) In equilibrium (clearly not thermal equilibrium), the pressures will be the same; from pV = NkT , kTISM ( N/V ) ISM = kTnebula ( N/V ) nebula and the result follows (f) With the result of part (e), ⎛ ( N/V )nebula ⎞ TISM = Tnebula ⎜ ⎟ = (20 K) ⎝ ( N/V ) ISM ⎠ 18.92 ⎛ 50 × 106 m3 ⎞ = × 105 K, ⎜⎜ −6 −1 ⎟ ⎟ ⎝ (200 × 10 m ) ⎠ more than three times the temperature of the sun This indicates a high average kinetic energy, but the thinness of the ISM means that a ship would not burn up EVALUATE: The temperature of a gas is determined by the average kinetic energy per atom of the gas The energy density for the gas also depends on the number of atoms per unit volume, and this is very small for the ISM IDENTIFY: Follow the procedure of Example 18.4, but use T = T0 − α y SET UP: ln(1 + x ) ≈ x when x is very small EXECUTE: (a) dp pM dp Mg dy , which in this case becomes =− =− This integrates to dy RT p R T0 − α y Mg/Rα ⎛ p ⎞ Mg ⎛ α y ⎞ ⎛ αy⎞ ln ⎜ ⎟ = ln ⎜1 − ⎟ , or p = p0 ⎜ − ⎟ T0 ⎠ T0 ⎠ ⎝ p0 ⎠ Rα ⎝ ⎝ ⎛ αy⎞ αy (b) For sufficiently small α , ln ⎜1 − , and this gives the expression derived in Example 18.4 ⎟≈− T0 ⎠ T0 ⎝ ⎛ (0.6 × 10−2 C°/m)(8863 m) ⎞ Mg (28.8 × 10−3 )(9.80 m/s ) (c) ⎜1 − = 0.8154, = = 5.6576 and ⎟ ⎜ ⎟ (288 K) Rα (8.3145 J/mol ⋅ K)(0.6 × 10−2 C°/m) ⎝ ⎠ p0 (0.8154)5.6576 = 0.315 atm, which is 0.95 of the result found in Example 18.4 EVALUATE: The pressure is calculated to decrease more rapidly with altitude when we assume that T also decreases with altitude 18.93 IDENTIFY and SET UP: For N particles, vav = EXECUTE: (a) vav = 12 (v1 + v2 ), vrms = ∑ vi and vrms = N ∑ vi2 N v12 + v22 and © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Thermal Properties of Matter 18-27 1 1 2 vrms − vav = (v12 + v22 ) − (v12 + v22 + 2v1v2 ) = (v12 + v22 − 2v1v2 ) = (v1 − v2 ) 2 4 This shows that vrms ≥ vav , with equality holding if and only if the particles have the same speeds 1 ( Nvrms ( Nvav + u ), and the given forms follow immediately + u ), v′ av = N +1 N +1 (c) The algebra is similar to that in part (a); it helps somewhat to express 2 v′av = ( N (( N + 1) − 1)vav + Nvavu + (( N + 1) − N )u ) ( N + 1)2 (b) v′ 2rms = v′av = N N (−vav vav + + 2vavu − u ) + u2 N +1 N + ( N + 1) Then, v′2rms − v′av = N N N N 2 2 (vrms )+ (v − 2vavu + u ) = (vrms )+ (vav − u ) − vav − vav av N + ( N + 1) ( N + 1) ( N + 1) If vrms > vav , then this difference is necessarily positive, and v′rms > v′av (d) The result has been shown for N = 1, and it has been shown that validity for N implies validity for N + 1; by induction, the result is true for all N EVALUATE: vrms > vav because vrms gives more weight to particles that have greater speed © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... from pV = NkT , kTISM ( N/V ) ISM = kTnebula ( N/V ) nebula and the result follows (f) With the result of part (e), ⎛ ( N/V )nebula ⎞ TISM = Tnebula ⎜ ⎟ = (20 K) ⎝ ( N/V ) ISM ⎠ 18.92 ⎛ 50 ×... = = 8.737 × 106 Pa V 3.10 × 10−3 m3 p2 = 100 atm(1. 013 × 105 Pa/1 atm) = 1. 013 × 107 Pa p/T = nR/V = constant, so p1/T1 = p2 /T2 18.16 ⎛ 1. 013 × 107 Pa ⎞ ⎛p ⎞ EXECUTE: T2 = T1 ⎜ ⎟ = (296.15 K)... Fav = Δt dt 4 .13 × 10−4 s (f) pressure = F/A = 1.24 × 10−19 N/(0.10 m) = 1.24 × 10−17 Pa (due to one molecule) (g) pressure = atm = 1. 013 × 105 Pa Number of molecules needed is 1. 013 × 105 Pa/(1.24