PROPERTIES THERMAL OF MATTER
18
18.1.
(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. m 0.225 kg = 56.2 mol EXECUTE: n = tot = M 4.00 10-3 kg/mol (b) pV = nRT implies p = nRT / V T must be in kelvins; T = (18 + 273) K = 291 K
18.2.
(56.2 mol)(8.3145 J/mol K)(291 K) = 6.80 106 Pa 20.0 10-3 m3 p = (6.80 106 Pa)(1.00 atm/1.013 105 Pa) = 67.1 atm EVALUATE: Example 18.1 shows that 1.0 mol of an ideal gas is about this volume at STP. Since there are 56.2 moles the pressure is about 60 times greater than 1 atm. IDENTIFY: pV = nRT . p=
SET UP: EXECUTE:
T1 = 41.0C = 314 K . R = 0.08206 L atm/mol K . n, R constant so pV pV pV = nR is constant. 1 1 = 2 2 . T1 T2 T
18.3.
18.4.
p V T2 = T1 2 2 = (314 K)(2)(2) = 1.256 103 K = 983C . p1 V1 (1.30 atm)(2.60 L) pV (b) n = = = 0.131 mol . mtot = nM = (0.131 mol)(4.00 g/mol) = 0.524 g . RT (0.08206 L atm/mol K)(314 K) 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 4. IDENTIFY: pV = nRT . SET UP: T is constant. V 0.110 m3 EXECUTE: nRT is constant so p1V1 = p2V2 . p2 = p1 1 = (3.40 atm) = 0.959 atm . 3 0.390 m V2 EVALUATE: For T constant, p decreases. IDENTIFY: pV = nRT .
SET UP: EXECUTE:
T1 = 20.0C = 293 K .
(a) n, R, and V are constant.
p p p nR = = constant . 1 = 2 . T1 T2 T V
p 1.00 atm T2 = T1 2 = (293 K) = 97.7 K = -175C . p1 3.00 atm (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 . p 1.00 atm V3 = V2 2 = (3.00 L) = 1.00 L . 3.00 atm p3 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.
18-1
18-2
Chapter 18
18.5.
IDENTIFY:
pV = nRT
SET UP: Assume a room size of 20 ft 20 ft 10 ft . V = 4000 ft 3 = 113 m3 . Assume a temperature of T = 20C = 293 K and a pressure of p = 1.01 105 Pa . 1 m3 = 106 cm3 . EXECUTE: (a) n =
pV (1.01 105 Pa)(113 m3 ) = = 4.68 103 mol . RT (8.315 J/mol K)(293 K)
N = nN A = (4.68 103 mol)(6.022 1023 molecules/mol) = 3 1027 molecules . N 3 1027 molecules = = 3 1025 molecules/m3 = 3 1019 molecules/cm3 V 113 m3 EVALUATE: The solution doesn't rely on the assumption that air is all N 2 .
(b) 18.6. IDENTIFY: SET UP:
pV = nRT and the mass of the gas is mtot = nM . The temperature is T = 22.0C = 295.15K. The average molar mass of air is M = 28.8 10-3 kg mol . pV (1.00 atm)(0.900 L)(28.8 10-3 kg/mol) M= = 1.07 10-3 kg. RT (0.08206 L atm/mol K)(295.15 K)
For helium M = 4.00 10-3 kg mol .
EXECUTE: (a) mtot = nM =
(b) mtot = nM = EVALUATE:
pV (1.00 atm)(0.900 L)(4.00 10-3 kg/mol) M= = 1.49 10-4 kg. RT (0.08206 L atm/mol K)(295.15 K)
18.7.
N pV = says that in each case the balloon contains the same number of molecules. The mass N A RT 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 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 n=
EXECUTE:
T1 = (27 + 273) K = 300 K
5
p1 = 1.01 10 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) p V 2.82 106 Pa 46.2 cm3 = 776 K T2 = T1 2 2 = 300 K 5 3 1.01 10 Pa 499 cm p1 V1 T2 = (776 - 273)C = 503C
EVALUATE: 18.8.
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. IDENTIFY: 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 , p1V1 (4.01 105 Pa)(0.075 m3 ) = = 11.7 mol . m = nM = (11.7 mol)(32.0 g/mol) = 374 g . RT1 (8.315 J/mol K)(310 K)
T2 = 295 K .
EXECUTE: (b) n2 = (a) n1 =
18.9.
p2V2 (2.81 105 Pa)(0.075 m 3 ) = = 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 . 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: EXECUTE:
T1 = 300 K , T2 = 430 K .
(a) n, R are constant so
pV pV pV = nR = constant . 1 1 = 2 2 . T1 T2 T
V T 0.750 m3 430 K 5 p2 = p1 1 2 = (1.50 105 Pa) = 3.36 10 Pa . V2 T1 0.480 m3 300 K EVALUATE: In pV = nRT , T must be in kelvins, even if we use a ratio of temperatures.
Thermal Properties of Matter
18-3
18.10.
IDENTIFY:
Use the ideal-gas equation to calculate the number of moles, n. The mass mtotal of the gas is The volume of the cylinder is V = r 2l , where r = 0.450 m and l = 1.50 m. T = 22.0C = 293.15 K. pV (21.0 atm)(1.013 105 Pa/atm) (0.450 m) 2 (1.50 m) = = 827 mol. RT (8.314 J/mol K)(295.15 K)
mtotal = nM .
SET UP:
1 atm = 1.013 105 Pa. M = 32.0 10-3 kg/mol. R = 8.314 J/mol K.
EXECUTE: (a) pV = nRT gives n =
18.11.
(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 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)
18.12.
V2 = V1 (T2 / T1 ) = (0.600 L)(77.3 K/292 K) = 0.159 L 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. (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: pV = nRT . SET UP: T is constant. EXECUTE: n, R, T are constant, so pV = nRT = constant. p1V1 = p2V2 .
18.13.
18.14.
V 6.00 L p2 = p1 1 = (1.00 atm) = 1.05 atm. 5.70 L V2 EVALUATE: For constant T, when V decreases, p increases. Since the volumes enter as a ratio we don't have to convert from L to m3 . IDENTIFY: pV = nRT .
SET UP: EXECUTE:
T1 = 277 K. T2 = 296 K. Assume the number of moles of gas in the bubble remains constant.
(a) n, R are constant so
pV pV pV = nR = constant. 1 1 = 2 2 and T1 T2 T
18.15.
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 + 273.13)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 p 1.013 107 Pa T2 = T1 2 = (296.15 K) = 343.4 K = 70.2C 6 8.737 10 Pa p1 (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.
EXECUTE:
18-4
Chapter 18
18.16.
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 = 20.0C = 293.15 K, nRT (3 mol) (8.3145 J mol K) (293.15 K) = = 3.66 104 N. L 0.200 m (b) For T = 100.00C = 373.15 K, F= F= nRT (3 mol)(8.3145 J mol K)(373.15 K) = = 4.65 104 N. L 0.200 m
18.17.
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 0C at all altitudes and neglects the variation of g with
elevation. With these approximations, p = p0e - Mgy / RT .
SET UP: EXECUTE:
ln(e - x ) = - x. For air, M = 28.8 10-3 kg/mol. We want y for p = 0.90 p0 so 0.90 = e - Mgy / RT and y = -
18.18.
RT ln(0.90) = 850 m. Mg EVALUATE: This is a commonly occurring elevation, so our calculation shows that 10% variations in atmospheric pressure occur at many locations. IDENTIFY: From Example 18.4, the pressure at elevation y above sea level is p = p0e - Mgy / RT .
SET UP: EXECUTE:
The average molar mass of air is M = 28.8 10-3 kg/mol. At an altitude of 100 m, Mgy1 (28.8 10-3 kg mol)(9.80 m s 2 )(100 m) = = 0.01243, and the percent RT (8.3145 J mol K)(273.15 K)
decrease in pressure is 1 - p p0 = 1 - e -0.01243 = 0.0124 = 1.24%. At an altitude of 1000 m, Mgy2 RT = 0.1243 and the percent decrease in pressure is 1 - 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 variation of pressure with altitude is exponential rather than linear. IDENTIFY: p = p0e - Myg RT from Example 18.4. Eq.(18.5) says p = ( M )RT. Example 18.4 assumes a constant T = 273 K, so p and are directly proportional and we can write = 0e - Mgy RT . Mgy = 1.10 when y = 8863 m. RT Mgy EXECUTE: For y = 100 m, = 0.0124, so = 0e-0.0124 = 0.988 0 . The density at sea level is 1.2% larger RT than the density at 100 m. m EVALUATE: The pressure decreases with altitude. pV = tot RT , so when the pressure decreases and T is M constant the volume of a given mass of gas increases and the density decreases. IDENTIFY: p = p0e - Mgy / RT from Example 18.4 gives the variation of air pressure with altitude. The density
SET UP:
18.19.
From Example 18.4,
18.20.
of the air is = pressure is p0 .
SET UP: EXECUTE:
pM , so is proportional to the pressure p. Let 0 be the density at the surface, where the RT Mg (28.8 10-3 kg/mol)(9.80 m/s 2 ) = = 1.244 10-4 m -1. RT (8.314 J/mol K)(273 K)
= 0.883 p0 .
From Example 18.4,
p = p0e - (1.24410
-4
p M = constant, so = 0 and = 0 = 0.883 0 . p RT p p0 p0 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.
m -1 )(1.00103 m)
=
Thermal Properties of Matter
18-5
18.21.
IDENTIFY: Use Eq.(18.5) and solve for p. SET UP: = pM / RT and p = RT / M T = ( -56.5 + 273.15) K = 216.6 K
For air M = 28.8 10-3 kg/mol (Example 18.3) (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: 18.22.
p=
SET UP: EXECUTE:
N A = 6.02 1023 molecules/mol . M = N A m = (6.02 1023 molecules mol)(1.41 10-21 kg molecule) = 849 kg/mol. For a carbon atom, M = 12 10-3 kg/mol . If this molecule is mostly carbon, so the average mass of 849 kg/mol = 71,000 atoms . 12 10-3 kg/mol
EVALUATE:
its atoms is the mass of carbon, the molecule would contain
18.23. IDENTIFY: = m /V . SET UP:
The mass mtot is related to the number of moles n by mtot = nM . Mass is related to volume by
4 For gold, M = 196.97 g/mol and = 19.3 103 kg/m3 . The volume of a sphere of radius r is V = 3 r 3 .
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
=
0.5909 kg 4 = 3.06 10-5 m3 . V = 3 r 3 gives 19.3 103 kg/m 3
1/ 3
18.24.
3[3.06 10-5 m3 ] 3V r = = 0.0194 m = 1.94 cm . The diameter is 2r = 3.88 cm . = 4 4 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 N = nN A .
SET UP: EXECUTE:
1/ 3
1 atm = 1.013 105 Pa . 1.00 cm3 = 1.00 10-6 m3 . N A = 6.022 1023 molecules/mol .
(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 =
N VN A N N pVN A = = constant and 1 = 2 . so p RT RT p1 p2
18.25.
p 1.00 atm 19 N 2 = N1 2 = (2.20 106 molecules) = 2.44 10 molecules . -14 9.00 10 atm 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= (7500 K) = 8.28 10 Pa 3 23 -6 1 10 m 6.022 10 molecules/mol p = 8.2 10-17 atm. This is much lower than the laboratory pressure of 1 10-13 atm in Exercise 18.24. (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.
18-6
Chapter 18
18.26.
IDENTIFY: SET UP:
pV = nRT = NkT At STP, T = 273 K , p = 1.01 105 Pa . N = 6 109 molecules . NkT (6 109 molecules)(1.381 10-23 J/molecule K)(273 K) = = 2.24 10-16 m3 . p 1.01 105 Pa
EXECUTE:. V =
18.27.
L3 = V so L = V 1/ 3 = 6.1 10-6 m . EVALUATE: This is a small cube. m N IDENTIFY: n = = M NA
SET UP:
N A = 6.022 1023 molecules/mol . For water, M = 18 10-3 kg/mol . m 1.00 kg = = 55.6 mol . M 18 10-3 kg/mol
EXECUTE:. n =
18.28.
N = nN A = (55.6 mol)(6.022 1023 molecules/mol) = 3.35 1025 molecules . EVALUATE: Note that we converted M to kg/mol. N IDENTIFY: Use pV = nRT and n = with N = 1 to calculate the volume V occupied by 1 molecule. The length NA l of the side of the cube with volume V is given by V = l 3 . SET UP: T = 27C = 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 NRT (1.00)(8.314 J/mol K)(300 K) = = 4.09 10-26 m3 . l = V 1/ 3 = 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. = m / V implies V = m / , where m = mtot , the mass of 5.00 mol of water. V=
EXECUTE:
18.29.
mtot = nM = (5.00 mol)(18.0 10-3 kg/mol) = 0.0900 kg
Then V =
m
=
0.0900 kg = 9.00 10-5 m 3 1000 kg/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 = 2.989 10-29 m3 / molecule (5.00 mol)(6.022 1023 molecules/mol) V = a 3 , where a is the length of each side of the cube occupied by a molecule. a 3 = 2.989 10-29 m3 , so a = 3.1 10-10 m. (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 = 3 kT . vrms = . 2 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 = 3 kT depends only on the temperature so it is the same for each species of atom in the 2 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
18.30.
vrms,Kr vrms,Rn
=
M Rn 222 g/mol = = 1.63 . M Kr 83.80 g/mol The average kinetic energies are the same. The gas atoms with smaller mass have larger vrms .
EVALUATE:
Thermal Properties of Matter
18-7
18.31.
IDENTIFY and SET UP: EXECUTE: (b)
3RT . M (a) vrms is different for the two different isotopes, so the 235 isotope diffuses more rapidly. vrms = M 238 0.352 kg/mol = = 1.004 . M 235 0.349 kg/mol The vrms values each depend on T but their ratio is independent of T. 1 With the multiplicity of each score denoted by ni , the average score is ni xi and 150
1/ 2
vrms,235 vrms,238
=
EVALUATE: 18.32.
IDENTIFY and SET UP:
18.33.
1 2 the rms score is 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 known 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 , where N A is Avogadro's number. We don't know how the pressures compare, (b) pV = RT so N = RT NA 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) 1 m ( v 2 ) = 3 kT . TA > TB and the average kinetic energy per molecule depends only on T, so the statement 2 2
av
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
18.34.
molecules could have a larger vrms . EVALUATE: Only statement (d) must be true. We need more information in order to determine whether the other statements are true or false. IDENTIFY: Use pV = nRT to solve for V. SET UP: Use R = 0.08206 L atm/mol K . T = 273.15 K . nRT (1.00 mol)(0.08206 L atm/mol K)(273.15 K) = = 22.4 L EXECUTE: (a) V = 1.00 atm p
18.35.
p 1.00 atm (b) pV = nRT = constant , so p1V1 = p2V2 . V2 = 1 V1 = (22.4 L) = 0.243 L . p2 92 atm EVALUATE: For constant T, the volume of 1.00 mol is inversely proportional to the pressure. 3kT IDENTIFY: vrms = m
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 . 3(1.381 10-23 J/molecule K)(300 106 K) v = 1.93 106 m/s . rms = 6.43 10-3 . -27 3.35 10 kg c
c = 3.00 108 m/s . k = 1.381 10-23 J/molecule K .
EXECUTE: (a) vrms =
3.35 10-27 kg m 7 2 10 (b) T = (vrms ) 2 = (3.0 10 m/s) = 7.3 10 K . 3k 3(1.381 10-23 J/molecule K) EVALUATE: Even at very high temperatures and for this light nucleus, vrms is a small fraction of the speed of light.
18-8
Chapter 18
18.36.
IDENTIFY: SET UP: EXECUTE:
vrms =
3RT n p , where T is in kelvins. pV = nRT gives = . M V RT 3(8.314 J/mol K)(273.15 K) = 393 m/s . For 44.0 10-3 kg/mol
R = 8.314 J/mol K . M = 44.0 10-3 kg/mol .
(a) For T = 0.0C = 273.15 K , vrms =
T = -100.0C = 173 K , vrms = 313 m/s . The range of speeds is 393 m/s to 313 m/s.
(b) For T = 273.15 K ,
n 650 Pa n = = 0.286 mol/m3 . For T = 173.15 K , = 0.452 mol/m 3 . V (8.314 J/mol K)(273.15 K) V
18.37.
The range of densities is 0.286 mol/m3 to 0.452 mol/m 3 . 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) 1 m(v 2 )av = 3 kT = 3 (1.38 10-23 J/molecule K)(300 K) = 6.21 10-21 J 2 2 2
(b) We need the mass m of one atom: m =
M 32.0 10-3 kg/mol = = 5.314 10-26 kg/molecule N A 6.022 1023 molecules/mol
2(6.21 10-21 J) 2(6.21 10-21 J) = = 2.34 105 m 2 / s 2 5.314 10-26 kg m
Then
1 2
m(v 2 )av = 6.21 10-21 J (from part (a)) gives (v 2 )av =
(c) vrms = (v 2 ) rms = 2.34 104 m 2 / s 2 = 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 =
! In a collision v changes direction, so p = 2mvrms = 2(2.57 10-23 kg m/s) = 5.14 10-23 kg m/s
0.20 m 0.20 m = = 4.13 10-4 s vrms 484 m/s
p 5.14 10-23 kg m/s dp = = 1.24 10-19 N so Fav = t 4.13 10-4 s dt (f ) pressure = F / A = 1.24 10-19 N/(0.10 m)2 = 1.24 10-17 Pa (due to one atom) F=
(g) pressure = 1 atm = 1.013 105 Pa
Number of atoms needed is 1.013 105 Pa/(1.24 10-17 Pa/atom) = 8.17 1021 atoms
(h) pV = NkT (Eq.18.18), so N =
pV (1.013 105 Pa)(0.10 m)3 = = 2.45 1022 atoms kT (1.381 10-23 J/molecule K)(300 K)
18.38.
2 (i) From the factor of 1 in (vx )av = 1 (v 2 )av . 3 3 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 SET UP: 1 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) = = 1.5 105 m 4 2r 2 4 2(2.0 10-10 m) 2 (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, then
EXECUTE:
=
tmean =
18.39.
v
= 330 s . Collisions are infrequent. Use equal vrms to relate T and M for the two gases. vrms = 3RT / M (Eq.18.19), so
IDENTIFY and SET UP:
v
2 rms
/ 3R = T / M , where T must be in kelvins. Same vrms so same T / M for the two gases and
M N2 28.014 g/mol 3 TN2 = TH2 = ((20 + 273) K) = 4.071 10 K MH 2.016 g/mol 2 = (4071 - 273)C = 3800C A N 2 molecule has more mass so N 2 gas must be at a higher temperature to have the same vrms .
TN2 / M N 2 = TH2 / M H2 .
EXECUTE:
TN2
EVALUATE:
Thermal Properties of Matter
18-9
18.40.
IDENTIFY: SET UP: EXECUTE:
3kT . m k = 1.381 10-23 J/molecule K. vrms =
(a) vrms =
3(1.381 10-23 J/molecule K)(300 K) = 6.44 10 -3 m/s = 6.44 mm/s 3.00 10 -16 kg
18.41.
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 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:
(b) Q = nCV T = n ( 3 R ) T 2
Q = (2.5 mol) ( 5 ) (8.3145 J/mol K)(30.0 K) = 1560 J 2 Q = (2.5 mol) ( 3 ) (8.3145 J/mol K)(30.0 K) = 935 J 2
(a) Q = nCV T = n ( 5 R ) T 2
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 T by Q = nCV T . SET UP: For H 2 , CV ,H 2 = 20.42 J/mol K and for Ne (a monatomic gas), CV , Ne = 12.47 J/mol K. Q EXECUTE: CV T = = constant , so CV ,H2 TH2 = CV ,Ne TNe . n CV ,H2 20.42 J/mol K TNe = TH 2 = (2.50 C) = 4.09 C. C 12.47 J/mol K V , Ne EVALUATE: The same amount of heat causes a smaller temperature increase for H 2 since some of the energy input goes into the internal degrees of freedom. m RT . IDENTIFY: C = Mc , where C is the molar heat capacity and c is the specific heat capacity. pV = nRT = M SET UP: M N2 = 2(14.007 g/mol) = 28.014 10-3 kg/mol . For water, cw = 4190 J/kg K . For N 2 ,
18.43.
CV = 20.76 J/mol K .
EXECUTE: (a) cN2 =
c 20.76 J/mol K C = = 741 J/kg K . w = 5.65 ; cw is over five time larger. -3 cN 2 M 28.014 10 kg/mol
(b) To warm the water, Q = mcw T = (1.00 kg)(4190 J/mol K)(10.0 K) = 4.19 104 J . For air,
m=
Q 4.19 104 J (5.65 kg)(8.314 J/mol K)(293 K) mRT = = 5.65 kg . V = = = 4.85 m3 . cN2 T (741 J/kg K)(10.0 K) Mp (28.014 10-3 kg/mol)(1.013 105 Pa) c is smaller for N 2 , so less heat is needed for 1.0 kg of N 2 than for 1.0 kg of water.
EVALUATE: 18.44.
(a) IDENTIFY and SET UP: 1 R contribution to CV for each degree of freedom. The molar heat capacity C is 2 related to the specific heat capacity c by C = Mc. EXECUTE: CV = 6 ( 1 R ) = 3R = 3(8.3145 J/mol K) = 24.9 J/mol K. The specific heat capacity is 2
cV = CV / M = (24.9 J/mol K)/(18.0 10-3 kg/mol) = 1380 J/kg K. (b) For water vapor the specific heat capacity is c = 2000 J/kg K. The molar heat capacity is C = Mc = (18.0 10-3 kg/mol)(2000 J/kg K) = 36.0 J/mol K.
EVALUATE: 18.45.
The difference is 36.0 J/mol K - 24.9 J/mol K = 11.1 J/mol K, which is about 2.7 ( 1 R ) ; the 2
vibrational degrees of freedom make a significant contribution. IDENTIFY: CV = 3R gives CV in units of J/mol K . The atomic mass M gives the mass of one mole.
SET UP: EXECUTE:
For aluminum, M = 26.982 10-3 kg/mol.
(a) CV = 3R = 24.9 J/mol K . cV =
24.9 J/mol K = 923 J/kg K . 26.982 10-3 kg/mol (b) Table 17.3 gives 910 J/kg K. The value from Eq.(18.28) is too large by about 1.4%. EVALUATE: As shown in Figure 18.21 in the textbook, CV approaches the value 3R as the temperature increases. The values in Table 17.3 are at room temperature and therefore are somewhat smaller than 3R.
18-10
Chapter 18
18.46.
IDENTIFY:
Table 18.2 gives the value of v / vrms for which 94.7% of the molecules have a smaller value of v / vrms .
3RT . vrms = M SET UP: For N 2 , M = 28.0 10-3 kg/mol . v / vrms = 1.60 .
EXECUTE:
v 3RT = , so the temperature is 1.60 M Mv 2 (28.0 10 -3 kg/mol) T= = v 2 = (4.385 10 -4 K s 2 /m 2 )v 2 . 3(1.60) 2 R 3(1.60) 2 (8.3145 J/mol K) vrms =
(a) T = (4.385 10-4 K s 2 /m 2 )(1500 m/s) 2 = 987 K (b) T = (4.385 10-4 K s 2 /m 2 )(1000 m/s) 2 = 438 K (c) T = (4.385 10-4 K s 2 /m 2 )(500 m/s) 2 = 110 K. EVALUATE: As T decreases the distribution of molecular speeds shifts to lower values. IDENTIFY and SET UP: Make the substitution P = 1 mv 2 in Eq.(18.32). 2
m 2P - P/kT 8 m - P/kT f ( v ) = 4 e . = Pe m 2kT 2kT m EVALUATE: The shape of the distribution of molecular speeds versus the temperature is a function only of the kinetic energy of the molecules. 3/2 8 m - P/kT IDENTIFY and SET UP: Eq.(18.33): f (v ) = Pe m 2 kT df At the maximum of f (P), = 0. dP 3/2 df 8 m d EXECUTE: (Pe - P/kT ) = 0 = d P m 2 kT d P d This requires that (Pe - P/kT ) = 0. dP e-P/kT - (P/kT )e -P/kT = 0
3/2 3/2
18.47.
EXECUTE:
18.48.
(1 - P/kT )e- P/kT = 0
This requires that 1 - P/kT = 0 so P = kT , as was to be shown. And then since P = 1 mv 2 , this gives 2 and vmp = 2kT/m , which is Eq.(18.34).
EVALUATE: 18.49. IDENTIFY:
vrms =
3 2 mp
1 2
2 mvmp = kT
v . The average of v 2 weights larger v.
Apply Eqs.(18.34) (18.35) and (18.36). k R/N A R = = SET UP: Note that . M = 44.0 10-3 kg/mol . m M/N A M
EXECUTE: (a) vmp = 2(8.3145 J/mol K)(300 K)/(44.0 10-3 kg/mol) = 3.37 102 m/s. (b) vav = 8(8.3145 J mol K)(300 K) ( (44.0 10-3 kg mol)) = 3.80 102 m s. (c) vrms = 3(8.3145 J mol K)(300 K) (44.0 10-3 kg mol) = 4.12 102 m s. EVALUATE: The average speed is greater than the most probable speed and the rms speed is greater than the average speed. IDENTIFY and SET UP: If the temperature at altitude y is below the freezing point only cirrus clouds can form. Use T = T0 - y to find the y that gives T = 0.0C. T - T 15.0C - 0.0C EXECUTE: y = 0 = = 2.5 km 6.0 C /km EVALUATE: The solid-liquid phase transition occurs at 0C only for p = 1.01 105 Pa. Use the results of Example 18.4 to estimate the pressure at an altitude of 2.5 km. p2 = p1e Mg ( y2 - y1 ) / RT Mg ( y2 - y1 ) / RT = 1.10(2500 m/8863 m) = 0.310 (using the calculation in Example 18.4)
18.50.
Then p2 = (1.01 105 Pa)e -0.31 = 0.74 105 Pa. This pressure is well above the triple point pressure for water. Figure 18.21 shows that the fusion curve has large slope and it takes a large change in pressure to change the phase transition temperature very much. Using 0.0C introduces little error.
Thermal Properties of Matter
18-11
18.51.
18.52.
18.53.
18.54.
18.55.
IDENTIFY: Refer to the phase diagram in Figure 18.24 in the textbook. SET UP: For water the triple-point pressure is 610 Pa and the critical-point pressure is 2.212 107 Pa . EXECUTE: (a) To observe a solid to liquid (melting) phase transition the pressure must be greater than the triplepoint pressure, so p1 = 610 Pa . For p < p1 the solid to vapor (sublimation) phase transition is observed. (b) No liquid to vapor (boiling) phase transition is observed if the pressure is greater than the critical-point pressure. p2 = 2.212 107 Pa . For p1 < p < p2 the sequence of phase transitions are solid to liquid and then liquid to vapor. EVALUATE: Normal atmospheric pressure is approximately 1.0 105 Pa , so the solid to liquid to vapor sequence of phase transitions is normally observed when the material is water. IDENTIFY: Refer to Figure 18.24 in the textbook. SET UP: The triple-point temperature for water is 273.16 K = 0.01C . EXECUTE: The temperature is less than the triple-point temperature so the solid and vapor phases are in equilibrium. The box contains ice and water vapor but no liquid water. EVALUATE: The fusion curve terminates at the triple point. IDENTIFY: Figure 18.24 in the textbook shows that there is no liquid phase below the triple point pressure. SET UP: Table 18.3 gives the triple point pressure to be 610 Pa for water and 5.17 105 Pa for CO2. EXECUTE: The atmospheric pressure is below the triple point pressure of water, and there can be no liquid water on Mars. The same holds true for CO2. EVALUATE: On earth patm = 1 105 Pa , so on the surface of the earth there can be liquid water but not liquid CO2. IDENTIFY: V = V0 T - V0 k p SET UP: For steel, = 3.6 10-5 K -1 and k = 6.25 10-12 Pa -1 . EXECUTE: V0 T = (3.6 10-5 K -1 )(11.0 L)(21 C) = 0.0083 L . -kVo p = (6.25 10-12 Pa)(11 L) (2.1 107 Pa) = -0.0014 L . The total change in volume is V = 0.0083 L - 0.0014 L = 0.0069 L. (b) Yes; V 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 = h r 2 = (1.00 m) (0.060 m) 2 = 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)
p 3.51 105 Pa Then m2 = m1 2 = (0.2845 kg) = 0.0713 kg. 6 1.40 10 Pa) p1 m2 is the mass that remains in the tank. The mass that has been used is m1 - m2 = 0.2848 kg - 0.0713 kg = 0.213 kg. 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 + 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 + gy T 280.15 K (1.013 105 Pa) h = (2.30 m) = 0.26 m . 5 3 2 (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 .
18.56.
18-12
Chapter 18
(b) The necessary gauge pressure is the term gy from the above calculation, pgauge = 7.37 105 Pa. . 18.57.
The gauge pressure required in part (b) is about 7 atm. N p IDENTIFY: pV = NkT gives = . V kT SET UP: 1 atm = 1.013 105 Pa . TK = TC + 273.15 . k = 1.381 10-23 J/molecule K .
EVALUATE: EXECUTE: (b) (a) TC = TK - 273.15 = 94 K - 273.15 = - 179C
(1.5 atm)(1.013 105 Pa/atm) N p = = = 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 = 22C = 295 K .
18.58.
(1.0 atm)(1.013 105 Pa/atm) N = = 2.5 1025 molecules/m3 . The atmosphere of Titan is about five times V (1.381 10-23 J/molecule K)(295 K) 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 to be 3RT p = p0e- Mgy / RT . vrms = . M
SET UP: EXECUTE:
g Earth = 9.80 m/s 2 . T = 460C = 733 K . M = 44.0 g/mol = 44.0 10-3 kg/mol .
(a)
Mgy (44.0 10-3 kg/mol)(0.894)(9.80 m/s 2 )(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.
(b) vrms =
3RT 3(8.314 J/mol K)(733 K) = = 645 m/s . vrms has this value both at the surface and at an altitude M 44.0 10-3 kg/mol
18.59.
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: EXECUTE:
In pV = nRT we must use the absolute pressure. T1 = 278 K . p1 = 2.72 atm . T2 = 318 K . n, R constant, so pV pV pV = nR = constant . 1 1 = 2 2 and T T1 T2
18.60.
V T 0.0150 m3 318 K p2 = p1 1 2 = (2.72 atm) = 2.94 atm . The final gauge pressure is 3 0.0159 m 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 =
SET UP: EXECUTE:
mtot RT to the ethane at its final temperature and pressure. M 1.50 L = 1.50 10-3 m3 . M = 30.1 10-3 kg/mol . Neglect the thermal expansion of the flask.
(a) p2 = p1 (T2 T1 ) = (1.013 105 Pa)(300 K 380 K) = 8.00 104 Pa.
pV (8.00 104 Pa)(1.50 10-3 m3 ) -3 (b) mtot = 2 M = (30.1 10 kg mol) = 1.45 g. (8.3145 J mol K)(300 K) RT2 EVALUATE: We could also calculate mtot with p = 1.013 105 Pa and T = 380 K , and we would obtain the same result. Originally, before the system was warmed, the mass of ethane in the flask was 1.013 105 Pa m = (1.45 g) = 1.84 g . 4 8.00 10 Pa
Thermal Properties of Matter
18-13
18.61.
(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 .
EXECUTE:
1.30 106 Pa 3 V2 = V1 ( p1 / p2 ) = (1.90 m3 ) = 24.46 m 5 1.01 10 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' principle (Chapter 14): B = weight of
air displaced by balloon = airVg. Apply Newton's 2nd 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 18.61. 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:
F
y
= ma y
B - mL g - mgas g = 0
Figure 18.61
airVg - mL g - mgas g = 0
mL = airV - mgas Calculate mgas , the mass of hydrogen that occupies 750 m3 at 15C 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 2 ) = 8420 N.
(c) mL = airV - 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 2 ) = 7800 N.
18.62. EVALUATE: A greater weight can be supported when hydrogen is used because its density is less. 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 = p r 2 , with r = 0.100 m and p = 1.00 atm = 1.013 105 Pa . For the cylinder, V = r 2 h . EXECUTE: (b) V = (a) p r 2 = mg and m =
p r 2 (1.013 105 Pa) (0.100 m) 2 = = 325 kg . g 9.80 m/s 2
4.33 10-2 m3 nRT (1.80 mol)(8.31 J/mol K)(293.15 K) V - = 4.33 10-2 m3 . h = 2 = = 1.38 m . 5 r (0.100 m) 2 p 1.013 10 Pa The calculation assumes a vacuum ( p = 0) in the tank above the piston.
EVALUATE:
18-14
Chapter 18
18.63.
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 1 and 2 are shown in Figure 18.63.
p1 = 4.20 105 Pa p2 = pair = 1.00 105 Pa large tank implies v1 0
Figure 18.63 EXECUTE:
1 2 2 p1 + gy1 + 1 v12 = p2 + gy2 + 1 v2 2 2 2 v2 = p1 - p2 + g ( y1 - y2 )
v2 = (2 / )( p1 - p2 ) + 2 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 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 / ) ( p1 - p2 ) + 2 g ( y1 - y2 )
v2 = 16.1 m/s h = 2.00 m 4.00 m - h0 4.00 m - 3.50 m 5 5 p = p0 = (4.20 10 Pa) = 1.05 10 Pa 4.00 m - h 4.00 m - 2.00 m v2 = (2 / )( p1 - p2 ) + 2 g ( y1 - y2 ) v2 = 5.44 m/s
(c) v2 = 0 means (2 / )( p1 - p2 ) + 2 g ( y1 - y2 ) = 0
p1 - p2 = - g ( y1 - y2 ) y1 - y2 = h - 1.00 m 0.50 m 0.50 m 5 p = p0 = (4.20 10 Pa) . This is p1 , so 4.00 m - h 4.00 m - h 0.50 m 5 2 3 (4.20 105 Pa) - 1.00 10 Pa = (9.80 m/s )(1000 kg/m )(1.00 m - h) 4.00 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 2 - 149h + 229.2 = 0 and h 2 - 15.20h + 23.39 = 0 quadratic formula: h = 1 15.20 (15.20) 2 - 4(23.39) = (7.60 5.86) m 2
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 + g ( y1 - y2 ) equals air pressure. For h = 1.74 m, p = 9.3 104 Pa and
g ( y1 - y2 ) = 0.7 104 Pa, so p + g ( y1 - y2 ) = 1.0 105 Pa, which is air pressure.
Thermal Properties of Matter
18-15
18.64.
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 . Mgy (28.8 10-3 kg/mol)(9.80 m/s 2 )(2000 m) = = 0.2316 . p = p0e - Mgy / RT = (1.00 atm)e -0.2316 = 0.793 atm . RT (8.314 J/mol K)(293.15 K) 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 2 is taken in with each breath at the higher altitude, so the person must take more breaths per 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
(b)
18.65.
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:
-26
18.66.
kg. Then N = mtot / m = 1.7 1027 molecules, which checks. N RT . Deviations will be noticeable when the volume V of a molecule is on the order IDENTIFY: pV = nRT = NA of 1% of the volume of gas that contains one molecule. 4 SET UP: The volume of a sphere of radius r is V = r 3 . 3 RT , and the volume of a molecule is about EXECUTE: The volume of gas per molecule is NA p 4 -10 3 -29 3 V0 = (2.0 10 m) = 3.4 10 m . Denoting the ratio of these volumes as f, 3 RT (8.3145 J mol K)(300 K) p= f = f = (1.2 108 Pa) f . N AV0 (6.023 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 3 kT . 2
vrms =
3kT . m
m = 2.99 10
18.67.
SET UP: k = 1.381 10-23 J/molecule K . EXECUTE: (a) 1 m(v 2 )av depends only on T and both gases have the same T, so both molecules have the same 2
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. m 5.34 10-26 kg T T T v 3 (c) = rms = constant , so A = B . TB = B TA = (283.15 K) = 4.53 10 K = 4250C . mA mB mA 3.34 10-27 kg m 3k
(d) TB > TA so the B molecules have greater translational kinetic energy per molecule. EVALUATE:
In
1 2
m(v 2 )av = 3 kT and vrms = 2
3kT the temperature T must be in kelvins. m
18-16
Chapter 18
18.68.
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, . The average time dN 1 between collisions is the mean free time, tmean . The number of collisions per second is = . dt tmean SET UP: EXECUTE:
V = (1.25 m)3 = 1.95 m3 . r = 0.750 10-2 m . v = 1.10 m/s . N = 2500 .
(a) =
V 1.95 m3 = = 0.780 m = 78.0 cm 4 2r 2 N 4 2(0.750 10-2 m) 2 (2500)
(b) = vtmean , so tmean = (c) 18.69.
v
=
0.780 m = 0.709 s . 1.10 m/s
dN 1 1 = = = 1.41 collisions/s dt tmean 0.709 s EVALUATE: The calculation is valid only if the motion of each bee is random. IDENTIFY: Apply the iteration procedure that is described in the problem. SET UP: Let x = n / V . T = 400.15 K . EXECUTE: (a) Dividing both sides of Eq.(18.7) by the product RTV gives the result. (b) The algorithm described is best implemented on a programmable calculator or computer; for a calculator, the numerical procedure is an iteration of (9.80 105 ) (0.448) + x= x 2 1 - (4.29 10-5 ) x . (8.3145)(400.15) (8.3145)(400.15) Starting at x = 0 gives a fixed point at x = 3.03 102 after four iterations. The number density is 3.03 102 mol m3 .
(c) The ideal-gas equation is the result after the first iteration, 295mol m3 . EVALUATE: The van der Waals density is larger. The term corresponding to a represents the attraction of the molecules, and hence more molecules will be in a given volume for a given pressure. IDENTIFY: Calculate vrms and use conservation of energy to relate the initial speed of the molecules (vrms ) to the maximum height they reach. SET UP: T = 298.15 K . M = 28.0 10-3 kg/mol . EXECUTE:
1 2
18.70.
vrms =
3RT 3(8.314 J/mol K)(298.15 K) = = 515 m/s . Conservation of energy gives 28.0 10-3 kg/mol M
18.71.
2 vrms (515 m/s) 2 = = 1.02 105 m = 102 km 2 g 2(1.30 m/s 2 ) EVALUATE: The result does not depend on the amount of gas in the canister. IDENTIFY: The mass of one molecule is the molar mass, M, divided by the number of molecules in a mole, N A . 2 mvrms = mgy and y =
The average translational kinetic energy of a single molecule is 1 m(v 2 )av = 3 kT . Use pV = NkT to calculate N, 2 2 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
4 V = 3 (0.250 m)3 = 0.0654 m3 . p = 1.25 atm = 1.27 105 Pa .
EXECUTE: (b)
1 2
(a) m =
M 28.0 10-3 kg/mol = = 4.65 10-26 kg N A 6.022 1023 molecules/mol
3 m(v 2 )av = 3 kT = 2 (1.381 10-23 J/molecule K)(295.15 K) = 6.11 10-21 J 2
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 N ( 1 m(v 2 )av ) = (2.04 1024 molecules)(6.11 10-21 J/molecule) = 1.25 104 J . 2
(c) N = EVALUATE:
The number of moles is n =
N 2.04 1024 molecules = = 3.39 mol . N A 6.022 1023 molecules/mol
K tr = 3 nRT = 3 (3.39 mol)(8.314 J/mol K)(295.15 K) = 1.25 104 J , which agrees with our results in part (d). 2 2
Thermal Properties of Matter
18-17
18.72.
IDENTIFY: SET UP:
U = mgy . The mass of one molecule is m = M/N A . K av = kT .
3 2
Let y = 0 at the surface of the earth and h = 400 m . N A = 6.023 1023 molecules/mol and k = 1.38 10 -23 J/K . 15.0C = 288 K . M 28.0 10-3 kg/mol 2 -22 EXECUTE: (a) U = mgh = gh = (9.80 m/s )(400 m) = 1.82 10 J. NA 6.023 1023 molecules/mol
3 2 1.82 10-22 J (b) Setting U = kT , T = = 8.80 K. 2 3 1.38 10 -23 J/K EVALUATE: (c) The average kinetic energy at 15.0C 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 ) = 0. The work done to increase the separation from r2 to is U ( ) - U ( r2 ).
18.73.
(a) EXECUTE:
U ( r ) = U 0 ( R0 / r )12 - 2( R0 / r )6
Eq.(13.26): F ( r ) = 12(U 0 / 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 = 0 implies ( R0 / r )12 - 2( R0 / r )6 = 0
(r1 / R0 )6 = 1/ 2 and r1 = R0 /(2)1/ 6 F = 0 implies ( R0 / r )13 - ( R0 / r )7 = 0 (r2 / R0 )6 = 1 and r2 = R0 Then r1 / r2 = ( R0 / 21/ 6 ) / R 0 = 2-1/ 6
(d) Wother = U
At r , U = 0, so W = -U ( R0 ) = -U 0 ( R0 / R0 )12 - 2( R0 / R0 )6 = +U 0
EVALUATE: 18.74. IDENTIFY: SET UP:
The answer to part (d), U 0 , is the depth of the potential well shown in the graph of U (r ). Use pV = nRT to calculate the number of moles, n. Then K tr = 3 nRT . The mass of the gas, mtot , is 2
given by mtot = nM . 5.00 L = 5.00 10-3 m 3 pV (1.01 105 Pa)(5.00 10 -3 m 3 ) EXECUTE: (a) n = = = 0.2025 moles . (8.314 J/mol K)(300 K) RT
K tr = 3 (0.2025 mol)(8.314 J/mol K)(300 K) = 758 J . 2
(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 = 1 mv 2 = 1 (4.08 10-4 kg)(300.0 m/s) 2 = 18.4 J . The total kinetic energy is 2 2
K 18.4 J 100% = 100% = 2.37% . 776 J K tot (c) No. The temperature is associated with the random translational motion, and that hasn't changed. EVALUATE: Eq.(18.13) gives K tr = 3 pV = 3 (1.01 105 Pa)(5.00 10-3 m3 ) = 758 J , which agrees with our result 2 2 K tot = K + K tr = 18.4 J + 758 J = 776 J . The percentage increase is
in part (a). vrms = 3RT = 1.93 103 m/s . vrms is a lot larger than the speed of the jet, so the percentage increase in M the total kinetic energy, calculated in part (b), is small.
18-18
Chapter 18
18.75.
IDENTIFY and SET UP: EXECUTE:
Apply Eq.(18.19) for vrms . The equation preceeding Eq.(18.12) relates vrms and (vx ) rms .
(a) vrms = 3RT / M
vrms =
3(8.3145 J/mol K)(300 K) = 517 m/s 28.0 10-3 kg/mol
2 (vx )av = 1/ 3
2 (b) (vx )av = 1 (v 2 )av so 3
(
)
(v 2 )av = 1/ 3 vrms = 1/ 3 (517 m/s) = 298 m/s
(
)
(
)
18.76.
EVALUATE: The speed of sound is approximately equal to (vx ) rms since it is the motion along the direction of propagation of the wave that transmits the wave. 3kT IDENTIFY: vrms = m SET UP: EXECUTE: (b) vescape =
M = 1.99 1030 kg , R = 6.96 108 m and G = 6.673 10-11 N m 2 /kg 2 .
(a) vrms =
3kT 3(1.38 10-23 J K) (5800 K) = = 1.20 104 m s. m (1.67 10-27 kg)
18.77.
2GM 2(6.673 10-11 N m 2 kg 2 ) (1.99 1030 kg) = = 6.18 105 m s. (6.96 108 m) R 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 2 + U 2 , where U = -Gmmp / r. Let point 1 be at the surface of the planet, where the projectile is launched, and let point 2 be far from the earth. Just barely escapes says v2 = 0.
EXECUTE:
Only gravity does work says Wother = 0.
U1 = -Gmmp / Rp ; r2 so U 2 = 0; v2 = 0 so K 2 = 0.
The conservation of energy equation becomes K1 - Gmmp / Rp = 0 and K1 = Gmmp / Rp .
2 But g = Gmp / Rp 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. Set K1 from part (a) equal to the average kinetic energy of a molecule as given by
1 2 1 2
(b) IDENTIFY and SET UP:
Eq.(18.16).
EXECUTE:
m(v )av = mgRp (from part (a)). But also,
2
m(v 2 )av = 3 kT , so mgRp = 3 kT 2 2
3k nitrogen mN2 = (28.0 10-3 kg/mol)/(6.022 1023 molecules/mol) = 4.65 10 -26 kg/molecule
T=
2mgRp
3k hydrogen mH2 = (2.02 10-3 kg/mol)/(6.022 1023 molecules/mol) = 3.354 10-27 kg/molecule
T=
2mgRp
=
2(4.65 10-26 kg/molecule)(9.80 m/s 2 )(6.38 106 m) = 1.40 105 K 3(1.381 10-23 J/molecule K)
T=
2mgRp 3k
=
2(3.354 10-27 kg/molecule)(9.80 m/s 2 )(6.38 106 m) = 1.01 104 K 3(1.381 10-23 J/molecule K)
3k nitrogen 2(4.65 10-26 kg/molecule)(1.63 m/s 2 )(1.74 106 m) T= = 6730 K 3(1.381 10-23 J/molecule K) hydrogen 2(3.354 10-27 kg/molecule)(1.63 m/s 2 )(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 the long time most of the molecules will have escaped from the moon.
(c) T =
2mgRp
Thermal Properties of Matter
18-19
18.78.
IDENTIFY: SET UP:
vrms =
3RT . M
M H2 = 2.02 10-3 kg/mol . M O2 = 32.0 10-3 kg/mol . For Earth, M = 5.97 1024 kg and
4 R = 6.38 106 m . For Jupiter, M = 1.90 1027 kg and R = 6.91 107 m . For a sphere, M = V = r 3 . The 3 escape speed is vescape =
EXECUTE:
2GM . R
(a) Jupiter: vrms = 3(8.3145J mol K)(140K) (2.02 10-3 kg mol) = 1.31 103 m s .
vescape = 6.06 104 m/s . vrms = 0.022vescape .
Earth: vrms = 3(8.3145J mol K)(220K) (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. (c) vrms = 3(8.3145J mol K)(200K) (32.0 10-3 kg mol) = 395m s. The radius of the asteroid is
R = (3M 4 )1/ 3 = 4.68 105 m, and the escape speed is vescape = 2GM R = 542m s . Over time the O 2
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. 3kT m IDENTIFY: vrms = . The number of molecules in an object of mass m is N = nN A = NA . M m 4 SET UP: The volume of a sphere of radius r is V = r 3 . 3 3kT 3(1.381 10-23 J K)(300K) EXECUTE: (a) m = 2 = = 1.24 10-14 kg. (0.0010m s) 2 vrms
(b) N = mN A M = (1.24 10-14 kg)(6.023 1023 molecules mol) (18.0 10-3 kg mol)
18.79.
N = 4.16 1011 molecules. 3V 3m/ (c) The diameter is D = 2r = 2 = 2 4 4 to see. EVALUATE: vrms decreases as m increases.
18.80. IDENTIFY: SET UP: EXECUTE:
1/3 1/3
3(1.24 10-14 kg) = 2 3 4 (920 kg/m )
1/3
= 2.95 10-6 m which is too small
For a simple harmonic oscillator, x = A cos t and vx = - A sin t , with = k / m . The average value of cos(2t ) over one period is zero, so (sin 2 t )av = (cos 2 t )av = 1 . 2
x = A cos t , vx = - A sin t , U av = 1 kA2 (cos 2 t )av , K av = 1 m 2 A2 (sin 2 t )av . Using 2 2
18.81.
(sin 2 t )av = (cos 2 t )av = 1 and m 2 = k shows that K av = U av . 2 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 1 kT for each degree 2 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 two-dimensional solid would be 2 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.
18-20
Chapter 18
18.82.
IDENTIFY:
The equipartition principle says that each molecule has average kinetic energy of 1 kT for each degree 2
of freedom. I = 2m( L / 2) 2 , where L is the distance between the two atoms in the molecule. K rot = 1 I 2 . 2
rms = ( 2 )av .
SET UP: The mass of one atom is m = M/N A = (16.0 10-3 kg/mol) /(6.02 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 twofifths of the total kinetic energy, so K rot = nRT = (1.00 mol)(8.3145 J mol K)(300 K) = 2.49 103 J .
16.0 10-3 kg mol -11 2 -46 2 (b) I = 2m( L 2) 2 = 2 (6.05 10 m) = 1.94 10 kg m 23 6.023 10 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
rms =
2 K rot N A 2(2.49 103 J) = = 6.52 1012 rad s . This is much larger -46 I (1.94 10 kg m 2 )(6.023 1023 molecules/mol)
than the typical value for a piece of rotating machinery. 2 rad EVALUATE: The average rotational period, T = , for molecules is very short.
18.83. IDENTIFY:
CV = N ( R ) , where N is the number of degrees of freedom.
1 2 5 2
rms
SET UP: There are three translational degrees of freedom. EXECUTE: For CO 2 , N = 5 and the contribution to CV other than from vibration is
5 2
R = 20.79 J/mol K and
CV - R = 0.270 CV . So 27% of CV is due to vibration. For both SO2 and H2S, N = 6 and the contribution to CV
other than from vibration is
EVALUATE: 18.84.
6 2
R = 24.94 J/mol K . The respective fractions of CV from vibration are 21% and 3.9%.
The vibrational contribution is much less for H 2S . In H 2S the vibrational energy steps are larger
because the two hydrogen atoms have small mass and = k / m . IDENTIFY: Evaluate the integral, as specified in the problem. SET UP: Use the integral formula given in Problem 18.85, with = m / 2kT .
1 m m 2 - mv 2 / 2 kT =1 dv = 4 f (v) dv = 4 2kT v e 2kT 4(m 2kT ) m 2kT 0 0 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 1. IDENTIFY and SET UP: Evaluate the integral in Eq.(18.31) as specified in the problem. 32 32
EXECUTE:
(a)
18.85.
EXECUTE:
0
v 2 f (v) dv = 4 (m / 2 kT )3/ 2 v 4e - mv / 2 kT dv
2
0
The integral formula with n = 2 gives Apply with a = m / 2kT ,
18.86.
0
v 4e - av dv = (3/8a 2 ) /a
2
0
v 2 f (v) dv = 4 (m/2 kT )3 / 2 (3/8)(2kT/m) 2 2 kT/m = (3/2)(2kT/m) = 3kT/m
EVALUATE: Equation (18.16) says 1 m(v 2 )av = 3kT / 2, so (v 2 )av = 3kT / m, in agreement with our calculation. 2 IDENTIFY: Follow the procedure specified in the problem. SET UP: If v 2 = x , then dx = 2vdv .
EXECUTE:
vf (v)dv = 4 2kT v e
0 0
m
32
3 - mv 2 2 kT
dv. Making the suggested change of variable, v 2 = x. 2vdv = dx,
v3dv = (1/2) x dx, and the integral becomes m vf (v)dv = 2 2kT 0
3/2
0
2 m 2kT xe- mx/ 2 kT dx = 2 = 2kT m
3/2
2
2 KT 8KT = m m
which is Eq. (18.35).
EVALUATE:
The integral vf (v)dv is the definition of vav .
0
Thermal Properties of Matter
18-21
18.87.
IDENTIFY:
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 O 2 , 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 N = N v m 2kT (b) Setting v = vmp = 2kT in f (v ) gives f (vmp ) = 4 m 2kT m vmp = 3.95 102 m/s and f (v) v = 0.0421.
3/2
v + v
f (v )dv.
4 -1 . For oxygen gas at 300 K, e = e vmp
(c) Increasing v by a factor of 7 changes f by a factor of 7 2e -48 , and f (v )v = 2.94 10-21. (d) Multiplying the temperature by a factor of 2 increases the most probable speed by a factor of
-21
2, and the
18.88.
answers are decreased by 2: 0.0297 and 2.08 10 . (e) Similarly, when the temperature is one-half what it was parts (b) and (c), the fractions increase by 2 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: (b) mtot = (a) The pressure due to water vapor is (0.60)(2.34 103 Pa) = 1.40 103 Pa.
18.89.
MpV (18.0 10-3 kg mol)(1.40 103 Pa)(1.00 m3 ) = = 10 g (8.3145 J mol K)(293.15 K) RT 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 16.0C, so the partial pressure of water vapor at 30.0C
is equal to the vapor pressure at 16.0C, which is 1.81 103 Pa. Thus the relative humidity = 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 28.0C 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 12C, which would be at an altitude
18.90.
18.91.
(30C - 12C) (0.6 C 100 m) = 3 km above the ground. For a relative humidity of 80%, the vapor pressure will be the same as the water pressure at around 24C, corresponding to an altitude of about 1 km. EVALUATE: Clouds form at a lower height when the relative humidity at the surface is larger. 3RT IDENTIFY: Eq.(18.21) gives the mean free path . In Eq.(18.20) use vrms = in place of v. M
pV = nRT = NkT . The escape speed is vescape =
SET UP: EXECUTE:
2GM . R
For atomic hydrogen, M = 1.008 10-3 kg/mol .
(a) From Eq.(18.21), = (4 2r 2 ( N V )) -1 = (4 2(5.0 10-11 m) 2 (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.
18-22
Chapter 18
(d) vescape =
2GM 2G ( Nm V )(4R 3 3) = = (8 3)G ( N V )mR 2 R R
vescape = (8/3)(6.673 10-11 N m 2 /kg 2 )(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),
(V N ) nebula 50 106 m3 5 TISM = Tnebula = (20 K) = 2 10 K, (V N ) ISM (200 10-6 m3 ) -1 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 - y . SET UP: ln(1 + x) x when x is very small.
EXECUTE: (a)
18.92.
dp Mg dy dp pM =- , which in this case becomes =- . This integrates to dy RT p R T0 - y
Mg / R
p Mg y y ln = ln 1 - , or p = p0 1 - . p0 R T0 T0 y y (b) For sufficiently small , 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 2 ) = = 5.6576 and (c) 1 - = 0.8154, R (8.3145 J/mol K)(0.6 10-2 C/m) (288 K)
18.93.
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. IDENTIFY and SET UP: The behavior of isotherms for a real gas above and below the critical point are shown in Figure 18.7 in the textbook. EXECUTE: (a) A positive slope P would mean that an increase in pressure causes an increase in volume, or V that decreasing volume results in a decrease in pressure, which cannot be the case for any real gas. (b) See Figure 18.7 in the textbook. From part (a), p cannot have a positive slope along an isotherm, and so can
have no extremes (maxima or minima) along an isotherm. When curve in a p -V diagram must be an inflection point, and p vanishes along an isotherm, the point on the V
2 p =0. V 2 nRT an 2 p nRT 2an 2 2 p 2nRT 6an 2 =- + 3 . = - 4 . Setting the last two of these equal to (c) p = - 2 . 2 2 3 V V V (V - nb) V (V - nb) V - nb V zero gives V 3nRT = 2an 2 (V - nb) 2 and V 4 nRT = 3an 2 (V - nb)3 .
(d) Following the hint, V = (3 2)(V - nb), which is solved for (V n)c = 3b. Substituting this into either of the last
two expressions in part (c) gives Tc = 8a 27 Rb.
(e) pc = (f )
RT a R (8a / 27 Rb) a a . - = - 2= (V n)c - b (V n)c 2b 9b 27b 2
RTc (8a / 27b) 8 = = . 2 pc (V n)c (a / 27b )3b 3
(g) H 2 : 3.28. N 2 : 3.44. H 2O : 4.35. EVALUATE: (h) While all are close to 8/3, the agreement is not good enough to be useful in predicting critical point data. The van der Waals equation models certain gases, and is not accurate for substances near critical points.
Thermal Properties of Matter
18-23
18.94.
IDENTIFY and SET UP: EXECUTE:
For N particles, vav =
v
N
i
and vrms =
v
N
2 i
.
1 2 v12 + v2 and 2 1 1 1 1 2 2 2 2 2 vrms - vav = (v12 + v2 ) - (v12 + v2 + 2v1v2 ) = (v12 + v2 - 2v1v2 ) = (v1 - v2 ) 2 2 4 4 4 This shows that vrms vav , with equality holding if and only if the particles have the same speeds.
(a) vav = 1 (v1 + v2 ) , vrms = 2
1 ( Nv 2 + u 2 ), v = 1 ( Nv + u ), and the given forms follow immediately. rms av av N +1 N +1 (c) The algebra is similar to that in part (a); it helps somewhat to express 1 2 2 ( N ((N + 1) - 1)vav + 2Nvavu + (( N + 1) - N )u 2 ) . vav = (N + 1) 2 2 (b) vrms = 2 vav = Then,
N 2 N 1 2 (-vav + 2vavu - u 2 ) + vav + u2 2 (N + 1) N +1 N +1
N N N N 2 2 2 2 2 (vrms - vav ) + (vav - 2vavu + u 2 ) = (vrms - vav ) + (vav - u ) 2 . If vrms > vav , then 2 (N + 1) (N + 1) (N + 1) 2 N +1 this difference is necessarily positive, and v > vav . rms
2 2 vrms - vav =
(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.