Solve: 16.1. The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have the
same mass its volume must be
Vwater =
mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 3
16.2. Solve: The volume of the uranium nucleus is
V = 4 R 3 = 4 7.5 10 -15 m 3 3
The density of the uranium nucleus is
(
)
3
= 1.767 10 -42 m 3
nucleus =
mnucleus 4.0 10 -25 kg = = 2.26 1017 kg m 3 Vnucleus 1.767 10 -42 m 3
Assess: This density is extremely large compared to the typical density of materials.
16.3. Solve: The volume of the aluminum cube is 10-3 m3 and its mass is
mAl = Al VAl = (2700 kg m 3 )(1.0 10 -3 m 3 ) = 2.7 kg
The volume of the copper sphere with this mass is
VCu =
m 4 2.7 kg (rCu )3 = Cu = 8920 kg m 3 = 3.027 10 -4 m 3 3 Cu 3(3.027 10 -4 m 3 ) rCu = 4
1 3
= 0.04165 m
The diameter of the copper sphere is 0.0833 m = 8.33 cm.
16.4. Solve: The volume of the hollow sphere is
V= 4 3 (rout - rin3 ) = 43 (0.050 m)3 - (0.040 m)3 = 2.555 10 -4 m 3 3
[
]
Thus, the density of the shell is
=
M 0.690 kg = = 2700 kg m 3 V 2.555 10 -4 m 3
Table 16.1 identifies this density with aluminum.
16.5. Solve: The volume of the aluminum cube V = 8.0 10-6 m3 and its mass is
M = V = (2700 kg/m3)(8.0 10-6 m3) = 0.0216 kg = 21.6 g One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is
6.02 10 23 atoms 1 mol 23 N= (21.6 g) = 4.82 10 atoms 1 mol 27 g
16.6. Solve: The volume of the copper cube is 8.0 10-6 m3 and its mass is
M = V = (8920 kg/m3)(8.0 10-6 m3) = 0.07136 kg = 71.36 g Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is
1 mol n= (71.36 g) = 1.12 mol 64 g
16.7. Solve: (a) The number density is defined as N V , where N is the number of particles occupying a volume
V. Because Al has a mass density of 2700 kg/m3, a volume of 1 m3 has a mass of 2700 kg. We also know that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is
1 mol 5 n = (2700 kg) = 1.00 10 mol 0.027 kg
The number of Al atoms in 1.00 105 mols is
N = nN A = 1.00 10 5 mol (6.02 10 23 atoms mol) = 6.02 10 28 atoms
(
)
Thus, the number density is
N 6.02 10 28 atoms = = 6.02 10 28 atoms m 3 V 1 m3
(b) Pb has a mass of 11,300 kg in a volume of 1 m3 . Since the atomic mass number of Pb is 207, the number of moles in 11,300 kg is
1 mole n = (11,300 kg) 0.207 kg
The number of Pb atoms is thus N = nNA, and hence the number density is
N nN A 11300 kg atoms 23 atoms (1 mol ) = 3.28 10 28 = = 6.02 10 1 m3 V V mol m3 0.207 kg
16.8. Solve: The mass density is = M/V. The mass M of the sample is related to the number of atoms N and
the mass m of each atom by M = Nm. Combining these, the atomic mass is
m=
1750 kg / m 3 M V = = = = 3.986 10 -26 kg / atom 4.39 10 28 atoms/ m 3 N N N /V
The atomic mass in m = A u, where A is the atomic mass number. Thus
A=
The element's atomic mass number is 24.
m 3.986 10 -26 kg = = 24 1 u 1.661 10 -27 kg
16.9. Solve: Because the atomic mass number of gold is 197, 1.0 mole of gold has a mass of 197 g or 0.197 kg.
The volume of 1.0 mole of gold is
V = L3 =
(0.197 kg) M L= 3 Au 19,300 kg m
13
= 2.17 cm
16.10. Solve: The mass of mercury is
10 -6 m 3 M = V = (13600 kg m 3 )(10 cm 3 ) = 0.136 kg = 136 g 1 cm 3
and the number of moles is
n=
M 0.136 g = = 0.6766 mol Mmol 201 g mol
The mass of aluminum with 0.6766 mol of Al is
27 g M = (0.6766 mol) Mmol = (0.6766 mol) = 18.27 g = 0.01827 kg mol
This mass M of aluminum corresponds to a volume of
V=
M 0.01827 kg = = 6.77 10 -6 m 3 = 6.77 cm 3 2700 kg m 3
16.11. Solve: The lowest temperature is
TF = 9 TC + 32 -127 = 9 TC + 32 TC = -88C Tk = ( -88.3 + 273) K = 185 K 5 5
In the same way, the highest temperature is 136 = 9 TC + 32 TC = 58C = 331 K 5
16.12. Solve: Let TF = TC = T:
TF = 9 TC + 32 T = 9 T + 32 T = -40 5 5
That is, the Fahrenheit and the Celsius scales give the same numerical value at -40.
16.13. Model: A temperature scale is a linear scale.
Solve: (a) We need a conversion formula for C to Z, analogous to the conversion of C to F. Since temperature scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling point of liquid nitrogen is 0Z and 196C. Similarly, the melting point of steel is 1000Z and 1538C. Thus
-196 = 0 a + b 1538 = 1000 a + b From the first, b = 196. Then from the second, a = (1538 + 196)/1000 = 1734/1000. Thus the conversion is TC = (1734/1000)TZ 196. Since the boiling point of water is TC = 100C, its temperature in Z is 1000 TZ = (100 + 196) = 171Z 1734 (b) A temperature TZ = 500Z is 1734 TC = 500 - 196 = 671C = 944 K 1000
16.14. Model: We assume that the absolute temperature in this problem is doubled.
Solve: Since 10C = 283 K, when the absolute temperature of the sample is doubled, the temperature becomes 2 283 K = 566 K. On the Celsius scale, the new temperature of the sample is 566 K 273 = 293C.
16.15. Solve: (a) The triple point of water is T = 0.01C and p = 0.006 atm. Thus
TF = 9 TC + 32 = 5
9 5
(0.01) + 32 = 32.02F
1.013 10 5 Pa = 608 Pa 1 atm (b) The triple point of carbon dioxide is T = 56C and p = 5 atm. Thus TF = 9 TC + 32 = 9 ( -56) + 32 = -68.8F 5 5 p = 0.006 atm
1.013 10 5 Pa 5 p = 5.0 atm = (5.0 atm) = 5.06 10 Pa 1 atm
16.16. Visualize: Please refer to the phase diagram for water in Figure 16.04.
Solve: (a) Yes. Because the water-ice phase boundary has a negative slope down to the triple point and the solid-gas phase boundary has a positive slope up to the triple point, ice does not exist for temperatures greater than the triple point. (b) No. The positive slope of the solid-vapor boundary begins at the point (0 K, 0 atm). This means that as long as the pressure is low enough, water can always exist in the vapor phase.
16.17. Solve: The pressure due to seawater at a depth d is
p = p0 + sea water gd = 1.013 10 5 Pa + (1030 kg m 3 )(9.8 m s 2 )(100 m ) = 1.1107 10 6 Pa 11 atm
From Figure 16.04, we see that the freezing temperature of water at p = 11 atm is below 0C and the boiling temperature is above 100C. This is because the solid-liquid transition line has a negative slope, but the liquid-gas transition line has a positive slope.
16.18. Solve: (a) 500 Pa corresponds to a pressure of
500 Pa 1 atm = 0.005 atm 1.013 10 5 Pa
This pressure and temperature (-0.01C) are to the left and below the triple point in Figure 16.04. The sample is only barely in the vapor phase. Increasing p first changes the phase from a gas to a solid, and then, at a pressure of slightly greater than 1 atm, from a solid to a liquid because of the negative slope of the solid-liquid transition line. (b) At 0.03C warmer, that is, at +0.02C, increasing the pressure changes the phase from gas to liquid. The sample never enters the solid phase.
16.19. Model: Treat the gas in the sealed container as an ideal gas.
Solve:
(a) From the ideal gas law equation pV = nRT, the volume V of the container is
V=
nRT (2.0 mol)(8.31 J mol K )[(273 + 30) K ] = = 0.0497 m 3 p 1.013 10 5 Pa
Note that pressure must be in Pa in the ideal gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is
p1V p2 V T (273 + 130) K = 1.33 atm = p2 = p1 2 = (1.0 atm) T1 T2 T1 (273 + 30) K
Note that gas-law calculations must use T in kelvins.
Solve: (a) The density before and after the compression are before = m1 V1 and after = m2 V2 . Noting that m1 = m2 and V2 = 1 V1 , 2
16.20. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas.
after m V1 = = 2 after = 2 before before V2 m
The density has changed by a factor of 2. (b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the same.
16.21. Model: Treat the gas in the container as an ideal gas.
Solve:
From the ideal-gas law pV = nRT, the pressure of the gas is
p=
nRT (3.0 mol)(8.31 J mol K )[(273 - 120) K ] = = 1.907 10 6 Pa = 18.8 atm V (2.0 10 -3 m 3 )
16.22. Model: Treat the gas as an ideal gas in the sealed container.
Solve:
(a) For p1 = p0, the before-and-after relationship of an ideal gas in a sealed container is
V1 V0 T 473 K = V1 = V0 1 = V0 = 1.27V0 373 K T1 T0 T0
where T0 = (273 + 100) K and T1 = (273 + 200) K. (b) When the Kelvin temperature is doubled, T1 = 2T0 = 2(373 K) = 746 K and the above equation becomes
V1 = V0
T1 746 K = V0 = 2V0 373 K T0
16.23. Model: Treat the air in the compressed-air tank as an ideal gas.
Solve: (a) From the ideal-gas law pV = nRT, the number of moles n is
n=
pV p(r h) = = RT RT
2
(150 atm)
1.013 10 5 Pa 2 (0.075 m ) (0.50 m ) 1 atm = 55.1 mol (8.31 J mol K)[(273 + 20) K]
[
]
(b) At STP, the ideal-gas law yields
V=
nRT (55.1 mol)(8.31 J mol K )(273 K ) = = 1.234 m 3 p 1.013 10 5 Pa
Assess: Because the volume of the compressed air tank is (r2)h = 8.8357 10-3 m3 , the volume at STP is 140 times the volume of the tank.
16.24. Model: Treat the oxygen gas in the cylinder as an ideal gas.
Solve: (a) The number of moles of oxygen is
n=
M 50 g = = 1.563 mol Mmol 32 g mol
2
(b) The number of molecules is N = nN A = (1.563 mol) 6.02 10 23 mol -1 = 9.41 10 23 .
(c) The volume of the cylinder V = r 2 L = (0.10 m ) (0.40 m ) = 1.257 10 -2 m 3 . Thus,
(
)
N 9.41 10 23 = = 7.49 10 25 m -3 V 1.257 10 -2 m 3
(d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be
p=
nRT (1.563 mol)(8.31 J mol K )(293 K ) = = 303 kPa V 1.257 10 -2 m 3
pg = p - 1 atm = 303 kPa - 101 kPa = 202 kPa
where we used T = 20C = 293 K. But a pressure gauge reads gauge pressure:
16.25. Model: Treat the helium gas in the sealed cylinder as an ideal gas.
Solve:
2
The volume of the cylinder is V = r 2 h = (0.05 m ) (0.30 m ) = 2.356 10 -3 m 3 . The gauge pressure of the
gas is 120 psi
1 atm 1.013 10 5 Pa = 8.269 10 5 Pa, so the absolute pressure of the gas is 8.269 10 5 Pa + 14.7 psi 1 atm 1.013 10 5 Pa = 9.282 10 5 Pa. The temperature of the gas is T = (273 + 20) K = 293 K. The number of moles of the gas in the cylinder is
n= 9.282 10 5 Pa (2.356 10 -3 m 3 ) pV = = 0.898 mol RT (8.31 J mol K)(293 K)
(
)
(a) The number of atoms is
N = nN A = (0.898 mol)(6.02 10 23 mol -1 ) = 5.41 10 23 atoms.
(b) The mass of the helium is
M = nMmol = (0.898 mol)( 4 g mol) = 3.59 g = 3.59 10 -3 kg
(c) The number density is
N 5.41 10 23 atoms = = 2.30 10 26 atoms m 3 V 2.356 10 -3 m 3
(d) The mass density is
=
M 3.59 10 -3 kg = = 1.52 kg m 3 V 2.356 10 -3 m 3
16.26. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 20C or 293 K. The number of
gas molecules can be found as
N = nN A =
1.013 10 5 Pa (1.0 m 3 ) pV NA = (6.02 10 23 mol -1 ) = 2.51 10 25 RT (8.31 J mol K)(293 K)
(
)
According to Figure 16.8, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and 900 m/s, and 3% between 900 and1000 m/s. Thus, the number of molecules in the cube with a speed between 700 m/s and 1000 m/s is (0.22)(2.51 1025) = 5.5 1024.
16.27. Model: The rigid sphere's volume does not change, so this is an isochoric process. The air is assumed
to be an ideal gas. Solve: (a) When the valve is closed, the air inside is at p1 = 1 atm and T1 = 100C. The before-and-after relationship of an ideal gas in the closed sphere (constant volume) is
T p1V p2 V (273 + 0) K = p2 = p1 2 = (1.0 atm) = 0.732 atm T1 T2 (273 + 100) K T1
(b) Dry ice is CO2. From Figure 16.04, we can see that the solid-gas transition line gives a temperature of -78C when p = 1 atm. Cooling the sphere to 78C gives
T (273 - 78) K p2 = p1 2 = (1.0 atm) = 0.523 atm T1 373 K
16.28. Model: The gas is assumed to be ideal and it expands isothermally.
Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1. (b) The before-and-after relationship of an ideal gas under isothermal conditions is
V p p1V1 p2 V2 V = p2 = p1 1 = p1 1 = 1 2 T1 T1 V2 2V1
16.29. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the container is
assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm 3 = 50 10 -6 m 3 , and T1 = 20C = 293 K. This produces a pressure
nRT (0.1 mol)(8.31 J mol K )(293 K ) = 4.87 10 6 Pa = 4870 kPa = V1 50 10 -6 m 3 An ideal gas process has p2V2/T2 = p1V1/T1. Isochoric heating to a final temperature T2 = 300C = 573 K has V2 = V1, so the final pressure is V T 573 p2 = 1 2 p1 = 1 4870 kPa = 9520 kPa 293 V2 T1 Note that it is essential to express temperatures in kelvins. (b) p1 =
16.30. Model: The isobaric heating means that the pressure of the argon gas stays unchanged.
nRT1 (0.1 mol)(8.31 J mol K )(293 K ) = = 4.87 10 6 Pa = 4870 kPa V1 50 10 -6 m 3
Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 10-6 m3 , and T1 = 20C = 293 K. This produces a pressure
p1 =
An ideal gas process has p2V2/T2 = p1V1/T1. Isobaric heating to a final temperature T2 = 300C = 573 K has p2 = p1, so the final volume is
V2 =
(b)
p1 T2 573 V1 = 1 50 cm 3 = 97.8 cm 3 p2 T1 293
16.31. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is
assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 106 m3 , and T1 = 20C = 293 K. This produces a pressure
p1 =
nRT1 (0.1 mol)(8.31 J / mol K)(293 K) = = 4.87 10 6 Pa = 12.02 atm 50 10 -6 Pa V1 T2 V1 200 12.02 atm = 48.1 atm p1 = 1 50 T1 V2
An ideal gas process obeys p2V2/T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure
p2 =
(b)
16.32. Model: Assume the gas to be an ideal gas.
Visualize: Please refer to Figure Ex16.32. Solve: (a) Because the volume stays unchanged, the process is isochoric. (b) The ideal-gas law p1V1 = nRT1 gives
T1 =
3 1.013 10 5 Pa (100 10 -6 m 3 ) p1V1 = 914 K = nR (0.0040 mol)(8.31 J mol K)
(
)
The final temperature T2 is calculated as follows for an isochoric process:
p1 p2 p 1 atm = = 305 K T2 = T1 2 = (914 K ) 3 atm T1 T2 p1
16.33. Model: Assume that the gas is an ideal gas.
Visualize: Please refer to Figure Ex16.33. Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal. (b) From the ideal-gas law,
T1 =
3 1.013 10 5 Pa (100 10 -6 m 3 ) p1V1 = 914 K = nR (0.0040 mol)(8.31 J mol K)
(
)
T2 is also 914 K, because the process is isothermal. (c) The before-and-after relationship of an ideal gas under isothermal conditions is
p1V1 = p2 V2 V2 = V1
p1 3 atm = 300 cm 3 = (100 cm 3 ) 1 atm p2
16.34. Model: Assume that the gas is ideal.
Visualize: Please refer to Figure Ex16.34. Solve: (a) Because the process is at a constant pressure, it is isobaric. (b) For an ideal gas at constant pressure,
V2 V1 V 100 cm 3 = 391 K = T2 = T1 2 = [(273 + 900) K ] T2 T1 300 cm 3 V1
(c) Using the ideal-gas law p2 V2 = nRT2 ,
n=
3 1.013 10 5 Pa (100 10 -6 m 3 ) p2 V2 = 9.35 10 -3 mol = RT2 (8.31 J mol K)(391 K)
(
)
16.35.
Visualize:
Solve: Suppose we have a 1 m 1 m 1 m block of copper of mass M containing N atoms. The atoms are spaced a distance L apart along all three axes of the cube. There are Nx atoms along the x-edge of the cube, Ny atoms along the y-edge, and Nz atoms along the z-edge. The total number of atoms is N = NxNyNz. If L is expressed in meters, then the number of atoms along the x-edge is Nx = (1 m)/L. Thus,
N=
1 m3 1 m 1 m 1 m 1 m3 = 3 L= L L L L N
13
This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of copper is
M = CuV = (8920 kg m 3 )(1 m 3 ) = 8920 kg
But M = mN, where m = 64 u = 64 1.661 10 -27 kg is the mass of an individual copper atom. Thus,
(
)
N=
M 8920 kg = = 8.39 10 28 atoms m 64 (1.661 10 -27 kg)
13
1 m3 L= 8.39 10 28
= 2.28 10 -10 m = 0.228 nm
16.36. Solve: The volume of the cube associated with each atom is
Vatom = (0.227 10 -9 m ) = 1.1697 10 -29 m 3
3
The volume of a mole of atoms is
V = Vatom N A = (1.1697 10 -29 m 3 )(6.02 10 23 mol -1 ) = 7.0416 10 -6 m 3 mol
Thus, the mass of a mole of atoms is
Mmol = V = (7.0416 m 3 mol)(7950 kg m 3 ) = 0.056 kg/mol = 56 g/mol
The atomic mass number of the element is 56.
16.37. Solve: (a) A nitrogen molecule N2 has an atomic mass number of 28, so the mass of the molecule is
m = 28 u = 28 (1.661 10 -27 kg) = 4.651 10 -26 kg
From Figure 16.8, the most likely speed of a room-temperature N2 molecule is v 550 m/s. Thus, the kinetic energy of a molecule is
K avg = 1 mv 2 2
1 2
(4.651 10
2
-26
kg)(550 m s) = 7.03 10 -21 J
2
(b) Room-temperature H2 molecules have the same kinetic energy as room-temperature N2 molecules. This means
1 2
mH ( vH ) = 1 mN ( v N ) vH = 2
2
mN (vN ) = mH
28 (550 m s) = 2060 m s 2
16.38. Model: Air is an ideal gas. The pressure at sea level is 1 atm.
Solve: From the ideal gas law pV = NkBT, the number density is
N p 1.013 10 5 Pa = = = 2.5 10 25 molecules / m 3 V kB T (1.38 10 -23 J / K) (293 K)
16.39. Model: Assume the gas in the solar corona is an ideal gas.
Solve: The number density of particles in the solar corona is N V . Using the ideal-gas equation,
pV = NkB T
N p = V kB T
N (0.03 Pa ) = = 1.1 1015 particles m 3 V (1.38 10 -23 J K )(2 10 6 K )
16.40. Model: Assume the gas in the evacuated volume is an ideal gas.
Solve: The number density of particles is N V . Using the ideal-gas equation,
pV = NkB T
The pressure is
N p = V kB T
p = (1 10 -10 mm of Hg)
1 atm 1.013 10 5 Pa = 1.33 10 -8 Pa 760 mm of Hg 1 atm
N 1.33 10 -8 Pa = = 3.3 1012 m -3 = 3.3 10 6 cm -3 V (1.38 10 -23 J / K)(293 K)
16.41. Model: Assume that the gas in the vacuum chamber is an ideal gas.
Solve: (a) The fraction is
pvacuum chamber 1.0 10 -10 mm of Hg = = 1.32 10 -13 760 mm of Hg patmosphere
(b) The volume of the chamber V = (0.20 m ) (0.30 m ) = 0.03770 m 3 . From the ideal-gas equation pV = NkB T , the number of molecules of gas in the chamber is
2
N=
-13 5 3 pV (1.32 10 ) 1.013 10 Pa (0.03770 m ) = = 1.24 1011 molecules kB T 1.38 10 -23 J K )(293 K ) (
(
)
16.42. Model: The carbon dioxide in the cube is an ideal gas.
Solve: Using the ideal gas equation and n = M/Mmol,
pV = nRT V =
The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus,
nRT MRT = p pMmol
V=
(10,000kg)(8.31 J mol K)(273 K) (1.013 10 5 Pa)(0.044 kg mol)
= 5090 m3
The length of the cube is L = (V) 1/3 = 17.2 m.
16.43. Model: Assume the gas is an ideal gas.
Solve: The before-and-after relationship of an ideal gas is
1 p V p1V1 p2 V2 p 3V = T2 = T1 2 2 = (298 K ) 2 1 1 = 447 K = 174C p1 V1 T1 T2 p1 V1
16.44. Model: Assume the water vapor is an ideal gas.
Solve: The volume of 1 mole of steam is obtained from the ideal-gas law:
V=
nRT (1.0 mol)(8.31 J mol K )[(273 + 300) K ] = 0.047 m3 = p 1.013 10 5 Pa
1.0 mole of water has a mass of 18 g or 0.018 kg. Because the density of water is 1000 kg/m3, the volume of 1.0 mole of water is 1.8 10 -5 m 3 . Thus, the ratio of the of volume steam at 300C to the volume of water (both at 1.0 atm pressure) is 0.047 m 3 1.8 10 -5 m 3 = 2610 .
(
)(
)
16.45. Model: Assume that the steam (as water vapor) is an ideal gas.
Solve: The volume of the liquid water is
V=
20 1.013 10 5 Pa (10, 000 10 -6 m 3 )(0.018 kg mol) nMmol m pV Mmol = = = RT (8.31 J mol K)(473 K)(1000 kg m 3 )
= 9.28 10-5 m3 = 92.8 cm3
(
)
16.46. Model: Assume that the steam is an ideal gas.
Solve: (a) The volume of water is
V=
50 1.013 10 5 Pa (5.0 m 3 )(0.018 kg mol) nMmol M pV Mmol = = 0.0815 m3 = 81.5 L = = RT (8.31 J mol K)(673 K)(1000 kg m 3 )
(
)
(b) Using the before-and-after relationship of an ideal gas,
(273 + 150) K 50 atm p2 V2 pV T p = 1 1 V2 = 2 1 V1 = (5.0 m 3 ) = 78.6 m3 2.0 atm 673 K T2 T1 T1 p2
16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant.
p1 p2 T 273 + 45 = p2 = 2 p1 = (44.7 psi) = 49.4 psi 273 + 15 T1 T2 T1
Your tire gauge will read a gauge pressure pg = 49.4 psi - 14.7 psi = 34.7 psi.
Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the before-and-after relationship of an ideal gas for an isochoric (constant volume) process,
16.48. Model: The air is assumed to be an ideal gas.
Solve: At 20C and 1 atm pressure, the number of moles in the container is
n1 =
1.013 10 5 Pa (10 -3 m 3 ) p1V1 = 0.0416 mol = RT1 (8.31 J mol K)(293 K) 1.013 10 5 Pa (10 -3 m 3 ) p2 V2 = 0.0327 mol = RT2 (8.31 J mol K)(373 K)
(
)
At 100C and 1 atm pressure, the number of moles is
n2 =
(
)
When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm as gas escapes. Thus, the number of moles of air that escape as the container is opened is n1 n2 = 0.0416 mol 0.0327 mol = 0.0089 mol.
16.49. Model: The gas's pressure does not change, so this is an isobaric process.
Solve:
The triple point of water is 0.01C or 273.16 K, so T1 = 273.16 K. Because the pressure is a constant,
V1 V2 V 1638 mL = 447.44 K = 174.3C = T2 = T1 2 = (273.16 K ) 1000 mL T1 T2 V1
16.50. Model: Assume the gas in the manometer is an ideal gas.
Visualize: Please refer to Figure P16.50. Solve: In the ice-water mixture the pressure is p1 = patmos + Hgg (0.120 m) = 1.013 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.120 m) = 1.173 105 Pa In the freezer the pressure is p2 = patm + Hgg (0.030 m) = 1.013 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.030 m) = 1.053 105 Pa Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the gas cell. This means the volume of the chamber can be considered constant. Hence,
p p1 p2 1.053 10 5 Pa = T2 = T1 2 = (273 K ) = 245 K = -28C 1.173 10 5 Pa T1 T2 p1
Assess: This is a reasonable temperature for an industrial freezer.
16.51. Model: The gas can be treated as an ideal gas.
Visualize:
Solve: (a) Because the liquid on the right is below that on the left, the "height" of the mercury column is h = -25 cm = -0.25 m. Thus, the gas pressure in the cell is pgas = 1 atm + gh = 101,300 Pa + (13,600 kg/m3)(9.8 m/s2)(-0.25 m) = 68,000 Pa We can find the number of atoms from the version of the ideal-gas law that reads pV = NkBT (Equation 16.12):
3 -6 3 3 pV (68,000 Pa )(200 cm 10 m cm ) = = 3.05 10 21 atoms kB T (1.38 10 -23 J K)(50C + 273C)
N=
Notice how all quantities were converted to SI units. (b) The mass of an atom is m = 4 u = 4 (1.661 10-27 kg) = 6.644 10-27 kg. The total mass is M = mN = (6.644 10-27 kg/atom)(3.05 1021 atoms) = 2.02 10-5 kg = 20.2 mg
16.52. Model: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and the
air in the bubble is an ideal gas. Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the pressure is p1 = p0 + watergd = 1.013 105 Pa + (1000 kg/m3)(9.8 m/s2)(50 m) = 5.913 105 Pa At the lake's surface, p2 = p0 = 1.013 105 Pa. Using the before-and-after relationship of an ideal gas,
p2 V2 pV p T = 1 1 V2 = V1 1 2 T2 T1 p2 T1
5.913 10 5 Pa 293 K 4 3 4 r2 = r2 = 0.0091 m (0.005 m)3 3 3 1.013 10 5 Pa 283 K
The diameter of the bubble is 2r2 = 0.0182 m = 1.82 cm.
16.53. Model: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in the
cylinder is a constant. Solve: Using the before-and-after relationship of an ideal gas,
p2 V2 pV T V 1223 K V1 = 104.4 atm = 1 1 p2 = p1 2 1 = (25 atm) 293 K V1 T2 T1 T1 V2
where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the compressed air cylinder does not blow.
16.54. Model: Assume that the gas is an ideal gas.
Solve:
Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is proportional to 1/V1 for isothermal processes.
16.55. Model: Assume that the gas is an ideal gas.
Solve:
Assess: For the isothermal process, the pressure must double as the volume is halved. This is because p is proportional to 1/V for isothermal processes.
16.56. Model: Assume that the helium gas is an ideal gas.
Visualize: Please refer to Figure P16.56. Process 1 2 is isochoric, process 2 3 is isothermal, and process 3 1 is isobaric. Solve: The number of moles of helium is 8.0 g M = = 2.0 mol n= Mmol 4 g mol
Using the ideal-gas equation,
V1 =
nRT1 (2.0 mol)(8.31 J mol K )[(273 + 37) K ] = = 0.0254 m3 p1 2 1.013 10 5 Pa
(
)
For the isochoric process V2 = V1, and
p1 p2 T 657 + 273 = p2 = p1 2 = (2 atm) = 6 atm 37 + 273 T1 T2 T1
For the isothermal process, the equation p3V3 = p2V2 is
V3 = V2
p2 6 atm = 0.0762 m 3 = (0.0254 m 3 ) 2 atm p3
For the isothermal process, T3 = T2 = 657C.
16.57. Model: Assume the nitrogen gas is an ideal gas.
Visualize: Please refer to Figure P16.57. Solve: (a) The number of moles of nitrogen is M 1g 1 n= = = mol Mmol 28 g mol 28 Using the ideal-gas equation,
p1 =
nRT1 (1 28 mol)(8.31 J mol K )(298 K ) = = 8.84 105 Pa = 884 kPa V1 (100 10 -6 m 3 )
(b) For the process from state1 to state 3:
1.5 p1 50 cm 3 p V p1V1 p3V3 = 223.5 K = -49.5C T3 = T1 3 3 = (298 K ) = 3 p1 V1 T1 T3 p1 100 cm
For the process from state 3 to state 2:
p V 2.0 p1 100 cm 3 pV p2 V2 = 596 K = 323C = 3 3 T2 = T3 2 2 = (223.5 K ) 3 T2 T3 1.5 p1 50 cm p3 V3
For the process from state1 to state 4:
1.5 p1 150 cm 3 p4 V4 pV p V = 670.5 K = 397.5C = 1 1 T4 = T1 4 4 = (298 K ) 3 T4 T1 p1 V1 p1 100 cm
16.58. Model: The gas is an ideal gas.
Visualize: Please refer to Figure P16.58. Solve: (a) Using the ideal-gas equation,
T1 =
1.0 10 5 Pa (2.0 m 3 ) p1V1 = 301 K = nR (80 mol)(8.31 J mol K)
(
)
Because points 1 and 2 lie on the isotherm, T2 = T1 = 301 K. The temperature of the isothermal process is 301 K. (b) The straight-line process 1 2 can be represented by the equation
p = (3 - V ) 10 5
where V is in m3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as
10 5 pV = (3V - V 2 ) nR nR We can maximize T by setting the derivative dT/dV to zero: T= 10 5 3 dT 2 = (3 - 2Vmax ) = 0 Vmax = m 3 = 1.50 m 3 2 dV nR
At this volume, the pressure is pmax = 1.5 10 5 Pa and the temperature is
Tmax =
pmax Vmax (1.50 10 5 Pa)(1.5 m 3 ) = = 338 K (80 mol)(8.31 J / mol K) nR
16.59. Model: Assume the gas is an ideal gas.
Visualize: Please refer to Figure P16.59. Solve: (a) We can find the temperatures directly from the ideal-gas law after we convert all quantities to SI units:
3 -6 3 3 p1V1 (3.0 atm 101,300 Pa atm)(1000 cm 10 m cm ) = 366 K = 93C = nR (0.10 mol)(8.31 J mol K) 3 -6 3 3 p2 V2 (1.0 atm 101,300 Pa atm)(3000 cm 10 m cm ) = 366 K = 93C = nR (0.10 mol)(8.31 J mol K)
T1 = T2 =
(b) T2 = T1, so this is an isothermal process. (c) A constant volume process has V3 = V2. Because p1 = 3p2, restoring the pressure to its original value means that p3 = 3p2. From the ideal gas law,
p V p3V3 pV = 2 2 T3 = 3 3 T2 = 3 1 T2 = 3 366 K = 1098 K = 825C T3 T2 p2 V2
16.60. Model: Assume the gas is an ideal gas.
Visualize: Please refer to Figure P16.60. Solve: (a) Using the ideal-gas law,
T1 =
1.013 10 5 Pa (100 10 -6 m 3 ) p1V1 = 244 K = -29C = nR (5.0 10 -3 mol)(8.31 J mol K)
(
)
(b) Using the before and after relationship of an ideal gas,
3 p1V1 p2 V2 T V 2926 K 100 cm = 4.00 atm = p2 = p1 2 1 = (1 atm) 244 K 300 cm 3 T1 T2 T1 V2
(c) Using the before and after relationship of an ideal gas,
p3V3 pV p T 4 atm 2438 K = 2 2 V3 = 2 3 V2 = 300 cm 3 ) = 500 cm3 2 atm 2926 K ( T3 T2 p3 T2
16.61. Model: Assume CO2 gas is an ideal gas. Solve: (a) The molar mass for CO2 is Mmol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes 0.227 mol of gas at 0C. With V1 = 10,000 cm3 = 0.010 m3 and T1 = 0C = 273 K, the pressure is
p1 = nRT1 (0.2273 mol)(8.31 J mol K )(273 K ) = = 5.156 104 Pa = 0.509 atm 0.010 m 3 V1
p1 0.509 atm = (0.010 m 3 ) = 1.70 10 -3 m 3 = 1700 cm 3 3.0 atm p2
(b) From the isothermal compression,
p2 V2 = p1V1 V2 = V1
From the isobaric compression,
T3 = T2
(c)
1000 cm 3 V3 = (273 K ) = 161 K = -112C 1700 cm 3 V2
Solve: (a) The gas starts at pressure p1, temperature T1, and volume V1 = 100 cm3. The gas is first pressurized at constant volume until p3 = 3p1. From the ideal-gas law,
16.62. Model: The gas in the container is assumed to be an ideal gas.
p2 V2 pV p V = 1 1 T2 = 2 2 T1 = 3 1 T1 = 3T1 T2 T1 p1 V1
Next, the gas is expanded at a constant temperature T3 = T2 = 3T1 until p3 = from another application of the ideal gas law:
1 2
p1 . We can find the final volume
p3V3 3p pV pV = 2 2 V3 = 2 2 = 1 1 V2 = 6(100 cm 3 ) = 600 cm 3 T3 T2 p3 2 p1
(b)
Solve: (a) The gas starts at pressure p1 = 2.0 atm, temperature T1 = 127C = (127 + 273) K = 400 K and volume V1. It is first compressed at a constant temperature T2 = T1 until V2 = 1 V1 and the pressure is p2 . It is then further 2 compressed at constant pressure p3 = p2 until V3 = 1 V2. From the ideal-gas law, 2
16.63. Model: The gas in the container is assumed to be an ideal gas.
p2 V2 pV V T V = 1 1 p2 = p1 1 2 = (2.0 atm) 1 1 1 = 4.0 atm T2 T1 V2 T1 2V 1
Note that T2 = T1 = 400 K. Using the ideal gas law once again,
1 p3V3 V p pV V = 2 2 T3 = T2 3 3 = ( 400 K ) 2 2 1 = 200 K = -73C T3 T2 V2 p2 V2
The final pressure and temperature are 4.0 atm and 73C. (b)
16.64. Model: Assume that the nitrogen gas is an ideal gas.
Solve: (a) The molar mass of N2 gas is 28 g/mol. The number of moles is n = (5 g)/(28 g/mol) = 0.1786 mol. The initial conditions are p1 = 3.0 atm and T1 = 293 K. We use the ideal gas law to find the initial volume as follows:
V1 =
nRT1 (0.1786 mol)(8.31 J mol K )(293 K ) = = 1.430 10 -3 m 3 = 1430 cm 3 3.0 atm 101,300 Pa atm p1
An isobaric expansion until the volume triples results in V2 = 3V1 = 4290 cm3. (b) After the expansion,
p2 V2 pV p V = 1 1 T2 = 2 2 T1 = 1 3 T1 = 3T1 = 879 K = 606C T2 T1 p1 V1
(c) A constant volume decrease at V3 = V2 = 4290 cm3 back to T3 = T1 = 1 T2 results in the following: 3
p3V3 T V pV 1 1 = 2 2 p3 = 3 2 p2 = 1 p2 = 3.0 atm = 1.0 atm 3 3 T3 T2 T2 V3
(d) An isothermal compression at T4 = T3 back to the initial volume V4 = V1 = 1 V3 results in the following: 3
pV p4 V4 T V 1 = 3 3 p4 = 4 3 p3 = 1 1 p3 = 3 1.0 atm = 3.0 atm T4 T3 T3 V4 3
(e)
16.65. Solve: (a) A gas is compressed isothermally from a volume 300 cm3 at 2 atm to a volume of 100 cm3. What is the final pressure? (b)
(c) The final pressure is p2 = 6 atm.
16.66. Solve: (a) A gas at 400C and 500 kPa is cooled at a constant volume to a pressure of 200 kPa. What is
the final temperature in C? (b)
(c) The final temperature is T2 = -3.8C.
16.67. Solve: (a) A gas expands at constant pressure from 200 cm3 at 50C until the temperature is 400C.
What is the final volume? (b)
(c) The final volume is V2 = 417 cm3.
16.68. Solve: (a) 0.12 g of neon gas at 2.0 atm and 100 cm3 expands isobarically to twice its initial volume.
What is the final temperature of the gas? (b)
(c) The number of moles of neon is
n=
The initial temperature is
M 0.12 g = = 0.006 mol Mmol 20 g
T1 =
The final temperature is
-4 3 5 p1V1 2 1.013 10 Pa (1.0 10 m ) = 406 K = nR (0.006 mol)(8.31 J mol K)
(
)
T2 =
p2 V2 p 2V1 T1 = (406 K) = 812 K p1 V1 p V1
16.69. Model: Assume that the compressed air is an ideal gas.
Visualize: Please refer to Figure CP16.69. Solve: (a) Because the piston is floating in equilibrium, Fnet = (p 1 patmos)A w = 0 N where the piston's cross-sectional area A = r 2 = (0.050 m ) = 7.854 10 -3 m 2 and the piston's weight w = (50 kg)(9.8) = 490 N. Thus, w 490 N p1 = + patmos = + 1.013 10 5 Pa = 1.637 105 Pa A 7.854 10 -3 m 2 Using the ideal-gas equation p1V1 = nRT1,
2
( )
(1.637 10 5 Pa)Ah1 = (0.12 mol)(8.31 J / mol K)[(273 + 30) K]
With the value of A given above, this equation yields h1 = 0.235 m = 23.5 cm. (b) When the temperature is increased from T1 = 303 K to T2 = (303 + 100) K = 403 K, the volume changes from V1 = Ah1 to V2 = Ah2 at a constant pressure p2 = p1. From the before-and-after relationship of the ideal gas: p1 ( Ah1 ) p2 ( Ah2 ) = T1 T2
h2 =
p1 T2 403 K h1 = (1) (0.235 m) = 0.313 m = 31.3 cm 303 K p2 T1
Thus, the piston moves h2 h1 = 7.8 cm.
16.70. Model: The air in the closed section of the U-tube is an ideal gas.
Visualize: Please refer to Figure CP16.70. The length of the tube is l = 1.0 m and its cross-section area is A. Solve: Initially, the pressure of the air in the tube is p1 = patmos and its volume is V1 = Al. After the mercury is poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed pressure equals the pressure at the bottom of the column: p2 = patmos + gL. The volume of the compressed air is V2 = A(l L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal equilibrium with the surrounding air, so T2 = T1. In an isothermal process, pressure and volume are related by p1V1 = patmos Al = p2 V2 = ( patmos + gL) A(l - L) Canceling the A, multiplying through, and solving for L gives
L=l-
patmos 101, 300 Pa = 1.00 m - = 0.240 m = 24.0 cm (13,600 kg / m 3 ) (9.8 m / s 2 ) g
16.71. Model: The air in the diving bell is an ideal gas.
Visualize:
Solve: (a) Initially p1 = p0 (atmospheric pressure), V1 = AL, and T1 = 293 K. When the diving bell is submerged to d = 100 m at the bottom edge, the water comes up height h inside. The volume is V2 = A(L h) and the temperature is T2 = 283 K. Like a barometer, the pressure at points a and b must be the same. Thus p2 + gh = p0 + gd, or p2 = p0 + g(d h). Using the before and after relationship of an ideal gas,
p AL [ p0 + g( d - h)] A( L - h) p1V1 p2 V2 = 0 = 293 K 283 K T1 T2
Multiplying this out gives the following quadratic equation for h:
283 gh 2 - [ p0 + g( d + L)]h + 1 - p0 L + gLd = 0 293
Inserting the known values (using = 1030 kg/m3 for seawater) and dividing by g gives
h 2 - 113.04h + 301.03 = 0 h = 110.3 m or 2.73 m
The first solution is not physically meaningful, so the water rises to height h = 2.73 m. (b) To expel all the water, the air pressure inside the bell must be increased to match the water pressure at the bottom edge of the bell, d = 100 m. The necessary pressure is p = p0 + gd = 101,300 Pa + (1030 kg/m3)(9.8 m/s2)(100 m) = 1111 kPa = 10.96 atm
16.72. Model: Assume the trapped air to be an ideal gas.
Visualize:
Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure p1 = patmos = 1 atm and the volume is V1 = L1A, where A is the cross-sectional area of the pipe. By pushing the pipe in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an isothermal compression of the gas with T2 = T1. Solve: From the ideal gas law,
p2 V2 = p1V1 p2 L2 A = p1 L1 A p2 L2 = patmos L1 p2 = patmos ( L1 L2 )
As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If the pressures at a and b weren't equal, the pressure difference would cause the liquid level in the pipe to move up or down.) The pressure at point a is just the gas pressure inside the pipe: pa = p2. The pressure at point b is the pressure at depth L2 in water: pb = patmos+ gL2. Equating these gives p2 = patmos + gL2 Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes
patmos L1 = patmos + gL2 gL2 + patmos L2 - patmos L1 = 0 2 L2
This is a quadratic equation for L2 with solutions
L2 =
- patmos
( patmos )2 + 4 gpatmos L1
2 g
Length has to be a positive quantity, so the one physically acceptable solution is
L2 =
-101, 300 Pa +
(101,300 Pa )2 + 4(1000 kg m 3 )(9.8 m s 2 )(101, 300 Pa )(3.0 m)
2(1000 kg m 3 )(9.8 m s 2 )
= 2.43 m
16.73. Model: The gas in the cylinder is assumed to be an ideal gas.
Visualize: Please refer to Figure CP16.73. Solve: The gas at T1 = 20C = 293 K has a pressure p1 = 1 atm and a volume V1 = AL0. (The pressure has to be 1 atm to balance the external force of the air on the piston.) At a temperature T2 = 100C = 373 K, its volume becomes V2 = A(L0 + x) and its pressure increases to
p2 = p1 +
F kx = p1 + A A
where F = kx is the spring force on the piston. Using the before-and-after relationship of an ideal-gas equation,
p L A ( p1 + kx A)( L0 + x ) A p1V1 p2 V2 = 1 0 = T1 T2 T1 T2 T2 ( p1 + kx A)( L0 + x ) x kx = = 1 + 1 + T1 p1 L0 p1 A L0
L0 can be obtained from the ideal-gas law as follows:
L0 =
V1 nRT1 1 (0.004 mol)(8.31 J mol K )(293 K ) = = 0.0961 m = 9.61 cm = A p1 A 1.013 10 5 Pa (10 10 -4 m 2 )
(
)
Substituting this value of L0 and the values for p1, T1, T2, and k, the above equation can be simplified to
154.02 x 2 + 25.210 x - 0.2730 = 0 x = 0.0102 m = 1.02 cm
The spring is compressed by 1.02 cm.
16.74. Model: The gas in containers A and B is assumed to be an ideal gas.
Visualize: Please refer to Figure CP16.74. Solve: (a) The number of moles of gas in containers A and B can be expressed as follows:
nA =
1.0 10 5 Pa VA pA VA V = = 333.3 A RTA R(300 K ) R
(
)
nB =
5.0 10 5 Pa ( 4VA ) pBVB V = 5000 A = RTB R( 400 K ) R
(
)
where nA and nB are expressed in moles. Let nA and nB be the number of moles after the valve is opened. Since the total number of moles is the same,
nA + nB = 5333.3
The pressure is now the same in both containers:
VA = n A + nB R
pA =
n A = nB
Solving the above equations,
nA RTA n RT = pB = B B VA VB
TB VA 400 K VA = nB 3nA = nB 300 K 4VA TA VB
5333.3
The new pressure is
V VA V = nA + 3nA = 4nA nA = 1333.3 A nB = 4000 A R R R
nA RTA 1333.3 VA = RTA = 4.0 10 5 Pa VA VA R nB RTB 4000 VA = RTB = 4.0 10 5 Pa VB VB R
pA = pB =
Gas flows from B to A until the pressure is 4.0 10 5 Pa. (b) The change is irreversible because opening the valve is like breaking a membrane. It is not a quasi-static process.