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4.A quantity of ideal gas at 7°C and 51 kPa occupies a volume of 2.6 m3. (a) How many moles of the gas are present? (b) If the pressure is now raised to 210 kPa and the temperature is raised to 27.0°C, how much volume does the gas occupy? Assume no leaks.
21.The lowest possible temperature in outer space is 2.75 K. What is the rms speed of hydrogen molecules at this temperature? (The molar mass of molecular hydrogen is 2.02 g/mol )
26. What is the average translational kinetic energy of nitrogen molecules at 2140 K?
30. The mean free path of nitrogen molecules at 24°C and 1.52 atm is 0.251 x 10-5 cm. At this temperature and pressure there are 3.76 x 1019 molecules/cm3. What is the molecular diameter, in cm?
33.The speeds of 10 molecules are 2.0, 3.0, 4.0, . . . , 11 km/s. In km/s, what are their (a) average speed and (b) rms speed?
34.In fig. 19-25, 1.43 mole of an ideal diatomic gas can go from a to c along either the direct (diagonal) path ac or the indirect path abc. The scale of the vertical axis is set by pab = 6.27 kPa and pc = 2.28 kPa, and the scale of the horizontal axis is set by Vbc = 4.67 m3 and Va = 2.34 m3. (The molecules rotate but do not oscillate.) What is the net transfer of energy as heat for the indirect path?
ANSWERS
4. (a) With T = 283 K, we obtain
pV (100´ 103Pa)(2.50m3)
n = = = 106mol.
RT (8.31J/mol×K)(283K)
(b) We can use the answer to part (a) with the new values of pressure and
temperature, and solve the ideal gas law for the new volume, or we could set up the
gas law in ratio form as:
p V T
f f f
pV T
i i i
(where ni = nf and thus cancels out), which yields a final volume of
T 3100kPa 303K 3
VfVipi f 2.50m 0.892 m .
pfTi 300kPa 283K
21. According to kinetic theory, the rms speed is
3RT
v
rms
M
where T is the temperature and M is the molar mass. See Eq. 19-34. According to
–3
Table 19-1, the molar mass of molecular hydrogen is 2.02 g/mol = 2.02 10
kg/mol, so
3 8.31J/mol K 2.7K
2
v 1.8 10 m/s.
rms 3
2.02 10 kg/mol
Note: The corresponding average speed and most probable speed are
RT 8 8.31J/mol K 2.7K
8 2
v 1.7 10 m/s
avg 3
M (2.02 10 kg/mol)
and
1
RT 2 8.31J/mol K 2.7K
2 2
vp 1.5 10 m/s ,
3
M 2.02 10 kg/mol
respectively.
26. The average translational kinetic energy is given by Kavg 32 kT , where k is the
–23
Boltzmann constant (1.38 10 J/K) and T is the temperature on the Kelvin scale.
Thus
3 23 20
K (1.38 10 J/K)(1600K) = 3.31 10 J .
avg
2
30. We solve Eq. 19-25 for d:
11
d
5 19 3
2(N/V) (0.80 10 cm) 2(2.7 10 /cm )
–
which yields d = 3.2 10 8 cm, or 0.32 nm.
v
33. (a) The average speed is v , where the sum is over the speeds of the
N
particles and N is the number of particles. Thus
(2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0)km/s
v 6.5km/s.
10
2
v
(b) The rms speed is given by vrms . Now
N
2 2 2 2 2 2
v [(2.0) (3.0) (4.0) (5.0) (6.0)
2 2 2 2 2 2 2 2 2
(7.0) (8.0) (9.0) (10.0) (11.0) ] km /s 505km /s
so
22
505km /s
v 7.1km/s.
rms
10
2
44. Two formulas (other than the first law of thermodynamics) will be of use to us. It
is straightforward to show, from Eq. 19-11, that for any process that is depicted as a
straight line on the pV diagram, the work is
pp
if
WV
straight
2
which includes, as special cases, W = p V for constant-pressure processes and W = 0
for constant-volume processes. Further, Eq. 19-44 with Eq. 19-51 gives
ff
E n RT pV
int
22
where we have used the ideal gas law in the last step. We emphasize that, in order to
obtain work and energy in joules, pressure should be in pascals (N / m2) and volume
should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5.
(a) The internal energy change is
553 3 3 3
E E pV pV 2.0 10 Pa 4.0m 5.0 10 Pa 2.0m
int c int a c c a a
22
3
5.0 10 J.
(b) The work done during the process represented by the diagonal path is
papc 33
W V V = 3.5 10 Pa 2.0m
diag ca
2
which yields Wdiag = 7.0×103 J. Consequently, the first law of thermodynamics gives
3 3 3
Q E W ( 5.0 10 7.0 10 ) J 2.0 10 J.
diag int diag
(c) The fact that Eint only depends on the initial and final states, and not on the
details of the “path” between them, means we can write
3
E E E 5.0 10 J for the indirect path, too. In this case, the work done
int int cint a
3
consists of that done during the constant pressure part (the horizontal line in the
graph) plus that done during the constant volume part (the vertical line):
3 3 4
W 5.0 10 Pa 2.0m 0 1.0 10 J.
indirect
Now, the first law of thermodynamics leads to
3 4 3
Q E W ( 5.0 10 1.0 10 ) J 5.0 10 J.
Monday, September 26, 2011
Wiley Plus Chapter 18
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4. At what temperature (degrees Celsius) is the Fahrenheit scale reading equal to (a) 4 times that of the Celsius and (b) 1/5 times that of the Celsius?
12.An aluminum-alloy rod has a length of 8.3939 cm at 13.000°C and a length of 8.4047 cm at the boiling point of water. (a) What is the length (in cm) of the rod at the freezing point of water? (b) What is the temperature (in Celsius) if the length of the rod is 8.4384 cm? Give your answers to five significant figures.
27.Calculate the minimum amount of energy, in joules, required to completely melt 228 g of silver initially at 18.5°C. The melting point of silver is at 962°C. Its specific heat capacity is 236 J/kg·K and its latent heat of fusion is 105 kJ/kg.
43. A sample of gas expands from 1.0 m3 to 4.0 m3 while its pressure decreases from 40 Pa to 10 Pa. How much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C in Figure 18-36?
53.Consider a slab of face area A and thickness L. Suppose that L = 17 cm, A = 68 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 123°C and TC = 38°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.
ANSWERS
5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale
be y. Then y95x32. If we require y = 2x, then we have 9 2x x 32 x (5)(32) 160 C
which yields y = 2x = 320°F.
(b) In this case, we require y12x and find
1 9 (10)(32)
x x 32 x 24.6 C
2 5 13
which yields y = x/2 = –12.3°F.
12. (a) The coefficient of linear expansion for the alloy is
L 10.015cm 10.000cm 5
1.88 10 /C .
LT (10.01cm)(100 C 20.000 C)
Thus, from 100°C to 0°C we have
52
L L T (10.015cm)(1.88 10 /C )(0 C 100 C) = 1.88 10 cm.
The length at 0°C is therefore L = L + L = (10.015 cm – 0.0188 cm) = 9.996 cm.
(b) Let the temperature be Tx. Then from 20°C to Tx we have
5
L 10.009cm 10.000cm = L T (1.88 10 /C )(10.000cm) T,
giving T = 48 °C. Thus, Tx = (20°C + 48 °C )= 68°C.
1
27. The melting point of silver is 1235 K, so the temperature of the silver must first
be raised from 15.0° C (= 288 K) to 1235 K. This requires heat
4
Q cm(TfTi) (236J/kg K)(0.130kg)(1235 C 288 C) 2.91 10 J.
Now the silver at its melting point must be melted. If LF is the heat of fusion for
silver, this requires
34
Q mLF 0.130kg 105 10 J/kg 1.36 10 J.
The total heat required is ( 2.91 104 J + 1.36 104 J ) = 4.27 104 J.
43. (a) One part of path A represents a constant pressure process. The volume changes
from 1.0 m3 to 4.0 m3 while the pressure remains at 40 Pa. The work done is
3 3 2
WA p V (40Pa)(4.0m 1.0m ) 1.2 10 J.
(b) The other part of the path represents a constant volume process. No work is done
during this process. The total work done over the entire path is 120 J. To find the
work done over path B we need to know the pressure as a function of volume. Then,
we can evaluate the integral W = p dV. According to the graph, the pressure is a
linear function of the volume, so we may write p = a + bV, where a and b are
constants. In order for the pressure to be 40 Pa when the volume is 1.0 m3 and 10 Pa
when the volume is 4.00 m3 the values of the constants must be a = 50 Pa and b = –10
Pa/m3. Thus,
p = 50 Pa – (10 Pa/m3)V
and
44
24
WB p dV 50 10V dV 50V 5V 200J 50J 80J + 5.0J = 75J.
11 1
(c) One part of path C represents a constant pressure process in which the volume
changes from 1.0 m3 to 4.0 m3 while p remains at 10 Pa. The work done is
33
WC p V (10Pa)(4.0m 1.0m ) 30J.
2
The other part of the process is at constant volume and no work is done. The total
work is 30 J. We note that the work is different for different paths.
53. The rate of heat flow is given by
TT
P kAHC,
cond
L
where k is the thermal conductivity of copper (401 W/m·K), A is the cross-sectional
area (in a plane perpendicular to the flow), L is the distance along the direction of
flow between the points where the temperature is TH and TC. Thus,
42
401W/m K 90.0 10 m 125 C 10.0 C
3
P 1.66 10 J/s.
cond
0.250m
The thermal conductivity is found in Table 18-6 of the text. Recall that a change in
Kelvin temperature is numerically equivalent to a change on the Celsius scale.
4. At what temperature (degrees Celsius) is the Fahrenheit scale reading equal to (a) 4 times that of the Celsius and (b) 1/5 times that of the Celsius?
12.An aluminum-alloy rod has a length of 8.3939 cm at 13.000°C and a length of 8.4047 cm at the boiling point of water. (a) What is the length (in cm) of the rod at the freezing point of water? (b) What is the temperature (in Celsius) if the length of the rod is 8.4384 cm? Give your answers to five significant figures.
27.Calculate the minimum amount of energy, in joules, required to completely melt 228 g of silver initially at 18.5°C. The melting point of silver is at 962°C. Its specific heat capacity is 236 J/kg·K and its latent heat of fusion is 105 kJ/kg.
43. A sample of gas expands from 1.0 m3 to 4.0 m3 while its pressure decreases from 40 Pa to 10 Pa. How much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C in Figure 18-36?
53.Consider a slab of face area A and thickness L. Suppose that L = 17 cm, A = 68 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 123°C and TC = 38°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.
ANSWERS
5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale
be y. Then y95x32. If we require y = 2x, then we have 9 2x x 32 x (5)(32) 160 C
which yields y = 2x = 320°F.
(b) In this case, we require y12x and find
1 9 (10)(32)
x x 32 x 24.6 C
2 5 13
which yields y = x/2 = –12.3°F.
12. (a) The coefficient of linear expansion for the alloy is
L 10.015cm 10.000cm 5
1.88 10 /C .
LT (10.01cm)(100 C 20.000 C)
Thus, from 100°C to 0°C we have
52
L L T (10.015cm)(1.88 10 /C )(0 C 100 C) = 1.88 10 cm.
The length at 0°C is therefore L = L + L = (10.015 cm – 0.0188 cm) = 9.996 cm.
(b) Let the temperature be Tx. Then from 20°C to Tx we have
5
L 10.009cm 10.000cm = L T (1.88 10 /C )(10.000cm) T,
giving T = 48 °C. Thus, Tx = (20°C + 48 °C )= 68°C.
1
27. The melting point of silver is 1235 K, so the temperature of the silver must first
be raised from 15.0° C (= 288 K) to 1235 K. This requires heat
4
Q cm(TfTi) (236J/kg K)(0.130kg)(1235 C 288 C) 2.91 10 J.
Now the silver at its melting point must be melted. If LF is the heat of fusion for
silver, this requires
34
Q mLF 0.130kg 105 10 J/kg 1.36 10 J.
The total heat required is ( 2.91 104 J + 1.36 104 J ) = 4.27 104 J.
43. (a) One part of path A represents a constant pressure process. The volume changes
from 1.0 m3 to 4.0 m3 while the pressure remains at 40 Pa. The work done is
3 3 2
WA p V (40Pa)(4.0m 1.0m ) 1.2 10 J.
(b) The other part of the path represents a constant volume process. No work is done
during this process. The total work done over the entire path is 120 J. To find the
work done over path B we need to know the pressure as a function of volume. Then,
we can evaluate the integral W = p dV. According to the graph, the pressure is a
linear function of the volume, so we may write p = a + bV, where a and b are
constants. In order for the pressure to be 40 Pa when the volume is 1.0 m3 and 10 Pa
when the volume is 4.00 m3 the values of the constants must be a = 50 Pa and b = –10
Pa/m3. Thus,
p = 50 Pa – (10 Pa/m3)V
and
44
24
WB p dV 50 10V dV 50V 5V 200J 50J 80J + 5.0J = 75J.
11 1
(c) One part of path C represents a constant pressure process in which the volume
changes from 1.0 m3 to 4.0 m3 while p remains at 10 Pa. The work done is
33
WC p V (10Pa)(4.0m 1.0m ) 30J.
2
The other part of the process is at constant volume and no work is done. The total
work is 30 J. We note that the work is different for different paths.
53. The rate of heat flow is given by
TT
P kAHC,
cond
L
where k is the thermal conductivity of copper (401 W/m·K), A is the cross-sectional
area (in a plane perpendicular to the flow), L is the distance along the direction of
flow between the points where the temperature is TH and TC. Thus,
42
401W/m K 90.0 10 m 125 C 10.0 C
3
P 1.66 10 J/s.
cond
0.250m
The thermal conductivity is found in Table 18-6 of the text. Recall that a change in
Kelvin temperature is numerically equivalent to a change on the Celsius scale.
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