Homework 11—help 16.12. Solve: Let F π π π π

Homework 11—help
16.8. (a) The ideal-gas law says
1
V
pV 1
2 2
3p
1
3
(b)
2
T2 T1 T1 T1.
pV
1 1 p1 V1
2
pV pV pV 3p 2V
T T T 6T
1 1 2 2 2 2 1 1
2 1 1 1
T1 T2 pV
1 1
p1 V1
16.12. Solve: Let TF TC T :
T T 32T T 32T
40
F
9 9
5 C
5
That is, the Fahrenheit and the Celsius scales give the same numerical value at 40 .
Assess: It is usually unnecessary to specify the scale when the temperature is reported as 40 .
16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas.
Solve: (a) The number of moles of oxygen is
M 50 g
n 1.563 mol 1.6 mol
M 32 g mol
(b) The number of molecules is
mol
N nN A
1.563 mol 6.0210 mol 9.4110 9.4
10
2
(c) The volume of the cylinder V r L
(d) From the ideal-gas law pV
23 1 23 23
2 2 3
0.10 m 0.40 m 1.257 10 m . Thus,
N
V
23
9.4110
7.510 m
2 3
1.257 10 m
25 3
nRT we can calculate the absolute pressure to be
1.563 mol8.31 J mol K293 K
nRT
p 303 kPa
2 3
V
1.257 10 m
where we used T 20C 293 K. But a pressure gauge reads gauge pressure:
pg p1 atm 303 kPa 101 kPa 202 kPa 200 kPa
16.69. Model: Assume that the compressed air is an ideal gas.
Solve: (a) Because the piston is floating in equilibrium,
where the piston’s cross-sectional area A
490 N. Thus,
F ( p p ) Aw
0 N
net 1 atoms
2 2 3 2
( r ) (0.050 m) 7.854 10 m and the piston’s weight w (50 kg)(9.8)
w
490 N
p1 patmos
3 2
A 7.85410 m
Using the ideal-gas equation 1 1 1
5 5
1.013 10 Pa=1.637 10 Pa

5
(1.637 10 Pa)Ah
1
(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
V Ah at a constant pressure p . 2
p 1
From the before-and-after relationship of the ideal gas:
to
2 2
p1( Ah1) p2( Ah2)
T T
1 2
p1 T2
403 K
h2 h1
1 0.235 m0.313 m 31.3 cm
p2 T1
303 K
Thus, the piston moves h2 h1 7.8 cm.
16.58. Model: The gas is an ideal gas.
Solve: (a) Using the ideal-gas equation,
pV
nR
5 3
1.010 Pa2.0 m
1 1
1
T
80 mol8.31 J mol K
301 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
where V is in m 3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as
5
pV
2 10
T (3 V V
)
nR
nR
We can maximize T by setting the derivative dT/dV to zero:
At this volume, the pressure is
dT
10 3
(32 V ) 0 m 1.50 m
dV
nR
2
5
2 3 3
max
Vmax
5
pmax = 1.5 10 Pa and the temperature is
5
T
p V
nR
(80 mol)(8.31 J/mol K)
5 3
max max
(1.50 10 Pa)(1.5 m )
max
338 K
Making Espresso
Espresso is a coffee beverage made by forcing steam through finely ground coffee beans.
Modern espresso makers generate steam at very high pressures and temperatures, but in this
problem we'll consider a low-tech espresso machine that only generates steam at 100C and
atomospheric pressure--not much good for making your favorite coffee beverage. The amount of
heat Q needed to turn a mass m of room temperature (T1) water into steam at 100C (T2) can be
found using the specific heat c of water and the heat of vaporization Hv of water at 1 atmosphere
of pressure.
Suppose that a commercial espresso machine in a coffee shop turns 1.50 Kg of water at 22.0 C
into steam at 100C . If and, how much heat Q is absorbed by the water from the heating resistor

inside the machine? Assume that this is a closed and isolated system. C(water) = 4187J/Kg C,
Hv = 2258 KJ/Kg.
PartA :
Q = 4187 ( J/Kg C) ×(100-22) C × 1.5Kg + 2258 ×10 3 J/Kg × 1.5 Kg = 3.88×10 6 J
PartB:
P·t = 1200W ·(time) = Q = 3.88×10 6 J
So t = 3233 s = 54 minutes
PartC
Suppose the mass of propane is m, so that the number of mole of propane gas is m/44.1
It can release the heat m/44.1 × 2219 KJ / mol
Solve equation m/44.1 × 2219 KJ / mol = 3.88×10 6 J
Get m = 77g
17.44. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperature from
90C to 60C requires four steps: (1) raise the temperature of ice from 20C to 0C, (2) change ice at 0C to water at
0C, (c) raise the water temperature from 0C to 60C, and (4) lower the coffee temperature from 90C to 60C.
Solve: For the closed coffee-ice system,
QQice Qcoffee Q1 Q2 Q3 Q4 0 J
2090 J kg K20 K 41,800 J kg
Q M c T M M
1 ice ice ice ice
Q2 MiceLf Mice 330,000 J kg
4190 J kg K60 K 251,400 J kg
Q M c T M M
3 ice water ice ice
6 3 3
Q4 McoffeeccoffeeT
30010 m 1000 kg m 4190 J kg K 30 K 37,000 J
The Q 0 J equation thus becomes
M
ice
41,800 330,000 251,400 J kg 37,710 J 0 J M
ice
0.061 kg 61 g
Assess: 61 g is the mass of approximately 1 ice cube.
17.59. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.
Solve: (a) For the isochoric process, V 2 V 1 800 10 6 m 3 , p 1 4.0 atm, p 2 2.0 atm. The temperature T 1 of the gas is
obtained from the ideal-gas equation as:
1 1
T1 pV 390 K
nR
where n 0.10 mol. T 2 can be obtained from the ideal-gas equation as follows:
pV
1 1
pV
2 2
2.0 atm
T2 T1p2 p1390 K
195 K
T1 T2
4.0 atm
The heat required for the process 1 2 is
QnCV T2 T1 0.10 mol 20.8 J/mol K 195 K 390 K 406 J 410 J
Because of the negative sign, this is the amount of heat removed from the gas.
(b) For this isobaric process, p 2 p 3 2.0 atm, V 2 800 10 6 m 3 , and V 3 1600 10 6 m 3 .
6 3
V
3
160010 m
T3 T2 T2
2T
6 3
2
390 K
V2
80010 m
Thus, the heat required for the process 2 3 is

QnCP T3 T2 0.10 mol 29.1 J mol K 195 K 567 J 570 J
This is heat transferred to the gas.
(c) The change in the thermal energy of the gas is
E Q Q W W 406 J 567 J 0 J + W 162 J p
V
th 12 23 12 23 23
Assess: This result was expected since T 3 T 1 .
162 J – (2.0 1.013 10 5 Pa)(1600 10 6 m 3 – 800 10 6 m 3 ) 0 J