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and CHAPTER
18
Temperature the Kinetic Theory of Gases
1* True or false:
(a) Two objects in thermal equilibrium with each other must be in thermal equilibrium with a third object. (b) The Fahrenheit and Celsius temperature scales differ only in the choice of the zero temperature. (c) The kelvin is the same size as the Celsius degree. (d) All thermometers give the same result when measuring the temperature of a particular system. (a) False (b) False (c) True (d) False 2 How can you determine if two bodies are in thermal equilibrium with each other if it is impossible to put them into thermal contact with each other? Put each in thermal equilibrium with a third body, e.g., a thermometer. 3 4 Which is greater, an increase in temperature of 1 C or of 1 F ? "One day I woke up and it was 20 F in my bedroom," said Mert to his old friend Mort. "That's nothing," replied tC = (5/9)(20 32) = 6.67oC; 20oF is colder An increase of 1oC is greater. Mort. "My room was once 5 C." Which room was colder? Convert 20oF to tC; Equ. 18-2 5*
A certain ski wax is rated for use between 12 and 7 C. What is this temperature range on the Fahrenheit scale? tF1 = 10.4oF; tF2 = 19.4oF; between 10.4oF and 19.4oF
Convert 12 and 7oC to tF using tF = (9/5)tC + 32 6
The melting point of gold (Au) is 1945.4 F. Express this temperature in degrees Celsius. tC = (5/9)(1954.4 32) = 1068oC
Use Equ. 18-2 7
The highest and lowest temperatures ever recorded in the United States are 134 F (in California in 1913) and 80 F tCH = (5/9)102 = 56.7oC; tCL = (5/9)( 112) = 62.2oC
(in Alaska in 1971). Express these temperatures using the Celsius scale. Use Equ. 18-2 8
What is the Celsius temperature corresponding to the normal temperature of the human body, 98.6 F? tC = (5/9)66.6 = 37.0oC
Use Equ. 18-2
Chapter 18
Temperature and the Kinetic Theory of Gases
9*
The length of the column of mercury in a thermometer is 4.0 cm when the thermometer is immersed in ice water and
24.0 cm when the thermometer is immersed in boiling water. (a) What should the length be at room temperature, 22.0 C? (b) If the mercury column is 25.4 cm long when the thermometer is immersed in a chemical solution, what is the temperature of the solution? (a), (b) L = [(20/100)tC + 4] cm 10 (a) L = 8.4 cm (b) tC = (5 21.4)oC = 107oC
The temperature of the interior of the sun is about 107 K. What is this temperature on (a) the Celsius scale, and (b) (a) tC tK = 107oC (b) tF (9/5)tK = 1.8 107oF
the Fahrenheit scale? Neglect 32 and 273 compared to 107 11
The boiling point of nitrogen N2 is 77.35 K. Express this temperature in degrees Fahrenheit. 77.35 K = 198.5oC = (1.8 = 320.4oF 198.5 + 32)oF
Convert 77.35 K to oC then to oF
12
The pressure of a constant-volume gas thermometer is 0.400 atm at the ice point and 0.546 atm at the steam point.
(a) When the pressure is 0.100 atm, what is the temperature? (b) What is the pressure at 444.6 C, the boiling point of sulfur? (a) P = CT; find C, then T for P = 0.100 atm (b) Convert to T, then determine P T = (273.15/0.4) 0.100 K = 68.3 K 717.75 atm = 1.05 atm T = 717.75 K; P = (0.4/273.15)
13* A constant-volume gas thermometer reads 50 torr at the triple point of water. (a) What will the pressure be when the thermometer measures a temperature of 300 K? (b) What ideal-gas temperature corresponds to a pressure of 678 torr? (a), (b) Use Equ. 18-4 14 (a) T = 50(300/273.16) torr = 54.9 torr (b) T = 3704 K
A constant-volume gas thermometer has a pressure of 30 torr when it reads a temperature of 373 K. (a) What is its (a) P3 = 21.97 torr (b) T = 2.176 K
triple-point pressure P3? (b) What temperature corresponds to a pressure of 0.175 torr? (a), (b) Use Equ. 18-4 15
At what temperature do the Fahrenheit and Celsius temperature scales give the same reading? 0.8tC = 32; tC = 40oC = 40oF
Set tF = tC in Equ. 18-2 and solve for tC 16
Sodium melts at 371 K. What is the melting point of sodium on the Celsius and Fahrenheit temperature scales? tC = 98oC; tF = 208.4oF
tC = T 273; then use Equ. 18-2
17* The boiling point of oxygen at one atmosphere is 90.2 K. What is the boiling point of oxygen on the Celsius and Fahrenheit scales? Proceed as in Problem 16 tC = 182.95oC; tF = 297.3oF
Chapter 18 18
Temperature and the Kinetic Theory of Gases
On the Raumur temperature scale, the melting point of ice is 0 R and the boiling point of water is 80 R. Derive
expressions for converting temperatures on the Raumur scale to the Celsius and Fahrenheit scales. 0oC = 0oR and 100oC = 80oR. Consequently, tR = 0.8tC and tC = 1.25tR = (5/4)tR. From Equ. 18-2 we have tF = 1.8tC + 32 = 2.25tR + 32 = (9/4)tR + 32. 19 A thermistor is a solid-state device whose resistance varies greatly with temperature. Its temperature dependence is at the ice given approximately by R = R0 eB/T 0, where R is in ohms ( ), T is in kelvins, and R0 and B are constants that can be determined by measuring R at calibration points such as the ice point and the steam point. (a) If R = 7360 point and 153 at the steam point, find R0 and B. (b) What is the resistance of the thermistor at t = 98.6 F? (c) What is
the rate of change of the resistance with temperature (dR/dT) at the ice point and the steam point? (d) At which temperature is the thermistor most sensitive?
7360 (a) Find B from R 273 = = 48.1 153 R 373
Evaluate R0 from R0 = RTe
B/T
e
B 273
B 373
= 48.1; B =
ln 48.1 = 3.94 103 K 1/273 1/373
(b) 98.6oF = 310 K (see Problem 8)
R0 = 7360 e
R = 3.91 10
3
3940/273
= 3.91 10
3
e3940/310 = 1.29 k /K; (dR/dT)steam = 4.33 /K
(c), (d)
dR = dT
R 0 B B/T = 2 e T
RB T
2
(dR/dT)ice = 389
Sensitivity is greater the lower the temperature 20 Two identical vessels contain different ideal gases at the same pressure and temperature. It follows that
(a) the number of gas molecules is the same in both vessels. (b) the total mass of gas is the same in both vessels. (c) the average speed of the gas molecules is the same in both vessels. (d) none of the above is correct. (a) 21* Figure 18-15 shows a plot of volume versus temperature for a process that takes an ideal gas from point A to point B. What happens to the pressure of the gas? The pressure increases. 22 Figure 18-16 shows a plot of pressure versus temperature for a process that takes an ideal gas from point A to point B. What happens to the volume of the gas? The volume decreases. 23 A gas is kept at constant pressure. If its temperature is changed from 50 to 100 C, by what factor does the volume V2/V1 = 373/323 = 1.15 change? At constant pressure, V2/V1 = T2/T1 24
A 10-L vessel contains gas at a temperature of 0 C and a pressure of 4 atm. How many moles of gas are in the
vessel? How many molecules?
Chapter 18 From Equ. 18-12, V/mol at STP = 22.4 L 25* A pressure as low as 1
Temperature and the Kinetic Theory of Gases n = 10 4/22.4 = 1.79; N = nNA = 1.08 1024
108 torr can be achieved using an oil diffusion pump. How many molecules are there in 1
cm3 of a gas at this pressure if its temperature is 300 K? 1. Convert torr to atm and cm3 to L 2. Use Equs. 18-12 and 18-13 P = 10 V=1 = 3.22 26
8
1.316
3
10 3 atm = 1.316 6.022
10
11
atm; 300
10 L 10
8 14
N = 1.316 10
1023/0.08206
A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is 8.0 C. When she
arrives at her destination, the tire pressure has increased to 245 kPa. What is the temperature of the tires if we assume that (a) the tires do not expand, or (b) that the tires expand by 7%? (a) From Equ. 18-13, T2 = T1(P2/P1) (b) T2 = T1(P2V2/P1V1) 27 T2 = 265 T2 = 360.7 245/180 K = 360.7 K = 87.7oC 1.07 = 386 K = 113oC
A room is 6 m by 5 m by 3 m. (a) If the air pressure in the room is 1 atm and the temperature is 300 K, find the
number of moles of air in the room. (b) If the temperature rises by 5 K and the pressure remains constant, how many moles of air leave the room? (a) V/mol at STP = 22.4 L (b) Find n = n(305 K) and n 28 n = 90 103 273/22.4 300 = 3.66 103 mol n = n(300/305); n = n(1 0.9836) = 60 mol
A seafood restaurant hires Lou to run its advertising campaign. Lou figures that snorklers are a great pool of
potential customers for seafood, so he prints ads on Mylar balloons that he ties to the coral of an underwater reef. Each balloon has a volume of 4 L and is filled with air at 20 C. At 15 m below the ocean surface, the volume has diminished to 1.60 L. What is the temperature of the water at this depth? 1. Find P2 = pressure at 15 m below surface 2. Use Equ. 18-13 29* P2 = (101 + 1.025 T2 = 293 251.8 9.81 1.6/101 15) kPa = 251.8 kPa 4 K = 292 K = 19oC
The boiling point of helium at one atmosphere is 4.2 K. What is the volume occupied by helium gas due to V4.2 = 2.5 0.08206 4.2/1 L = 0.862 L
evaporation of 10 g of liquid helium at 1 atm pressure and a temperature of (a) 4.2 K, and (b) 293 K? (a) 10 g = 2.5 mol; V = nRT/P (b) V2 = V1(T2/T1) 30 V293 = 0.862 293/4.2 = 60.1 L
A container with a volume of 6.0 L holds 10 g of liquid helium. As the container warms to room temperature, what P=1 60.1/6 atm = 10.0 atm
is the pressure exerted by the gas on its walls? From Problem 29, P = 1 atm for V = 60.1 L 31
An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20 C. (Gauge pressure is the
Chapter 18
Temperature and the Kinetic Theory of Gases
difference between the actual pressure and atmospheric pressure.) After the car has been driven at high speeds, the tire temperature increases to 50 C. (a) Assuming that the volume of the tire does not change, and that air behaves as an ideal gas, find the gauge pressure of the air in the tire. (b) Calculate the gauge pressure if the volume of the tire expands by 10%. (a) P2 = P1(T2/T1) (b) P2' = P2(V1/V2) 32 P2 = 301 323/293 kPa = 332 kPa; P2gauge = 231 kPa P2' = 332/1.1 = 302 kPa; P2gauge = 201 kPa
A scuba diver is 40 m below the surface of a lake, where the temperature is 5 C. He releases an air bubble with a
volume of 15 cm3. The bubble rises to the surface, where the temperature is 25 C. What is the volume of the bubble right before it breaks the surface? Hint: Remember that the pressure also changes. 1. Determine P1 2. V2 = V1(T2/T1)(P1/P2); P2 = 101 kPa 33* P1 = 101 + 1.0 9.81 40 kPa = 493 kPa V2 = 15(298/278)(493/101) cm3 = 78.5 cm3
A helium balloon is used to lift a load of 110 N. The weight of the balloon's skin is 50 N, and the volume of the
balloon when fully inflated is 32 m3. The temperature of the air is 0 C and the atmospheric pressure is 1 atm. The balloon is inflated with sufficient helium gas so that the net buoyant force on the balloon and its load is 30 N. Neglect changes of temperature with altitude. (a) How many moles of helium gas are contained in the balloon? (b) At what altitude will the balloon be fully inflated? (c) Does the balloon ever reach the altitude at which it is fully inflated? (d) If the answer to (c) is affirmative, what is the maximum altitude attained by the balloon? (a) Find V from FB = mg + 30 N n = PV/RT (b) Find P for V = 32 m
3 airVg
= 190 N +
HeVg;
V = 17.38 m3 273 = 776
n=1
17.38/0.08206
h/7.93
P = 17.32/32 atm = 0.543 atm 0.543 = e 0.543 ; h = [7.93 ln(1/.543)] km = 4.84 km 32 776 9.81 N = 220.4 N 9.81) N = 190.5 N
air
Determine h from Problem 13-96 (c) Determine FB at 4.84 km; Find mtotg Note that FB at 4.84 km is greater than mtotg (d) Find
air
= 1.293
FB = 1.293
0.543
mtotg = (160 + 0.004
air h/7.93
Yes; the balloon will rise above 4.84 km = 190.5/32 9.81 kg/m3 = 0.6068 kg/m3 h = [7.93 ln(1/0.469)] km = 6.0 km
such that FB = 190.5 N
Note that P = (0.6068/1.293) atm = e 34
True or false: The absolute temperature of a gas is a measure of the average translational kinetic energy of the gas
molecules. True 35 36 By what factor must the absolute temperature of a gas be increased to double the rms speed of its molecules? 4; vrms
T.
How does the average translational kinetic energy of a molecule of a gas change if the pressure is doubled while the
volume is kept constant? If the volume is doubled while the pressure is kept constant? If P is doubled at constant V, T is doubled. Consequently, Kav increases by a factor of 2.
Chapter 18
Temperature and the Kinetic Theory of Gases
If V is doubled at constant P, T is halved. Consequently, Kav is reduced by a factor 1/2. 37* A mole of He molecules is in one container and a mole of CH4 molecules is in a second container, both at standard conditions. Which molecules have the greater mean free path? From Equ. 18-25, it follows that the He atoms have the greater mean free path since the diameter of He is smaller than that of CH4. 38 A vessel holds an equal number of moles of helium and methane, CH4. The ratio of the rms speeds of the helium atoms to the CH4 molecules is ___. (a) 1 (b) 2 (c) 4 (d) 16 (b) 39 (a) Find vrms for an argon atom if 1 mol of the gas is confined to a 1-L container at a pressure of 10 atm. 103 kg/mol.) (b) Compare this with vrms for a helium atom under the same conditions. 103 kg/mol.) (For argon, M = 40 (For helium, M = 4
(a) From Equs. 18-23 and 18-13, v rms =
3PV nM
v rms
=
3 1.01 106 10 1 40 10 3
3
m/s = 275 m/s
(b) v rms (He) = v rms (Ar )
M M
Ar He
vrms(He) = 870 m/s
40
Find the total translational kinetic energy of 1 L of oxygen gas held at a temperature of 0 C and a pressure of 1 K = 1.5 1.01 105 10 3 J = 152 J
atm. Use Equs. 18-22 and 18-13; K = (3/2)PV 41*
Find the rms speed and the average kinetic energy of a hydrogen atom at a temperature of 10 7 K. (At this
temperature, which is of the order of the temperature in the interior of a star, the hydrogen is ionized and consists of a single proton.) Use Equs. 18-22 and 18-23; M = 10 3 kg/mol K = 1.5kT J = 2.07 1016 J
v rms =
3 8.314 107 m/s = 499 m/s 3 10
42
In one model of a solid, the material is assumed to consist of a regular array of atoms in which each atom has a
fixed equilibrium position and is connected by springs to its neighbors. Each atom can vibrate in the x, y, and z directions. The total energy of an atom in this model is
1 1 1 1 1 E = m v2 + m v2 + m v2 + K x2 + K y 2 + K z 2 x y z 2 2 2 2 2
Chapter 18
Temperature and the Kinetic Theory of Gases
What is the average energy of an atom in the solid when the temperature is T? What is the total energy of one mole of such a solid? There are six degrees of freedom. Consequently, Eav = 6(1/2kT) = 3kT and E/mol = 3RT. 43 Show that the mean free path for a molecule in an ideal gas at temperature T and pressure P is given by
=
kT 2 P d2
= 1 2 nv d
2
We have
(Equ. 18-25). Here nv = N/V = nNA/V. From the ideal gas law,
V =
n RT n kT . = NA V V P . One obtains the expression given, namely, kT
= kT 2P d
2
Thus, n v =
.
44
The escape velocity on Mars is 5.0 km/s, and the surface temperature is typically 0 C. Calculate the rms speeds for
(a) H2, (b) O2, and (c) CO2 at this temperature. (d) If the rms speed of a gas is greater than about 15% to 20% of the escape velocity of a planet, virtually all of the molecules of that gas will escape the atmosphere of the planet. Based on this criterion, are H2, O2, and CO2 likely to be found in Mars's atmosphere?
3RT ; M of H2 = 2 M
(a) vrms =
10 3 kg/mol
v rms =
3 8.314 273 m/s = 1845 m/s 2 10 3
(b) M of O2 = 32 (c) M of CO2 = 44
10 3 kg/mol 10 3 kg/mol
vrms = 1845/4 m/s = 461 m/s vrms = 1845/ 22 m/s = 393 m/s No H2, but O2 and CO2 should be present
(d) vesc/5 = 1000 m/s
45*
Repeat Problem 44 for Jupiter, whose escape velocity is 60 km/s and whose temperature is typically (a) vrms = 1368 m/s (b) vrms = 342 m/s (c) vrms = 291 m/s H2, and O2, CO2 should all be found on Jupiter 1011 Pa has been obtained. Suppose a chamber contains helium at this pressure and and the collision time for helium in the chamber. Take
10
150 C. (a), (b), (c) See Problem 44; use T = 123 K (d) vesc/5 = 12000 m/s 46 A pressure as low as P = 7
at room temperature (300 K). Estimate the mean free path the diameter of a helium molecule to be 10 1. Use the result of Problem 43 for m.
=
1.38 10 23 300 1.41 7 10 11 10
20
m = 1.33 109 m
Chapter 18
Temperature and the Kinetic Theory of Gases
2. = /vav =
M 3RT
= 1.33 109
3
4 10 3 s = 9.72 105 s 8.314 300
47
Oxygen (O2) is confined to a cubic container 15 cm on a side at a temperature of 300 K. Compare the average
kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom. The average K = 1.5kT = 1.5 mgh = 32 48 10
3
1.38
5
10
23
300 J = 6.21 10
23
10 10
26
21
J. The change in potential energy is U =
9.81
0.15/6.022
J = 7.82
J. The average kinetic energy of the molecules is
the same (within 1 part in 10 ) for all molecules in the container. The class in Room 101 prepares their traditional greeting for a substitute teacher. Ten toy cars are wound up and released as the teacher arrives. The cars have the following speeds. Speed, m/s Number of cars 2 3 5 3 6 3 8 1
Calculate (a) the average speed, and (b) the rms speed of the cars. (a) vav =
i
ni v i /
i
ni
vav = [(6 + 15 + 18 + 8)/10] m/s = 4.7 m/s vrms =
(12 + 75 + 108 + 64)/10 m/s = 5.09 m/s
(b) vrms =
i
2 ni vi / i
ni
49*
Show that f(v) given by Equation 18-37 is maximum when v = 2kT/m. Hint: Set df/dv = 0 and solve
3/ 2
for v.
df The derivative of f with respect to v is dv
The result is v = 50
4
m 2kT
2v
m v3 . Set this equal to zero and solve for v. kT
2kT/m .
Since f(v) dv gives the fraction of molecules that have speeds in the range dv, the integral of f(v) dv over all the
possible ranges of speeds must equal 1. Given the integral
v e
0
2
av2
dv =
4
a
3/ 2
show that
0
f(v) dv = 1, where f(v) is given by Equation 18-37.
Follow the procedure indicated.
0
f(v) dv =
4
a
3/ 2 0
v e
2
av2
dv , where a =
m . The integral has the value 2kT
Chapter 18
Temperature and the Kinetic Theory of Gases
4
51
a
3/ 2
, and it follows that
0
f(v) dv = 1.
Given the integral
2
v3 e
0
av2
dv = a 2
calculate the average speed vav of molecules in a gas using the Maxwell-Boltzmann distribution function. From Problem 50 we know that f(v) is normalized. Therefore, vav =
0v
f(v) dv. We again set a = (m/2kT) and
perform the integration using the result given in the problem and obtain;
4 m 2kT
3/ 2
vav
=
1 2kT 2 m
2
=
2
2kT . m
52 In Chapter 11, we found that the escape speed at the surface of a planet of radius R is v e = 2 gR , where g is the acceleration due to gravity. (a) At what temperature is vrms for O2 equal to the escape speed for the earth? (b) At what temperature is vrms for H2 equal to the escape speed for the earth? (c) Temperatures in the upper atmosphere reach 1000 K. How does this account for the low abundance of hydrogen in the earth's atmosphere? (d) Compute the temperatures for which the rms speeds of O2 and H2 are equal to the escape velocity at the surface of the moon, where g is about one-sixth of its value on earth and R = 1738 km. How does this account for the absence of an atmosphere on the moon? (a) vrms2 = vesc2 = 2gRE; vrms2 = 3RT/M T = 2MgRE/3R (b) For H2, M(H2) = M(O2)/16 (c) If v > vesc/5 or T/25 (d) T = 2MgMRM/3R The temperature on the moon with atmosphere might be about 1000 K, so all O2 and H2 molecules would escape in the time since the formation of the moon to today. 53* 54
o
T = (2 T = 1.0
32
10
5
3
9.81
6.37
106/3
8.314) K
T = 1.60 Tatm, molecules escape
10 K 104 K
The more energetic H2 molecules escape from the upper atmosphere T(O2) = [1.6 T(H2) =(7.28 105(1/6)(1738/6370)] K = 7.28 103/16) K = 455 K 103 K
True or false: If the pressure of a gas increases, the temperature must increase. What is the difference between 1 C and 1 C ?
False 1 C is a specific temperature; 1 Co is a temperature interval.
Chapter 18 55
Temperature and the Kinetic Theory of Gases
Why might the Celsius and Fahrenheit scales be more convenient than the absolute scale for ordinary, nonscientific
purposes? For the Celsius scale, the ice point (0oC) and the boiling point of water at 1 atm (100oC) are more convenient than 273 K and 373 K; temperatures in roughly this range are normally encountered. On the Fahrenheit scale, the temperature of warm-blooded animals is roughly 100oF; this may be a more convenient reference than approximately 300 K. Throughout most of the world, the Celsius scale is the standard for nonscientific purposes. 56 The temperature of the interior of the sun is said to be about 107 degrees. Do you think that this is degrees Celsius or kelvins, or does it matter? It does not matter; 107 >> 273. 57* If the temperature of an ideal gas is doubled while maintaining constant pressure, the average speed of the molecules (a) remains constant. (b) increases by a factor of 4. (c) increases by a factor of 2. (d) increases by a factor of (d) 58 If both temperature and volume of an ideal gas are halved, the pressure (a) diminishes by a factor of 2. (b) remains constant. (c) increases by a factor of 2. (d) diminishes by a factor of (b) 59 The average translational kinetic energy of the molecules of an ideal gas depends on (a) the number of moles of the gas and its temperature. (b) the pressure of the gas and its temperature. (c) the pressure of the gas only. (d) the temperature of the gas only. (d) 60 If a vessel contains equal amounts, by weight, of helium and argon, which of the following are true? (a) The pressure exerted by the two gases on the walls of the container is the same. (b) The average speed of a helium atom is the same as that of an argon atom. (c) The number of helium atoms and argon atoms in the vessel are equal. (d) None of the above statements is correct. (d) 61* Two different gases are at the same temperature. What can you say about the rms speeds of the gas molecules? What can you say about the average kinetic energies of the molecules? The rms speeds are inversely proportional to the square root of the molecular masses. The average kinetic energies of the molecules are the same. 62 Explain in terms of molecular motion why the pressure on the walls of a container increases when a gas is heated at
2.
2.
Chapter 18 constant volume.
Temperature and the Kinetic Theory of Gases
The pressure is a measure of the change in momentum per second of a gas molecule on collision with the wall of the container. When the gas is heated, the average velocity and, hence, the average momentum of the molecules increases, and so does the pressure. 63 Explain in terms of molecular motion why the pressure on the walls of a container increases when the volume of a gas is reduced at constant temperature. The molecule's time between collisions with the walls of the container is reduced. Consequently, each molecule makes more collisions with the walls per second, and so p/ t increases. 64 Oxygen has a molar mass of 32 g/mol, and nitrogen has a molar mass of 28 g/mol. The oxygen and nitrogen molecules in a room have (a) equal average kinetic energies, but the oxygen molecules are faster. (b) equal average kinetic energies, but the oxygen molecules are slower. (c) equal average kinetic energies and speeds. (d) equal average speeds, but the oxygen molecules have a higher average kinetic energy. (e) equal average speeds, but the oxygen molecules have a lower average kinetic energy. (f) None of the above is correct. (b) 65* At what temperature will the rms speed of an H2 molecule equal 331 m/s? T = (2 10
3
From Equ. 18-23, T = Mvrms2/3R 66
3312/3
8.314) K = 8.79 K
A solid-state temperature transducer is essentially a linear amplifier whose amplification is linearly temperature
dependent. If the amplification is 25 times at 20 C and 60 times at 70 C, what is the temperature when the amplification is 45 times? 1. Amplification, A = a + btC; find a and b 2. Find tC for A = 45 67 b = 35/50 = 0.7; a = 25 tC = [(45
o
0.7
o
20 = 11
11)/0.7] C = 48.6 C
(a) If 1 mol of a gas in a container occupies a volume of 10 L at a pressure of 1 atm, what is the temperature of the
gas in kelvins? (b) The container is fitted with a piston so that the volume can change. When the gas is heated at constant pressure, it expands to a volume of 20 L. What is the temperature of the gas in kelvins? (c) The volume is fixed at 20 L, and the gas is heated at constant volume until its temperature is 350 K. What is the pressure of the gas? (a) At STP, V/mol = 22.4 L (b) At constant P, T (c) At constant V, P 68 V T T = (10/22.4)273 K = 122 K T = 244 K P=1 350/244 atm = 1.43 atm
A cubic metal box with sides of 20 cm contains air at a pressure of 1 atm and a temperature of 300 K. The box is
sealed so that the volume is constant, and it is heated to a temperature of 400 K. Find the net force on each wall of the box. F = A P; P = Pinside Poutside P = 101/3 kPa = 33.7 kPa; F = 33.7 0.04 kN = 1.35 kN
Chapter 18
Temperature and the Kinetic Theory of Gases
69*
Water, H2O, can be converted into H2 and O2 gas by electrolysis. How many moles of these gases result from the n(H2) + 2n(O2); M(H2O) = 18 n(H2O) = 2000/18 = 111; n(H2) = 111, n(O2) = 55.5
electrolysis of 2 L of water? n(H2O) 70
A massless cylinder 40 cm long rests on a horizontal frictionless table. The cylinder is divided into two equal
sections by a membrane. One section contains nitrogen and the other contains oxygen. The pressure of the nitrogen is twice that of the oxygen. How far will the cylinder move if the membrane is removed? The Figure shows the cylinder before removal of the membrane. The approximate location of the center of mass (CM) is indicated. The temperatures are the same on both sides of the membrane. Consequently, n(N2) = 2n(O2). The mass of O2 = n(O2)M(O2) and the mass of N2 = 2n(O2)M(N2).
1. Locate the center of mass 2. After the membrane is removed, the CM is at the center of the cylinder 71
xCM = (2 = 17.3 cm
10
28 + 30
32)/ (2
28 + 32) cm
Cylinder moves (20 - 17.3) cm = 2.7 cm to the left
A cylinder contains a mixture of nitrogen gas (N2) and hydrogen gas (H2). At a temperature T1 the nitrogen is
completely dissociated but the hydrogen does not dissociate at all, and the pressure is P1. If the temperature is doubled to T2 = 2T1, the pressure is tripled due to complete dissociation of hydrogen. If the mass of hydrogen is mH, find the mass of nitrogen mN. 1. Write the ideal gas law for both cases 2. Find n(H2) in terms of n(N2) 3. Note that mN = 28n(N2) and mH = 2n(H2) 72 P1V = [2n(N2) + n(H2)]RT1; 3P1 = [2n(N2) + 2n(H2)]2RT1 n(H2) = 2n(N2) mN = 7mH
A vertical closed cylinder of cross-sectionaal area A is divided into two equal parts by a heavy insulating movable
piston of mass mp. The top part contains nitrogen at a temperature T1 and pressure P1, and the bottom part is filled with oxygen at a temperature 2T1. The cylinder is turned upside-down. To keep the piston in the middle, the oxygen must be cooled to T2 = T1/3, with the temperature of the nitrogen remaining at T1. Find the initial pressure of oxygen Pi.
Chapter 18
Temperature and the Kinetic Theory of Gases
The initial and final configurations of the cylinder and piston of mass mP are shown. We will set the volume of each half equal to 1 for convenience and let n be the number of moles of O2. To support the piston at the middle we must have Pi = P1 + mPg/A, where A is the area of the cylinder. Applying the ideal gas law, 2nRT1 = P1 + mPg/A. In the final position of the cylinder, the pressure exerted by the oxygen is nRT1/3. Since the pressure exerted by N2 is again P1 we have nRT1/3 + mPg/A = P1. We can now solve for Pi in terms of P1 and mPg/A and find Pi = (P1 + 13mPg/A)/6. 73* Three insulated vessels of equal volumes V are connected by thin tubes that can transfer gas but do not transfer
heat. Initially all vessels are filled with the same type of gas at a temperature T0 and pressure P0. Then the temperature in the first vessel is doubled and the temperature in the second vessel is tripled. The temperature in the third vessel remains unchanged. Find the final pressure P in the system in terms of the initial pressure P0. Initially, we have 3P0V = n0RT0. Later, the pressures in the three vessels, each of volume V, are still equal, but the number of moles are not. We can now write
R 2T 0 R 3T 0 R P' = P1= P2 = P3 = n1 = n2 = n3 T 0 . Also, V V V
3
ni = n0 =
i=1
3 P0 V RT 0
=
PV 1 1 P V 11 18 + + 1 = , and solving for P , we find P = P0 . 3 11 RT 0 2 RT 0 6
74
At the surface of the sun, the temperature is about 6000 K, and all the substances present are gaseous. From data
given by the light spectrum of the sun, it is known that most elements are present. (a) What is the average kinetic energy of translation of an atom at the surface of the sun? (b) What is the range of rms speeds at the surface of the sun if the atoms present range from hydrogen (M = 1 g/mol) to uranium (M = 238 g/mol)? (a) Kav = (3/2)kT (b) vrms = 75 Kav = 1.5 = 0.776 eV 1.22 1.38 10 vrms
23
6
103 J = 1.24 102 m/s
10
19
J
3R T/M ; M = 0.001 to 0.238 kg/mol
104 m/s
7.92
A constant-volume gas thermometer with a triple-point pressure P3 = 500 torr is used to measure the boiling point
of some substance. When the thermometer is placed in thermal contact with the boiling substance, its pressure is 734 torr. Some of the gas in the thermometer is then allowed to escape so that its triple-point pressure is 200 torr. When it is again placed in thermal contact with the boiling substance, its pressure is 293.4 torr. Again, some of the gas is removed from the thermometer so that its triple-point pressure is 100 torr. When the thermometer is placed in thermal contact with the boiling substance once again, its pressure is 146.65 torr. Find the ideal-gas temperature of the boiling
Chapter 18 substance.
Temperature and the Kinetic Theory of Gases
1. Find the temperature for each measurement 2. Fit the two pressure ratios to a linear equation Then extrapolate to zero gas pressure
T1 = 273.16(734/500) K = 401 K; T2 = 400.73 K; T3 = 400.59 K 1.468 = a + 500b; 1.4665 = a + 100b; a is ratio for P = 0 a = 1.46613; so T = 273.16 1.46613 = 400.49 K
76 A cylinder 2.4 m tall is filled with an ideal gas at standard temperature and pressure (Figure 18-17). The top of the cylinder is then closed with a tight-fitting piston whose mass is 1.4 kg and the piston is allowed to drop until it is in equilibrium. (a) Find the height of the piston, assuming that the temperature of the gas does not change as it is compressed. (b) Suppose that the piston is pushed down below its equilibrium position by a small amount and then released. Assuming that the temperature of the gas remains constant, find the frequency of vibration of the piston. (a) Let A be the area of the cylinder; find A P(inside) = Pin = 1 atm + mg/A = nRT/hA (b) At equilibrium, Fnet on m is zero Let y be displacement from equilibrium Write Fnet acting on m and spring constant k Evaluate f = ( 1/2 ) k/m 0 At STP, 0.1mol = 0.00224 m3 = 2.4A m3; A = 9.33 10 4 m2 105/h Pa; 105A y/h) if y << h nRT/hA = (2.4/h)(1 atm) = 2.424 h = 2.095 m For y = 0, Fnet = PinA - mg
2
1.01
2
Pin(y) = Pin(0)/(1 + y/h) = Pin(0)(1 Fnet(y) = y(nRT/h ); k = nRT/h
f = (1/ 2)
0.1 8.314 300 Hz = 1.014 Hz 2.0952 1.4

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