12 Thermal Properties of Matter · 416 CHAPTER 12 THERMAL PROPERTIES OF MATTER 12 Thermal Properties of Matter Answers to Discussion Questions •12.1 temperature —– The degree

416 CHAPTER 12 THERMAL PROPERTIES OF MATTER

12 Thermal Properties of Matter

Answers to Discussion Questions

•12.1temperature —– The degree of hotness or coldness of a body or environment. More scien-tifically, it is the measure of the average kinetic energy of the particles in a sample of matter,expressed in terms of units or degrees designated on a standard scale.

Fahrenheit scale —– A temperature scale that registers the freezing point of water as 32 andthe boiling point as 212 at one atmosphere of pressure.

Celsius scale —– A temperature scale that registers the freezing point of water as 0 and theboiling point as 100 under normal atmospheric pressure.

thermodynamic temperature —– Also known as the absolute temperature. It is measuredor calculated in kelvin, starting from Absolute Zero.

Absolute Zero —– The theoretical temperature at which substances possess no thermalenergy, equal to −273.15◦C, or −459.67◦F.

zero-point energy —– the minimum energy possessed by atoms and molecules at AbsoluteZero due to quantum effects.

temperature coefficient of linear expansion —– the amount of change in the linear sizeof a substance per unit length per unit temperature change.

temperature coefficient of volume expansion —– the amount of change in the volumeof a substance per unit length per unit temperature change.

Boyle’s Law —– For a certain amount of ideal gas at constant temperature, the product ofits pressure and its volume is a constant: PV = constant.

Charles’s Law —– For a certain amount of ideal gas at constant pressure, its volume isproportional to its absolute temperature: V/T = constant.

CHAPTER 12 THERMAL PROPERTIES OF MATTER 417

Gay-Lussac’s Law —– For a certain amount of ideal gas at constant volume, its pressure isproportional to its absolute temperature: P/T = constant.

Ideal Gas Law —– For a certain amount (n moles) of ideal gas, its pressure, volume, and itsabsolute temperature are related by PV = nRT .

Universal Gas Constant —– R = 8.314 4 J/mol·K.

Boltzmann’s Constant —– kB = 1.380 662 × 10−23 J/K.

critical point —– The point on the phase diagram where liquid and gas coexist at the samedensity and are indistinguishable.

vapor —– A gaseous state, especially when diffused in the atmosphere and at a temperaturebelow boiling point. Unlike gases, vapors can be condensed via compression.

equilibrium vapor pressure —– The pressure of the vapor at which the rates of condensa-tion and evaporation are equal, whereupon the vapor-liquid system reaches equilibrium.

triple point —– The point in the phase diagram at which all three phases (liquid, gas andsolid) coexist in equilibrium.

sublimation —– Raising temperature of a solid at a fixed pressure below that of the triplepoint will cause it to pass directly from solid to vapor. This process is known as sublimation.

Kinetic Theory —– The theory that depicts the thermal properties of gases as a result ofthe random motion of the gas molecules.

mean free path —– The mean distance a molecule travels in between two successive collisions.

Maxwell-Boltzmann distribution —– The classical probability distribution of states as afunction of energy for a system in thermal equilibrium at a given temperature.

•12.2Here are a few: (a) The bore hole in the glass tube must be uniform. (b) All of the mercury,including the stuff way up in the stem, must be at the same temperature. (c) Generally, to speedup its response the walls of the bulb are made thin and that thinness makes it vulnerable topressure variations, which change its volume (via barometric changes or hydrostatic pressure,if it’s immersed in a liquid). (d) There is a variation in pressure in the mercury due to thedifferent heights of the column. (e) There is a difference in internal pressure if it’s held verticallyas opposed to horizontally. (f) There are errors associated with the softness of the glass. If thethermometer is raised to a high temperature and then cooled rapidly, it might take weeks forthe glass to return to its original volume. Try measuring the freezing point of water before andimmediately after reading its boiling point — the difference can be as great as 1 C◦. (g) Themere presence of the thermometer in a small system may change the temperature of the system.(h) When measuring a changing temperature at any moment, the thermometer will always readwarmer if the bath temperature is falling and vice versa.

•12.3The glass will expand first, dropping the mercury level until it too becomes heated and expands.


•12.4We know that the antimony expands on solidifying, as does water. Since that would be a veryhelpful trait for a casting material to have (it would fill all the fine details in the mold), it’sreasonable to expect that’s the answer to this question.

•12.5The water inside the bomb freezes and expands and the bomb explodes. When I was anundergraduate I foolishly put one of these in a mix of dry ice and alcohol. The resulting violentexplosion blew the thing to bits. It’s common for water collecting in cracks in rocks to splitthem when it freezes.

•12.6Besides being more malleable when hot they contracted on cooling, pulling the plates together.

•12.7No. It wouldn’t work at all. The thermometer relies on a difference between the two β valuesto produce a net excess expansion which sends the fluid out of the storage bulb.

•12.8Being a poor conductor, the center of a boulder so treated would remain quite hot while theoutside was cooled and contracted rapidly. Pressure would build up, there would be considerableinternal stress, and the thing would rupture at any flaw or weak spot.

•12.9Chemists do a lot of heating with open flames and so the coefficient of expansion must be assmall as possible. The walls are kept thin to insure that they will heat up almost uniformlyand therefore not suffer appreciably different amounts of expansion and/or contraction and soshatter.

•12.10Bulbs are cheaper thinner and can tolerate the changes in temperature associated with ordinaryoperation, which are fairly gradual. They are made of inexpensive glass with a relatively largeβ, so a drop of water (or latex paint) can cause enough contraction and stress to shatter a hotlight bulb. Clearly, outdoor lamps have to be protected from rain and snow. The heating in aflashbulb is so rapid even the thin walls tend to burst and they therefore are usually enclosedin a tough plastic film to keep them from shattering.


•12.11The increased pressure at the points of contact melts the ice which, as the pressure is subse-quently removed, refreezes. This process called regelation, was named by the English physicistJohn Tyndall who used it to explain the making of snowballs.

•12.12At a temperature of even −1◦C it takes about 140 atm of pressure (≈ 2000 lb/in2) to melt ice,so the problem was probably that the snow was just too cold.

•12.13If the curves intersect as in Figure Q13 the region PQR is below the fusion curve and so mustrepresent a solid; it’s above the vaporization curve and so must represent a liquid; it’s below thesublimation curve and so must represent a vapor. These conclusions are contradictory hencethe configuration cannot exist.

•12.14The liquid will expand rapidly: its density decreasing as the density of the vapor increases. Thesurface meniscus will flatten out and then disappear altogether when the density of liquid andvapor are equal at a pressure of 7.38 MPa.

•12.15At the triple point (273.16 K, 0.61 kPa) water can boil and freeze simultaneously. Place somewater in a dish positioned above a bowl of strong sulfuric acid and seal the whole thing in avacuum chamber. If you have a good high capacity pump the acid isn’t necessary. On evacuatingthe chamber the water will evaporate rapidly and the vapor will be absorbed by the acid. At alow enough pressure the water will begin to boil and becoming cooled by evaporation will soonfreeze.

•12.16Yes. They were called permanent because they could not at first be liquified, which suggests aweak intermolecular cohesive force and that, in turn, suggests ideal behavior.

•12.17Equation (12.15), KEav = 3

2kBT , describing the average translational KE of each molecule iswhat Jeans is talking about but he overlooks the quantum mechanical zero-point energy — at0 K the energy of motion is not nil.

•12.18PV = nRT , so both pressures must be equal since everything else is the same. The speeds ofthe hydrogen molecules must be greater than those of the nitrogen because the average KE is


the same from Eq. (12.15). The pressures can be equal because the lighter hydrogens hit thewalls at greater speeds and they do it more frequently because they traverse the chamber morequickly. The pressure is proportional to the average KE via Eq. (12.14).

•12.19It can be expected that they have discovered the same absolute zero we know and would beequally familiar with low-temperature physics. But it’s quite unlikely that they too happened touse the freezing and boiling points of water at the surface pressure of planet Earth as referencelevels. It’s even more unlikely that they might use human body temperature as a reference asdid Fahrenheit. So they’ll probably not have thermometers graduated in Fahrenheit or Celsiusdegrees or even in kelvins, since the latter matches up with the Celsius scale. Still, it wouldn’ttake much effort to understand their temperature system and they will surely be as concernedwith the energy concentration of their environment as we are.

•12.20Remember that the pressure in the room is more or less constant. Increasing T increases theaverage KE, which increases the net KE and the P but leads to an over-pressure and an outwardcurrent of air. The room leaks warm air to the outside. The temperature goes up because it’sdependent on the average KE of each molecule, which is higher. The pressure remains the samebecause it depends on both the number of molecules per unit volume and their average KE.Compare Eqs. (12.14) and (12.15).

•12.21According to Figure (12.18), the phase diagram for water, the answers are (a) vapor (b)liquid (c) vapor (d) vapor (e) vapor.

Answers to Multiple Choice Questions

1. b 2. d 3. a 4. c 5. c 6. c 7. c8. c 9. c 10. b 11. b 12. d 13. a 14. a

15. b 16. d 17. d 18. d 19. a 20. e 21. a22. b 23. b 24. c 25. e 26. a 27. a


Solutions to Problems

12.1Use Eq. (12.1), with TF = 98.6 ◦F:

TC =59(TF − 32◦C) =

59(98.6 − 32)◦C = 37.0◦C .

(Note that we needn’t be concerned with the significant number in the figure 32◦C above — itis defined to be exact.)

12.2Use Eq. (12.1), with TF = 70.0 ◦F:

TC =59(TF − 32◦) =

59(70.0 − 32)◦C = 21.1◦C .

12.3Fig. P3 is a plot of the Celsius temperature TC versus the Fahrenheit temperature TF , whichare related by Eq. (12.1): TC = 5

9 (TF − 32◦). Since TC is a linear function of TF the plot is astraight line.In Problem (12.1) TF = 98.6 ◦F and in Problem (12.2) TF = 70.0 ◦F. You can verify that thecorresponding temperature readings in Celsius should be 37.0◦ and 21.1◦C, respectively, byexamining the plot.

12.4The slope of a straight line in a plot of y vs x is k, if y = kx+ b. In our case the equation readsTC = 5

9TF − 17.8◦, so in a plot of TC vs TF the slope is 59 , as you should verify by applying

slope = rise/run directly over the plot.

12.5Read the value of TC1 directly from the plot. It is about −18◦C. For more precision, notethat the slope of the line is 5/9, so |TF1/TC1 | = 5/9 in the small triangle formed by the linewith the two axes. This gives |TC1 | = 5

9 |TF1 | = 59 (32.0◦) = 17.8◦; and, since TC1 < 0,

TC1 = −|TC1 | = −17.8◦C. Alternatively, you can also plug TF1 = 0 ◦F into Eq. (12.1) to obtainTC1 :

TC1 =59(TF1 − 32◦) =

59(0 − 32.0)◦C = −17.8◦C .


12.6The coordinates of the leftmost point is (−40◦C,−40 ◦F). You can verify from Eq. (12.1) that−40◦C = −40 ◦F. In fact this is the only temperature at which the two scales (TF and TC) giveidentical readings.

12.7From Eq. (12.2), with T = 2.726 K,

TC = T − 273.15 = 2.726 − 273.15 = −270.42◦C .

12.8Use Eq. (12.1), with TC = 39.0◦C:

TF =95TC + 32◦ =

95× 39.0◦ + 32◦ = 102.2 ◦F .

12.9From Eq. (12.2), with T = 58 K,

TC = T − 273.15 = 58 − 273.15 = −215◦C .

12.10Since T = TC + 273.15, ∆T = ∆(TC + 273.15) = ∆TC . This means that if a certain changein temperature measures n units in kelvin it should also measure n degrees in Celsius. so theratio in question is 1.

12.11First, convert the temperature from kelvin to Celsius degrees: TC = T−273 = 5.6×103−273 =5327◦. Now convert it into Fahrenheit:

TF =95TC + 32◦ =

95× 5327◦ + 32◦ = 9.6 × 103 ◦F .

12.12Use Eq. (12.2). For TC = −268.9◦C, T = TC + 273.15 = −268.9 + 273.15 = 4.25 K; and forTC = −272◦C, T = −272 + 273.15 = 1 K.


12.13First convert the temperatures to Celsius scale. For TF = 0 ◦F, TC = 5

9 (TF − 32◦C) =59 (0 − 32)◦C = −17.8◦C; and for TF = 100 ◦F, TC = 5

9 (TF − 32) = 59 (100 − 32)◦C = 37.8◦C.

Thus the temperature change, in Celsius, is ∆TC = 37.8◦C − (−17.8◦C) = 55.6 C◦. Thecorresponding temperature change in Kelvins is the same as that in Celsius: ∆T = 55.6 K.

12.14The melting temperature in Celsius is TC = T − 273.15 = 14.01 − 273.15 = −259.14◦C.

12.15According tot he result of Problem (12.10) ∆T (in kelvin units) = ∆TC(in Celsius degrees).Since ∆TC = 42 C◦, ∆T = 42 K.

12.16Since TC = T − 273.15, doubling T to 2T would result in a new reading in Celsius degree of

T ′C

= 2T − 273.15 = T + (T − 273.15) = T + TC ,

which is not the same as 2TC . Similarly, from TF = 95TC + 32, if TC is doubled to 2TC the

resulting new reading in Fahrenheit degree would be

T ′F

=95(2TC) + 32 = TC +

(95TC + 32

)= TC + TF ,

which is generally not the same as 2TF .

12.17Use Eq. (12.2) to convert T to TC : TC = T − 273.15 = 10× 106 − 273.15 ≈ 10× 106 ◦C. Nowuse Eq. (12.1) to convert T to TF :

TF = 32◦ +95

TC ≈ 32◦ +95

(10 × 106

)◦ ≈ 1.8 × 107 ◦F .

12.18The temperature in Celsius is TC = 5

9 (TF − 32◦C) = 59 (4000 − 32)◦C = 2204◦C, which in

Kelvins is T = TC + 273.15 = 2477 K.

12.19The temperature in Celsius is TC = T − 273.15 = 770 − 273.15 = 497◦C, which is equivalentto

TF = 32◦ +95

TC = 32◦ +95

(497)◦ = 927 ◦F .


12.20From Table (12.1), the ratio is 1893 K/600 K = 3.16 when the temperatures are measured inKelvins; and 1620◦C/327◦C = 4.95 when they are measured in Celsius.

12.21Since the TC vs TF curve is a straight line, the equation it represents can be written as TC =k(TF − C), where k is the slope of the line and C is a constant. To determine C, note thatTC = 0 when TF = C. Thus C = 32 degrees. Next, use the two points (TF2 , TC2) = (212, 100)and (32, 0) to determine the slope k:

k =TC2 − 0TF2 − 32

=100 − 0212 − 32

=100

212 − 32.

Thus

TC = k(TF − C) = (TF − 32)(

100212 − 32

).

If you work out the algebra, you’ll find that this is the same as Eq. (12.1): 100/(212 − 32) =100/180 = 5/9; so

TC = (TF − 32)(

100212 − 32

)=

59(TF − 32) .

12.22With the previous problem in mind, write TC = kTF + b and determine k and b. For b, letTF = TF1 = 0 to obtain TC = TC1 = kTF1 + b = k× 0 + b = b, or b = TC1 . The slope k can beobtained from the two points, (TF2 , TC2) and (TF1 , TC1): k = (TC2 − TC1)/(TF2 − TF1). Thus

TC = kTF + b =(

TC2 − TC1

TF2

)TF + TC1 ,

where in the last step we noted TF1 = 0. Now plug in TC1 = −17.8◦C [see Problem (12.5)],TC2 = 100◦C, TF2 = 212 ◦F, and TF = 68 ◦F to obtain

TC =[100 − (−17.8)

212

](68) − 17.8 = 20◦C .

12.23Use Eq. (12.3). For Pyrex glass α = 3 × 10−6 K−1, and the temperature change is ∆T =80.0 C◦ = 80.0 K; so the length of the rod will increase by

∆L = αL0 ∆T = (3 × 10−6 K−1)(1.00 m)(80.0 K) = 2 × 10−4 m = 0.2 mm .


(A note in notation: in many books, ◦C is used for the actual temperature in degrees Celsiuswhereas C◦ denotes a change in temperature, measured in Celsius degrees. A similar distinctionexists between ◦F and F◦. For example, although 9 ◦F �= 5◦C, we do have 9 F◦ = 5 C◦, asa change of 9 Fahrenheit degrees is equivalent to a change of 5 Celsius degrees. We will befollowing this convention in later problems.)

12.24Start from Eq.(12.3), ∆L = αL0 ∆T , with ∆L = 2.00 mm = 2.00×10−3 m, α = 12×10−6/K,and L = 10.0 m. Solve for ∆T :

∆T =∆L

αL0

=2.00 × 10−3 m

(12 × 10−6 K−1)(10.0 m)= 17 K = 17 C◦ .

12.25Use Eq. (12.3). For aluminum α = 25 × 10−6 K−1, and the temperature change is ∆T =1.00 C◦ = 1.00 K; so the length of the rod will increase by

∆L = αL0 ∆T = (25 × 10−6 K−1)(1.00 m)(1.00 K) = 2.5 × 10−5 m = 25µm .

12.26Again, use Eq. (12.3). For brass α = 18.9 × 10−6 K−1, and the temperature change is ∆T =−1.00 C◦ = −1.00 K, so the length of the rod will change by

∆L = αL0 ∆T = (18.9 × 10−6 K−1)(1.00 m)(−1.00 K) = −1.89 × 10−5 m = −18.9 µm ,

where the minus sign indicates that the length of the rod has decreased.

12.27Apply Eq. (12.3), ∆L = αL0 ∆T , which gives the change in length of an object of length L0

as a result of ∆T , a change in its temperature. Here α is the coefficient of linear expansion ofthe material of which the object is made. In this case we are dealing with an aluminum bar,for which α = 25 × 10−6 K−1, ∆T = 50◦C − 30◦C = 20 C◦ = 20 K, and L0 = 10 m. So thelength of the bar will increase by

∆L = αL0 ∆T = (25 × 10−6 K−1)(10 m)(20 K) = 5.0 × 10−3 m = 5.0 mm .

12.28Similar to the previous problem, apply ∆L = αL0 ∆T . Here α = 12 × 10−6 K−1 for steel,∆T = 35◦C − 5◦C = 30 C◦ = 30 K, and L0 = 10 m. So the length of the rod will increase by

∆L = αL0 ∆T = (12 × 10−6 K−1)(10 m)(30 K) = 3.6 × 10−3 m = 3.6 mm .


12.29Apply ∆L = αL0 ∆T . Here α = 12 × 10−6 K−1 for steel, the temperature change is ∆T =35◦C − 0◦C = 35 C◦ = 35 K, and L0 = 1280 m. So the length of the roadway will increase by

∆L = αL0 ∆T = (12 × 10−6 K−1)(1280 m)(35 K) = 0.54 m .

12.30We are looking for the percentage change in length for the rod, i.e., ∆L/L0 . Convert ∆Tto kelvin: ∆T = 168 ◦F − 68 ◦F = 100 F◦ = 5

9 (100) C◦ = 55.56 C◦ = 55.56 K. Thus fromEq. (12.3)

∆L

L0

= α∆T = (17.3 × 10−6 K−1)(55.56 K) = 9.61 × 10−4 = 0.096 1% .

12.31

(a) The relationship in question is given by Eq. (12.3), ∆L = αL0∆T .

(b) Each slab expands by 12∆L on either side, so the spacing between two adjacent slabs are

shortened by 12∆L + 1

2∆L = ∆L. The minimum initial spacing dmin must then be equal to∆L, whereupon the gap between the slabs would completely close as temperature rises to thehighest value.

(c) ∆T = Tf − Ti = 46◦C − 10◦C = 36 C◦ = 36 K.

(d) Take α = 12 × 10−6 K−1 to obtain

dmin = ∆L = αL0∆T = (12 × 10−6 K−1)(2.4 m)(36 C◦) = 1.0 × 103 m = 1.0 mm .

12.32Use Eq.(12.3) to find the new length and width of the sheet. If the original length and width aredenoted as a0 and b0 , respectively, then their new values will be a = a0 + ∆a = a0 + a0α ∆Tand b = b0 + ∆b = b0 + b0α ∆T , respectively. Plug in a0 = 50.00 cm, b0 = 20.00 cm,α = 16.8× 10−6 K−1, and ∆T = 60.00◦C− 30.00◦C = 30.00 C◦ = 30.00 K to obtain the newarea as a result of the temperature change:

A = ab = (a0 + a0α ∆T )(b0 + b0α ∆T ) = a0b0(1 + α ∆T )2

= (50.00 cm)(20.00 cm)[1 + (16.8 × 10−6 K−1)(30.00 K)

]2= 1001 cm2 .


12.33According to Problem (12.28) the change in area for the sheet, whose original area is A0 = a0b0 ,is

∆A = 2αa0b0 ∆T = 2(16.8 × 10−6 K−1)(50.00 cm)(20.00 cm)(30.00 K) = 1.008 cm2 ,

which gives the new area to be A = a0b0 + ∆A = (50.00 cm)(20.00 cm) + 1.008 cm2 =1001 cm2, in good agreement with the result of Problem (12.32).

12.34Use Eq. (12.3), ∆L = αL0 ∆T , to find ∆L, the change in height for the steel tower. Here α =12×10−6 K−1 for steel, L0 = 150.00 m, and ∆T = ∆TC = 5

9 ∆TF = 59 [95− (−10)] = 58.3 K.

Thus the new height is

L = L0 + ∆L = L0(1 + α ∆T ) = (150.00 m)[1 + (12 × 10−6 K−1)(58.3 K)

]= 150.11 m .

12.35

(a) For the pipe of diameter d0 (= 2.00 cm) the area is

A =14πd2

0=

14π(2.00 cm)2 = 3.14 cm2 .

(b) The new diameter d is found from Eq. (12.3) (with L replaced by d and L0 by d0):

d = d0(1 + α ∆T ) = (2.00 cm)[1 + (16.6 × 10−6 K−1)(88.0 − 20.0)K

]= 2.002 27 cm .

(c) The new area is A = 14πd2, so the change in area is

∆A = A − A0 =14πd2 − 1

4πd2

0

=14πd2

0

[(1 + α∆T )2 − 1

]= A0

[2α∆T + (α∆T )2

]≈ 2αA0∆T

= 2(16.6 × 10−6 K−1)(88.0 − 20.0)K = 7.1 × 10−3 cm2 ,

where we neglected the term (α∆T )2 since it is much less than α∆T .

12.36

(a) Ice and water: T1 = 0.0◦C, boiling water: T2 = 100.0◦C.

(b) The assumption is that the thermal expansion of the rod is linear, i.e., the length of therod increases in proportion to the temperature change: ∆L ∝ ∆T .


(c) The slope is ∆L/∆T , which by definition equals αL, with α the coefficient of thermalexpansion and L the length of the rod.

(d) Find the slope of the line using rise-over-run:

∆L

∆T=

4.007 6 cm − 4.000 0 cm

100.0◦C − 0.00◦C= 7.6 × 10−5 cm/C◦ .

Thus α = (∆L/∆T )/L = (7.6 × 10−5 cm/C◦)/4.00 cm = 19 × 10−6/C◦.

(e) Checking Table 12.2 for α, we find the closest match to be brass.

(f) As the temperature changes from 0.0◦C to 60.0◦C we have ∆T = 60.0◦C−0.0◦C = 60.0 C◦,and so

∆L = αL∆T = (19 × 10−6/C◦)(4.000 0 cm)(60.0 C◦) = 0.004 6 cm ,

and so its length at 60.0◦C is 4.000 0 cm + 0.004 6 cm = 4.004 6 cm.

12.37

(a) We may heat up the brass tube to increase its diameter.

(b) ∆d = αd0∆T , where d0 (= 3.995 cm) is the initial diameter of the tube.

(c) The diameter of the rod is greater than that of the tube by 4.000 cm−3.995 cm = 0.005 cm,so we need to increase d0 by ∆d = 0.005 cm. The corresponding change in temperature of thetube is then

∆T =∆d

αd0

=0.005 cm

(18.9 × 10−6 K−1)(3.995 cm)= 66.2 K = 66.2 C◦ .

(d) The new temperature of the tube should be Ti + ∆T = 20◦C + 66.2 C◦ = 86◦C.

12.38

(a) The density is the mass per unit volume: ρ = m/V .

(b) The volume V increases as temperature rises and decreases at T falls, due to thermalexpansion.

(c) ρ0V0 is the initial amount of mass of the sample before its temperature changes, whileρV is its mass afterwards. Since the mass m of the sample is independent of the temperature,ρ0V0 = ρV = m.

(d) If the initial volume is V0 before the temperature change, then afterwards V = V0(1+β∆T ).Thus

ρ =m

V=

ρ0V0

V0(1 + β∆T )=

ρ0

1 + β∆T≈ ρ0(1 − β∆T ) ,

where in the last step we made use of the approximation (1+x)−1 ≈ 1−x for |x| = β∆T � 1.


(e) In this case ∆T = 120◦C − 20◦C = 100 C◦ = 100 K and β = 72.0 × 10−6 K−1, so

ρ ≈ ρ0(1 − β∆T )

= (2.70 × 103 kg/m3)[1 − (72.0 × 10−6 K−1)(100 K)

]= 2.68 × 103 kg/m3 .

12.39Similar to the previous problem, we use Eq. (12.3), ∆L = αL0 ∆T , to find ∆L, the change inlength for the steel tape. Here α = 12×10−6 K−1 for steel, L0 = 10.000 m, and ∆T = Tf −Ti ,where Ti = 20◦C and Tf = 5

9 (96.8 − 32.0) = 36.0◦C. Thus the new length of the tape is

L = L0(1 + α ∆T ) = (10.000 m)[1 + (12.15 × 10−6 K−1)(36.0 − 20)K

]= 10.002 m ,

which is longer than the correct length by 0.002 m, or 2 mm. This is not likely to trouble thecarpenter. (Note that we used 1 K = 1 C◦ in finding ∆T .)

12.40For simplicity, imagine that the sheet of area A0 assumes the shape of a rectangle of sidelengths a0 and b0 . Then A0 = a0b0 . As a result of a temperature change ∆T the sidelengths increase by ∆a = αa0 ∆T and ∆b = αb0 ∆T , respectively; and so the new area ofthe sheet is A = (a0 + ∆a)(b0 + ∆b) = (a0 + αa0 ∆T )(b0 + αb0 ∆T ) = a0b0(1 + α ∆T )2 =A0 [1 + 2α ∆T + α2( ∆T )2] ≈ A0(1 + 2α ∆T ), where in the last step we neglected the termproportional to (α ∆T )2 since α ∆T � 1. The change in area for the sheet is then

∆A = A − A0 ≈ A0(1 + 2α ∆T ) − A0 = 2αA0 ∆T .

12.41Apply Eq. (12.4) for volume expansion: ∆V = βV0 ∆T . Here β = 182 × 10−6 K−1 is thecoefficient of volume expansion for mercury, V0 = 0.50 cm2 is its initial volume at 32 ◦F, and∆T = ∆TC = 5

9 ∆TF = 59 (TFf − TFi) = 5

9 (212 − 32) = 100 K. (Note that 1 K = 95 F◦ for

temperature change measured in K and in F◦.) Thus the new volume at 212 ◦F is given by

V = V0 + ∆V = V0(1 + β ∆T )

= (0.50 cm3)[1 + (182 × 10−6 K−1)(100 K)

]= 0.51 cm3 .

12.42Suppose that the initial volume of the mercury in the Pyrex glass is V0 . As a result of atemperature change ∆T the mercury (M) expands by ∆VM = βMV0 ∆T , causing the mercury


level to rise. Here βM is the coefficient of volume expansion for mercury. Meanwhile the Pyrexglass (P) holding the mercury also expands by ∆VP = βPV0 ∆T , causing the mercury level tofall. Here βP is the coefficient of volume expansion for Pyrex glass. Since these two expansionshave opposite effects, the apparent change in volume for the mercury is

∆V = ∆VM − ∆VP = (βM − βP)V0 ∆T = βeffV0 ∆T ,

whereβeff = βM − βP = 182 × 10−6 K−1 − 9 × 10−6 K−1 = 173 × 10−6 K−1

is the apparent coefficient of volume expansion.

12.43If the inside diameter of the beaker is D and the height of the mercury column is H , then theoriginal volume of the mercury inside the beaker is V0 = 1

4πD2H . The apparent change involume for the mercury as a result of the temperature change ∆T is ∆V = (βM − βP)V0 ∆T ,following the result of the previous problem. Thus ∆V = 1

4πD2 ∆H = (βM − βP)V0 ∆T , andso the change in height for the mercury column is

∆H =(βM − βP)V0 ∆T

πD2/4

=(182 × 10−6 K−1 − 26 × 10−6 K−1)(800.0 cm3)(0 − 95)K

π(10.00 cm)2/4= −0.15 cm = −1.5 mm .

12.44Let the original volume of the mercury inside the thermometer be V0 . Then when the tem-perature increases by ∆T the change in volume for the mercury is ∆V = βMV0 ∆T , whereβM is the coefficient of volume expansion for mercury. Similar to the previous problem, if thebore diameter of hole in the stem containing the mercury is D, then ∆V = 1

4πD2 ∆H , where∆H is the change in height for the mercury column. Equate the two expressions for ∆V :βMV0 ∆T = 1

4πD2 ∆H , which gives

∆H =βMV0 ∆T

πD2/4=

(182 × 10−6 K−1)(0.400 cm3)(90 − 10)Kπ(0.010 cm)2/4

= 74 cm .

12.45Let the initial volume of the water (W) in the bowl be V0 . As a result of a temperature change∆T the water expands by ∆VW = βWV0 ∆T , where βW is the coefficient of volume expansionfor water. Meanwhile the Pyrex bowl (P) also expands by ∆VP = βPV0 ∆T , where βP is thecoefficient of volume expansion for Pyrex. The new volume of water is therefore V0 + ∆VW ,


which is greater than that of the bowl by ∆V = (V0 + ∆VW)− (V0 + ∆VP) = ∆VW − ∆VP =(βW − βP)V0 ∆T , which in turn is equal to the amount of water that will overflow:

∆V = (βW − βP)V0 ∆T

= (207 × 10−6 K−1 − 9 × 10−6 K−1)(100 cm3)(50 − 10)K

= 0.79 cm3 .

12.46The diameter D of the hole expands in accordance with Eq. (12.3): ∆D = αD0 ∆T . HereD0 = 2.000 cm is the original diameter, α = 20 × 10−6 K−1, and ∆T = ∆TC = 100 − 20 =80 K. Thus

∆D = αD0 ∆T = (20 × 10−6 K−1)(2.000 cm)(80 K) = 3.2 × 10−3 cm ,

and the new diameter is D = D0 + ∆D = 2.000 cm + 0.003 2 cm = 2.003 cm.

12.47The initial density of the iron block of mass m and volume Vi is ρi = m/Vi . When thetemperature of the iron block increases by ∆T its final volume is Vf = Vi + βVi ∆T , so its newdensity is

ρf =m

Vf

=m

Vi(1 + β ∆T )=

ρi

1 + β ∆T

=7.85 g/cm3

1 + (36 × 10−6 K−1)(0 − 20)K= 7.86 g/cm3 = 7.86 × 103 kg/m3 .

12.48At 273 K (0◦C) the density of water (W) is ρW = 999.8 kg/m3, while that of ice (I) is ρI =917 kg/m3. Since the mass m is unchanged as water freezes into ice, m = ρWVW = ρIVI , whichgives the volume of the resulting ice to be

VI =ρWVW

ρI

=(999.8 kg/m3)(1.00 m3)

917 kg/m3= 1.09 m3 .

12.49The period τ0 of the pendulum of length L0 at 20.00◦C is given by

τ0 = 2π

√L0

g= 2π

√1.000 m

9.81 m/s2= 2.006 s .


When the temperature drops by ∆T = 0.000◦C − 20.00◦C = −20.00 C◦ = −20.00 K, thelength of the pendulum changes to L = L0 + ∆L = L0 + αL0 ∆T (where α is the coefficientof linear expansion for aluminum); and so the new period is

τ = 2π

√L

g= 2π

√L0(1 + α ∆T )

g

= 2π

√1.000 m [1 + (25 × 10−6 K−1)(−20 K)]

9.81 m/s2

= 2.006 s < τ0 ,

meaning that the clock runs slightly faster.

12.50Similar to the previous problem, at 20.000◦C the period τ0 of the pendulum of length L0 is

given by τ0 = 2π√

L0/g; and after a temperature change ∆T the new period is

τ = 2π

√L

g= 2π

√L0(1 + α ∆T )

g=

(2π

√L0

g

)√

1 + α ∆T = τ0

√1 + α ∆T

= (1.000 s)√

1 + (12.113 5 × 10−6 K−1)(40.000 − 20.000)K= 1.000 1 s > τ0 ,

meaning that the clock will run slightly slower.

12.51The initial density ρ0 of the specimen of mass m and volume V0 is given by ρ0 = m/V0 . Aftera temperature change ∆T , V0 becomes V = V0 + ∆V = V0 + βV0 ∆T , and the new density is

ρ =m

V=

m

V0(1 + β ∆T )=

(m

V0

)1

1 + β ∆T=

ρ0

1 + β ∆T.

Since β is small, |β ∆T | � 1; so we may apply the Binomial Theorem, which states that(1+x)n ≈ 1+nx for |x| � 1. With x = β ∆T and n = −1, we have (1+β ∆T )−1 ≈ 1−β ∆T .So finally

ρ = ρ0(1 + β ∆T )−1 ≈ ρ0(1 − β ∆T ) .

12.52Consider an object with initial length L0 undergoing a length change ∆L as a result of atemperature change ∆T : ∆L = αL0 ∆T . The Young’s modulus Y of the object is defined asY = stress/strain, so the Thermal stress in question is given by

Thermal stress = Y × strain = Y

(∆L

L0

)= Y

(αL0 ∆T

L0

)= Y α ∆T .


12.53Use the result of the previous problem to find the stress in the concrete slabs due to thermalexpansion:

stress = Y α ∆T = (25 × 109 Pa)(10 × 10−6 K−1)(42 − 5)K = 9.3 MPa .

Concrete is strong in compression, with an ultimate compressional strength of 30 - 40 MPa (seeTable 10.1), so it is more likely to buckle than to break.

12.54

(a) The relationship between P and V of an ideal gas in an isothermal process (during whichT = constant) is Boyle’s Law: PV = constant.

(b) Since the product of P and V is unchanged, P increases as V decreases.

(c) From P1V1 = P2V2 and P2 = 3P1 we get

V2 =P1V1

P2

=P1V1

3P1

=13V1 .

12.55The relationship between T and V of an ideal gas in an isobaric process (during which P =constant) is Charles’s Law: V/T = constant.

(b) Faster.

(c) As the temperature increases while the pressure is kept as a constant, the volume of thegas must increase. So the piston moves out.

(d) From V1/T1 = V2/T2 and T2 = 2T1 we get

V2 =V1T2

T1

=V1(2T1)

T1

= 2V1 .

12.56The relationship between T and P of an ideal gas in an constant-volume process (during whichV = constant) is Gay-Lussac’s Law: P/T = constant.

(b) Slower.

(c) The piston will not move, since the volume of the cylinder is held fixed.

(d) From P1/T1 = P2/T2 and T2 = 12T1 we get

P2 =P1T2

T1

=P1(T1/2)

T1

=P1

2.


12.57The law relating P , T and V of an ideal gas is the Ideal Gas Law: PV/T = constant.

(b) Slower, since the temperature decreases.

(c) The volume of the gas can be solved from the Ideal Gas Law as V = constant × T/P .When both T and P are halved the ratio T/P remains the same, so the volume is unchangedand the piston will not move.

(d) From the analysis in part (c) above we know that V2 = V1 .

12.58Since the container is sealed ∆V = 0, so Eq. (12.7) applies: P/T = constant. Initially,Pi = 1.00 atm = 0.101 MPa and Ti = 273 K, and finally Tf = 730 K. The final pressure Pf canthen be obtained from Pi/Ti = Pf/Tf , as

Pf =PiTf

Ti

=(0.101 MPa)(730 K)

273 K= 0.271 MPa .

12.59According to the hint given in the problem statement we may use Eq. (12.5), PV = constant.Initially, P = Pi = 0.101 MPa and V = Vi , and finally Vf = 1

10Vi . Thus from PiVi = PfVf

Pf =PiVi

Vf

=(0.101 MPa)Vi

110

Vi

= 1.0 MPa .

12.60Apply Boyle’s Law, Eq. (12.5), for constant temperature: PV = constant. Let the absolutepressure at the bottom of the lake be P and the volume of the bubble down there be V . Asthe bubble rises to the surface of the lake its pressure becomes Ps (= 1.0 atm) and its volumebecomes 3V . Thus PV = Ps(3V ), or

P =Ps(3V )

V= 3Ps = 3(1.0 atm) = 3.0 atm .

12.61The initial absolute pressure of the gas is Pi = PA + 20.0 atm = 21.0 atm, and the initialvolume is Vi = 1.00 m3. After the transfer the absolute pressure becomes Pf , and the volume


becomes Vf = Vi +9.00 m3 = 10.00 m3 (since the gas is uniformly present in both the tank andthe chamber). For T =constant we have PiVi = PfVf , so

Pf =PiVi

Vf

=(21.0 atm)(1.00 m3)

10.00 m3= 2.10 atm .

The final reading of the gauge pressure is then PG = 2.10 atm − 1.0 atm = 1.1 atm.

12.62Since the pressure is a constant, Eq. (12.6) applies: Vi/Ti = Vf/Tf = constant. With Vi =500 × 10−6 m3, Ti = 273 K, and Tf = 300 K, the final volume is

Vf =ViTf

Ti

=(500 × 10−6 m3)(300 K)

273 K= 549 × 10−6 m3 .

12.63Similar to the previous problem, the pressure of the gas is kept constant so Vi/Ti = Vf/Tf =constant, where Vi = 300 cm3, Ti = 273.15 + 25.0 = 298.15 K, and Vf = 200 cm3. Thus thefinal temperature is

Tf =TiVf

Vi

=(298.15 K)(200 cm3)

300 cm3= 199 K .

12.64Apply Eq. (12.8) to two states of the gas, labeled i and f, respectively: PiVi/Ti = PfVf/Tf =constant, where Pi = 81.3 kPa, Vi = 5.50 liters = 5.50×10−3 m3, Ti = 273.15+28 = 301.15 K,Pf = 1.00 atm = 101 kPa, and Tf = 273.15 K. Thus the final volume of the gas is

Vf = Vi

(Pi

Pf

)(Tf

Ti

)= (5.50 × 10−3 m3)

(81.3 kPa

101 kPa

)((273.15 K

301.15 K

)= 4.0 × 10−3 m3 ,

or 4.0 liters.

12.65Apply Eq.(12.8) to the initial state of the helium gas (labeled with subscript i) and the final stateat STP (labeled with f): PiVi/Ti = PfVf/Tf = constant. Here Pi = 99 kPa, Vi = 1200 cm3,Ti = 273.15+15 = 288.15 K, Pf = 1.00 atm = 101 kPa, and Tf = 273.15 K. Solve for the finalvolume of the gas:

Vf = Vi

(Pi

Pf

)(Tf

Ti

)= (1200 cm3)

(99 kPa

101 kPa

)((273.15 K

288.15 K

)= 1.1 × 103 cm3 .


12.66One mole of oxygen gas occupies a volume of V = 22.4 liters = 22.4 × 10−3 m3 at STP. Sincethe molecular mass of an O2 molecule is 2 × 16.0 u = 32.0 u, the mass of one mole of oxygengas is m = 32.0 g = 32.0 × 10−3 kg. Thus its density is

ρ =m

V=

32.0 × 10−3 kg

22.4 × 10−3 m3= 1.43 kg/m3 .

12.67A water molecule, H2O, is made of two hydrogen atoms and one oxygen atom. Thus its mass is

m = [2(1.008 u) + 15.999 u] (1.660 6 × 10−27 kg/u) = 2.992 × 10−26 kg .

12.68The volume occupied by one mole of ammonia gas at STP is V = 22.4 liters = 22.4× 10−3 m3.Since the mass of this much ammonia gas is m = 17.03 g = 17.03 × 10−3 kg, its density is

ρ =m

V=

17.03 × 10−3 kg

22.4 × 10−3 m3= 0.760 kg/m3 .

12.69Each oxygen molecule, O2 , is made of two oxygen atoms, each of mass 15.999 u [see Prob-lem (12.67)]. Thus its molecular mass is

m = [2(15.999 u/molecule)] (1.660 6 × 10−27 kg/u) = 5.314 × 10−26 kg/molecule ,

and the number of oxygen molecules with a total mass of M = 16.0 kg is then

N =M

m=

16.0 kg

5.314 × 10−26 kg/molecule= 3.01 × 1026 molecules ,

which corresponds to

n =N

NA

=3.011 × 1026 molecules

6.022 × 1023 molecules/mol= 500 mol .

12.70The density of oxygen gas at STP is found in Problem (12.66) to be ρ = 1.43 kg/m3. Thevolume V occupied by a total of m = 16.0 kg of oxygen gas at STP is therefore

V =m

ρ=

16.0 kg

1.43 kg/m3= 11.2 m3 .


12.71The pressure of the balloon is always equal to the atmospheric pressure, which is assumed tobe a constant. Thus Eq. (12.6) applies: Vi/Ti = Vf/Tf , where in our case Vi = 1.000 m3,Ti = 0 + 273.15 = 273.15 K, and Tf = 20 + 273.15 = 293.15 K. Thus the final volume of theballoon at 20◦C is

Vf =ViTf

Ti

=(1.000 m3)(293.15 K)

273.15 K= 1.1 m3 .

12.72Apply Eq. (12.8) to the initial and final states of the hydrogen gas: PiVi/Ti = PfVf/Tf =constant. Here Pi = 2.00 atm, Vi = 30.0 liters, Ti = 273 K, Pf = 3.00 atm, and Vf = 15.0 liters.Solve for the final temperature of the gas:

Tf = Ti

(Pf

Pi

)(Vf

Vi

)= (273 K)

(3.00 atm

2.00 atm

)(15.0 liters

30.0 liters

)= 205 K .

12.73Apply Eq. (12.8) to states 1 and 2 of a given amount of ideal gas: P1V1/T1 = P2V2/T2 . Letthe mass of the gas in question be m, then m = ρ1V1 = ρ2V2 , or V1 = m/ρ1 , V2 = m/ρ2 .Substitute these expressions for V1 and V2 into the first equation above to yield

P1(m/ρ1)T1

=P2(m/ρ2)

T2

.

Cancel m from both sides and the identity in the problem statement follows.

12.74One mole of hydrogen gas occupies a volume of V = 22.4 liters = 22.4×10−3 m3 at STP. Sincethe molecular mass of an H2 molecule is 2 × 1.0 u = 2.0 u, the mass of one mole of hydrogengas is m = 2.0 g = 2.0 × 10−3 kg. Thus its density is

ρ =m

V=

2.0 × 10−3 kg

22.4 × 10−3 m3= 0.089 kg/m3 .

As the pressure remains at 1.00 atm while the temperature increases to T = 273◦C = (273 +273)K = 546 K, i.e., double the value at STP, the volume of the gas doubles while its densityis halved: ρ = 1

2 (0.089 kg/m3) = 0.045 kg/m3.

12.75

(a) For a given amount of ideal gas at constant pressure, its volume V is proportional to itsabsolute temperature: V/T = constant.


(b) The device measures the temperature of the gas (which equals that of the environmentupon thermal equilibrium) by varying its volume V while keeping its pressure fixed (as themercury drop remains in the same position). First, calibrate the thermometer by submerging itin a thermal bath of known absolute temperature, say T0 , at which the volume of the gas readsV0 . Then we may use it to measure the absolute temperature T of another thermal bath byfinding the new volume V of the gas which keeps the mercury drop in the same position (whichindicates that the pressure remains the same): V0/T0 = V/T , or

T = T0

(V

V0

).

(c) Since the capillary tube is open and the mercury drop does not move, the pressure of thegas remains the same.

(d) Since the mercury drop does not move, the pressure P exerted on it from the gas sealedby it must equal to that form the outside air. So P = PA = 1.0 atm.

(e) Apply the formula in part (b) above, with the reference volume at V0 = 1.50 × 103 cm3,the reference temperature at T0 = 0◦C = (0 + 273.15) K = 273.15 K, and the new volume atV = 1.80 × 103 cm3:

T = T0

(V

V0

)= (273.15 K)

(1.80 × 103 cm3

1.50 × 103 cm3

)= 328 K ,

or (328 − 273)◦C = 55◦C.

12.76

(a) The relationship between P and V of an ideal gas in an isothermal process (during whichT = constant) is Charles’s Law: PV = constant.

(b) As the gas is compressed V decreases. Since PV remains the same P must increase.

(c) Use P1V1 = P2V2 to find P2 . Note that P1 = 1.9 atm + 1.0 atm = 2.9 atm is the absolutepressure of the gas before the compression. Thus

P2 =P1V1

V2

=(2.9 atm)(590 cm3)

181 cm3= 9.5 atm ,

and so the final gauge pressure is 9.5 atm − 1.0 atm = 8.5 atm. (Note that in the Ideal GasLaw, PV/T = constant, P is the absolute pressure, not the gauge pressure.)

12.77The absolute pressure at a depth h below the surface of the liquid is Pi = Ps + ρgh, wherePs = 1.0 atm. As the bubble rises to the surface of the liquid the final absolute pressure isPf = PA . Apply Eq. (12.5) for constant temperature: PiVi = PfVf . This gives

Vf =PiVi

Pf

=(PA + ρgh)Vi

PA

=(

1 +ρgh

PA

)Vi .


12.78The volume of the gas inside the tire is unchanged, so Eq. (12.7) applies: P/T = constant.Initially, Pi = (33 lb/in.2)

[6.895 × 103 Pa/(lb/in.2)

]= 2.28×105 Pa and Ti = 273+ 5

9 (40.0−32.0) = 277.6 K, and finally Tf = 273 + 5

9 (120 − 32.0) = 321.9 K. Thus the final pressure Pf

follows from Pi/Ti = Pf/Tf to be

Pf =PiTf

Ti

=(2.28 × 105 Pa)(321.9 K)

277.6 K= 2.6 × 105 Pa .

12.79

(a) At a depth h below the mercury level the gauge pressure caused by the weight of themercury column is PG = ρHggh.

(b) The barometer reading gives the air pressure PA in the room: PA = ρHggH The absolutepressure P at a depth h below the mercury level is the sum of the air pressure and the gaugepressure due to the mercury column:

P = PA + PG = ρHggH + ρHggh = ρHgg(H + h) .

(c) The volume of the gas trapped in the tube on the right is V = AL.

(d) A reasonable assumption in this problem is that the temperature of the gas in the tube onthe right is constant. Thus its pressure P ′ and its volume V are related by P ′V = C, where Cis a constant. Since V = AL, P ′ = C/V = C/AL. To achieve mechanical equilibrium in thetwo arms of the U-tube we must require P = P ′, i.e., ρHgg(H + h) = C/AL, which we rewriteas

H + h =(

C

ρHggA

)1L

,

which indicates that H + h varies linearly as a function of 1/L. This explains the plot inFig. P79(b). The slope of the straight line should be C/ρHggA.

12.80

(a) The relationship is the Ideal Gas Law: PV/T = constant.

(b) and (c) Use P1V1/T1 = P2V2/T2 to find the new volume V2 of the gas as its pressure dropsfrom P1 = 49.8 kPa to P2 = 39.8 kPa while its temperature drops from T1 to 1

2T1 :

V2 =P1V1T2

P2T1

=(49.8 kPa)(0.58 m3)( 1

2T1)(39.8 kPa)T1

= 0.36 m3 .

Since V2 < V1 the piston moves in.

Pi

PA

h

mercury


12.81The situation described in the problem statement isdepicted in the diagram to the right. To achieve me-chanical balance the gas pressure Pi in the graduatedcylinder before cooling must satisfy Pi + ρMgh =PA , where ρM is the density of mercury, h is theheight of the mercury column (= 3.90 cm), and PA =101.05 kPa is the air pressure outside. The tempera-ture of the gas then is Ti = 20 + 273 = 293 K (roomtemperature), and its volume is Vi = 214 cm3. Aftercooling the new pressure is Pf = 101.32 kPa, the newtemperature is Tf = 0.00 + 273.15 = 273.15 K, andthe new volume of the gas Vf must satisfy Eq. (12.8):PiVi/Ti = PfVf/Tf , or

Vf = Vi

(Pi

Pf

)(Tf

Ti

)= Vi

(PA − ρMgh

Pf

)(Tf

Ti

)

= (214 cm3)[101.05 kPa − (13.6 × 103 kg/m3)(9.81 m/s2)(0.0390 m)

101.32 kPa

](273.15 K

293 K

)= 189 cm3 .

12.82With the previous problem in mind, the total pressure P of the water vapor-hydrogen gasmixture inside the beaker satisfies P +ρWgh = PA , where ρW is the density of water (= 1.00×103 kg/m3), h = 5.00 cm, and PA = 759 mmHg = (0.759 m)(13.6 × 103 kg/m3)(9.81 m/s2) =1.012 6 × 105 Pa. Since the total pressure P is the sum of PH , the pressure due to hydrogenalone; and PV , the pressure of the water vapor (= 1.92 kPa),

PH = P − PV = PA − ρWgh − PV

= 1.012 6 × 105 Pa − (1.00 × 103 kg/m3)(9.81 m/s2)(0.0500 m) − 1.92 × 103 Pa

= 9.88 × 104 Pa = 98.8 kPa .

12.83Label the oxygen gas with subscript O and the nitrogen with N, respectively, and consider thetwo gases separately. The expansion process for either gas is isothermal (i.e., T = constant.),


so for the oxygen gas Eq. (12.5) gives PiOViO = PfOVfO , which yields PfO , the final pressure inthe chamber due to the oxygen gas alone:

PfO =PiOViO

VfO

=(5.0 atm)(3.0 liters)

4.0 liters= 3.75 atm .

Similarly for the nitrogen gas PiNViN = PfNVfN , so PfN , the contribution from the nitrogen tothe final pressure in the chamber, is

PfN =PiNViN

VfN

=(2.0 atm)(1.0 liters)

4.0 liters= 0.50 atm .

The total pressure Pf in the chamber is the sum of the contributions from both gases: Pf =PfO + PfN = 3.75 atm + 0.50 atm = 4.3 atm.

12.84Apply Eq. (12.10), PV = nRT (the Ideal Gas Law), to solve for V . Here P = 93.0 × 103 Pa,n = 3.0 mol, and T = 60 + 273 = 333 K. Thus

V =nRT

P=

(3.0 mol)(8.314 4 J/mol·K)(333 K)93.0 × 103 Pa

= 0.089 m3 .

12.85Use Eq. (12.10), PV = nRT , to solve for T . Here P = 202.6 × 103 Pa, V = 10.0 × 10−3 m3,and n = 1.40 mol. Thus

T =PV

nR=

(202.6 × 103 Pa)(10.0 × 10−3 m3)(1.40 mol)(8.314 4 J/mol·K)

= 174 K .

12.86Use Eq.(12.11), PV = NkBT , to solve for N , the number of molecules. Here P = (10−15 atm)×(1.01 × 105 Pa/atm) = 1 × 10−10 Pa, V = 1.0 cm3 = 1.0 × 10−6 m3, and T = 273 K. Thus

N =PV

kBT=

(1 × 10−10 Pa)(1.0 × 10−6 m3)(1.380 662 × 10−23 J/K)(273 K)

= 3 × 104 .

That’s still about 30,000 molecules left per cubic centimeters.

12.87Similar to the previous problem, use PV = NKBT to solve for N . Here P = 1.00 atm =1.013 × 105 Pa, V = 1.00 cm3 = 1.00 × 10−6 m3, and T = 273 K. Thus

N =PV

kBT=

(1.01 × 105 Pa)(1.00 × 10−6 m3)(1.380 662 × 10−23 J/K)(273 K)

= 2.68 × 1019 .


12.88Apply Eq. (12.11), PV = NkBT , to solve for the number of air molecules, with P = 0.01 ×106 Pa, V = 0.10 m3, and T = 273 + 20 = 293 K. Thus

N =PV

kBT=

(0.01 × 104 Pa)(0.10 m3)(1.380 662 × 10−23 J/K)(293 K)

= 2.5 × 1023 .

12.89Apply the Ideal Gas law, Eq. (12.10): PV = nRT . Here P = 2.00 × 106 Pa, V = 1.00 ×10−2 m3, and T = 273.15 + TC = 273.15 + 20 = 293 K. Thus

n =PV

RT=

(2.00 × 106 Pa)(1.00 × 10−2 m3)(8.314 J/mol·k)(293 K)

= 8.21 mol .

12.90Use Eq. (12.11), PV = NkBT , to solve for N , the number of air molecules. Here P =(1.00 atm)(1.013 × 105 Pa/atm) = 1.013 × 105 Pa, V = 1.00 m3, and T = 300 K. Thus

N =PV

kBT=

(1.013 × 105 atm)(1.00 m3)(1.380 6 × 10−23 J/K)(300 K)

= 2.45 × 1025 .

12.91Apply Eq. (12.10), PV = nRT , with V = 5.00 × 10−4 m3, n = 0.050 mol, and T = 100 +273.15 = 373 K. Solve for P :

P =nRT

V=

(0.050 mol)(8.314 J/mol·K)(373 K)5.00 × 10−4 m3

= 3.1 × 105 Pa ,

which is about 3.1 atm.

12.92

(a) It is the Ideal Gas Law: PV = NkBT , where N is the number of gas molecules and kB

is the Boltzmann Constant. This is equivalent to PV = nRT , with n = N/NA the number ofmoles of the gas and R = kBNA the Universal Gas Constant.

(b) T = (20 + 273.15) K = 293 K.

(c) V = 2.0 liters = 2.0 × 103 cm3 = (2.0 × 103 cm3)(10−2 m/cm)3 = 2.0 × 10−3 m3.

(d) Solve for P from the Ideal Gas Law:

P =nRT

V=

(4.0 mol)(8.314 4 J/mol·K)(293 K)2.0 × 10−3 m3

= 4.872 × 106 Pa .


Thus the gauge pressure is P − P0 = 4.872 × 106 Pa − 1.013 × 105 Pa = 4.8 × 106 Pa.

12.93

(a) It is the Ideal Gas Law: PV = NkBT , where N is the number of gas molecules and kB

is the Boltzmann Constant. This is equivalent to PV = nRT , with n = N/NA the number ofmoles of the gas and R = kBNA the Universal Gas Constant.

(b) T = (74 + 273.15) K = 347 K.

(c) V = 4.0 liters = 4.0 × 103 cm3 = (4.0 × 103 cm3)(10−2 m/cm)3 = 4.0 × 10−3 m3.

(d) P = PG + P0 = 2.0 atm + 1.0 atm = 3.0 atm = (3.0 atm)(1.012 × 105 Pa/atm) =3.0 × 105 Pa.

(d) Solve for n, the number of moles, from the Ideal Gas Law:

n =PV

RT=

(3.03 × 105 Pa)(4.0 × 10−3 m3)(8.314 4 J/mol·K)(347 K)

= 0.42 mol .

Multiply this by the Avogadro’s Number to obtain the number of gas molecules: N = nNA =(0.42 mol)(6.02 × 1023/mol) = 2.5 × 1023.

12.94According to the result of Problem (12.90) there are N = 2.45×1025 air molecules in a volumeof V = 1.00 m3. Thus the average volume occupied by each molecule is v = V/N . If thisvolume is cubic, then the side length l of the cube, which is also the center-to-center distancebetween air molecules, satisfies v = l3. Solve for l:

l = v1/3 =(

V

N

)1/3

=(

1.00 m3

2.45 × 1025

)1/3

= 3.45 × 10−9 m = 3.45 nm ,

which is about an order of magnitude greater than the size of individual air molecules.

12.95Apply Eq. (12.11), PV = NkBT , to estimate N , the number of air molecules per breath.Here P = 761 mmHg = (0.761 m)(13.6 × 103 kg/m3)(9.81 m/s2) = 1.015 × 105 Pa, V ≈500 × 10−6 m3, and T = 273 + 37 = 310 K. Thus

N =PV

kBT≈ (1.015 × 105 Pa)(500 × 10−6 m3)

(1.380 662 × 10−23 J/K)(310 K)= 1.2 × 1022 ,

which is about the same as the number of stars there are in the entire Universe.

12.96From Eq. (12.10), PV = nRT , we find the number of moles per unit volume to be n/V =P/RT . Let the molar mass of nitrogen be mN (= 28.0 g/mol), then the mass of n moles of


nitrogen is nmN , and the density of nitrogen is given by ρN = nmN/V = mNP/RT . Solve forT :

T =mNP

ρNR=

(28.0 × 10−3 kg/mol)(1.01 × 105 Pa)(1.245 kg/m3)(8.314 4 J/mol·K)

= 273 K .

12.97First find n, the number of moles of helium gas, from Eq. (12.10):

n =PV

RT=

(2.0 × 1.013 × 105 Pa)(1000 × 10−6 m3)(8.314 J/mol·K)(273.15 + 23.0)K

= 0.082 3 mol .

Since the molar mass of helium gas is 4.0 g/mol, this amounts to (4.0 g/mol)(0.082 3 mol) =0.33 g of helium.

12.98For constant pressure, apply Eq. (12.6): Vi/Ti = Vf/Tf . Here Vi = 600 cm3, Ti = 23.0 +273.15 = 296.15 K, and Vf = 750 cm3. Solve for Tf :

Tf =TiVf

Vi

=(296.15 K)(750 cm3)

600 cm3= 370 K .

12.99First find n, the number of moles of hydrogen gas needed, from PV = nRT . Here P =19.4 × 103 Pa, V = 2000 m3, and T = 273 + (−55) = 218 K; so

n =PV

RT=

(19.4 × 103 Pa)(2000 m3)(8.314 4 J/mol·K)(218 K)

= 2.14 × 104 mol .

Since the molar mass of hydrogen is 2.0 × 103 kg/mol, the total mass of hydrogen needed ism = (2.0 × 103 kg/mol)(2.14 × 104 mol) = 43 kg.

12.100The volume coefficient β for a substance of volume V at temperature T is defined in the equation∆V = βV ∆T . In this case the substance we are dealing with is the ideal gas, which satisfiesPV = nRT , or V = nRT/P = (nR/P )T . Since P and n are both constants, so is nR/P .Thus if V changes by ∆V due to a temperature change ∆T , then ∆V = ∆ [(nR/P )T ] =(nR/P ) ∆T ; hence

β =∆V

V ∆T=

(nR/P ) ∆T

V ∆T=

nR

PV=

1T

,

where in the last step we again used PV = nRT .


12.101The average molecular speed is

vav =15(1.0 m/s + 2.0 m/s + 3.0 m/s + 4.0 m/s + 5.0 m/s) = 3.0 m/s .

12.102The rms -speed of the molecules is

vrms =

√15

[(1.0 m/s)2 + (2.0 m/s)2 + (3.0 m/s)2 + (4.0 m/s)2 + (5.0 m/s)2] = 3.3 m/s ,

which is greater than vav computed in the previous problem. In fact this is an example of thegeneral result vrms ≥ vav , with the equality valid only when all of the individual speeds areidentical to each other.

12.103Use Eq. (12.16), vrms =

√3kBT/m, to find vrms . For oxygen molecules with a molecular mass

of m = 5.313 6 × 10−26 kg at T = 293.15 K,

vrms =

√3kBT

m=

√3(1.380 662 × 10−23 J/K)(293.15 K)

5.313 6 × 10−26 kg= 478.03 m/s .

12.104Use Eq. (12.15), KEav = 3

2kBT , with T = 0 + 273.15 = 273.15 K:

KEav =32kBT =

32(1.380 662 × 10−23 J/K)(273.15 K) = 5.66 × 10−21 J .

12.105Apply Eq. (12.15), with T = 20 + 273 = 293 K:

KEav =32kBT =

32(1.38 × 10−23 J/K)(293 K) = 6.1 × 10−21 J .

12.106Use Eq. (12.16), with T = 0.000 + 273.15 = 273.15 K and m = (44.0 g/mol)/NA = (4.40 ×10−2 kg/mol)/(6.022 × 1023 mol) = 7.306 5 × 10−26 kg (note that the molar mass of CO2 is44.0 g/mol). Thus

vrms =

√3kBT

m=

√3(1.380 6 × 10−23 J/K)(273.15 K)

7.306 5 × 10−26 kg= 393 m/s .


12.107

(a) The relationship is given by Eq. (12.14): KEav = 32kBT .

(b) Since KEav ∝ T , as T is doubled so is KEav .

(c) Solve for T from Eq. (12.14):

T =2KEav

3kB

=2(7.0 × 10−22 J)

3(1.380 662 × 10−23 J/K)= 34 K .

(d) TC = T + 273.15 = 34 + 273.15 = 307◦C.

(e) The average molecular kinetic energy is directly proportional to the absolute temperatureof an ideal gas: KEav = 3

2kBT . As long as KEav is fixed, so is T , regardless of whether the gasis nitrogen or hydrogen.

12.108

(a) The relationship is given by Eq. (12.14): KEav = 32kBT .

(b)

KEesc =12mv2

esc=

12(3.3 × 10−27 kg)(11 × 103 m/s)2 = 2.0 × 10−19 J .

(c) Let KEav = 32kBT = KEesc and solve for the absolute temperature T :

T =2KEesc

3kB

=2(2.0 × 10−19 J)

3(1.380 662 × 10−23 J/K)= 9.7 × 103 K ,

(d) TC = (9657 − 273.15)◦C = 9384◦C, or 9.4 × 103 ◦C to 2 significant figures.

(e)

TF =95TC + 32 =

95(9384) + 32 = 1.7 × 104 ◦F .

(f) At a given temperature the average translational KE of different kinds of molecules is thesame. Hydrogen has the smallest value of m so for the same average KE it has the greatestvalue of average speed, which makes it more likely for hydrogen to reach the escape speed thanany other type of molecules.

12.109The average translational KE per oxygen molecule at T = 273.15 K is given by Eq. (12.14):KEav = 3

2kBT . The number of such molecules, N , can be found from Eq.(12.11), PV = NkBT ,

where P = 1.00 atm = 1.01 × 105 Pa and V = 0.50 m3. Combine these two equations to findthe total translational KE:

KEtotal = NKEav =(

PV

kBT

)(32

kBT

)=

32PV

=32(1.01 × 105 Pa)(0.50 m3)

= 7.6 × 104 J = 76 kJ .


12.110The average translational KE per molecule is given by Eq.(12.15), with T = 100+273 = 373 K.The total KE of N = 12.0 × 1023 molecules is then

( KE)total = N( KE)av =32NkBT

=32(12.0 × 1023)(1.380 6 × 10−23 J/K)(373 K)

= 9.27 × 103 J = 9.27 kJ .

12.111The number of gas molecules in the chamber is N = nNA , where n = 25.0 mol. The averagetranslational KE for each molecule is ( KE)av = 3

2kT , where T = 200 + 273 = 473 K. Thusthe total translational KE is

( KE)total = N( KE)av =32nNAkBT =

32nRT

=32(25.0 mol)(8.31 J/mol·K)(473 K)

= 1.47 × 105 J = 0.147 MJ .

Here we noted that kB = R/NA , so NAkB = R.

12.112Divide both sides of Eq. (12.14), PV = 2

3N(KE)av , by V :

P =23

(N

V

)(KE)av ,

where N/V on the RHS is the number of molecules per unit volume. This is the desired result.

12.113Start with Eq. (12.14), PV = 2

3NKEav , by rewriting KEav on the RHS as KEav = 12mv 2

rms,

where m is the molecular mass:

PV =23NKEav =

23N

(12mv 2

rms

)=

13(Nm)v 2

rms.

Since Nm is the total mass of the N molecules, the density of the gas of volume V is given byρ = mN/V . Use this result to solve for vrms from the equation above:

vrms =

√3PV

mN=

√3P

mN/V=

√3P

ρ.


12.114According to the Ideal Gas Law [Eq. (12.11)] PV = NkBT . However, from Eq. (12.15) theaverage molecular translational KE is ( KE)av = 3

2kBT . Thus the internal energy, which comesfrom translational KE alone in a monatomic ideal gas, is given by

Internal energy = N( KE)av =32NkBT =

32(nNA)kBT =

32nRT =

32PV ,

where we used the identities N = nNA and NAkB = R.

12.115Use the result of the previous problem. Here P = 1.00 atm = 1.01 × 105 Pa and V = 3.5 m ×5.0 m × 6.0 m = 105 m3, so

Internal energy =32PV =

32(1.01 × 105 Pa)(105 m3) = 1.6 × 107 J = 16 MJ .

12.116According to Eq. (12.16) the rms -speed varies as a function of T as vrms =

√3kBT/m ∝

√T .

Thus at T1 = 0◦C = 273 K and T2 = 100◦C = 100+273 = 373 K the ratio of the correspondingrms -speeds is

v1 rms

v2 rms

=

√T1

T2

=

√273 K

373 K= 0.86 = 86% .

12.117According to Eq. (12.16) the rms -speed varies as a function of m as vrms =

√3kBT/m ∝

1/√

m. For 238U the mass is m238 = 238 u + 6(18.998 u) = 351.99 u, and for 235U the mass ism235 = 235 u + 6(18.998 u) = 348.99 u. Thus the percent difference in the rms -speeds is

v 235rms

− v 238rms

v 235rms

= 1 − v 238rms

v 235rms

= 1 −√

m235

m238

= 1 −√

348.99 u

351.99 u= 0.004 27 = 0.427% .

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12 Thermal Properties of Matter · 416 CHAPTER 12 THERMAL PROPERTIES OF MATTER 12 Thermal Properties of Matter Answers to Discussion Questions •12.1 temperature —– The degree