12 Thermal Properties of Matter

Transcription

1 416 CHAPTER 12 THERMAL PROPERTIES OF MATTER 12 Thermal Properties of Matter Answers to Discussion Questions 12.1 temperature The degree of hotness or coldness of a body or environment. More scientifically, 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 and the boiling point as 212 at one atmosphere of pressure. Celsius scale A temperature scale that registers the freezing point of water as 0 and the boiling point as 100 under normal atmospheric pressure. thermodynamic temperature Also known as the absolute temperature. It is measured or calculated in kelvin, starting from Absolute Zero. Absolute Zero The theoretical temperature at which substances possess no thermal energy, equal to C, or F. zero-point energy the minimum energy possessed by atoms and molecules at Absolute Zero due to quantum effects. temperature coefficient of linear expansion the amount of change in the linear size of a substance per unit length per unit temperature change. temperature coefficient of volume expansion the amount of change in the volume of a substance per unit length per unit temperature change. Boyle s Law For a certain amount of ideal gas at constant temperature, the product of its pressure and its volume is a constant: PV constant. Charles s Law For a certain amount of ideal gas at constant pressure, its volume is proportional to its absolute temperature: V/T constant.

2 CHAPTER 12 THERMAL PROPERTIES OF MATTER 417 Gay-Lussac s Law For a certain amount of ideal gas at constant volume, its pressure is proportional to its absolute temperature: P/T constant. Ideal Gas Law For a certain amount (n moles) of ideal gas, its pressure, volume, and its absolute temperature are related by PV nrt. Universal Gas Constant R J/mol K. Boltzmann s Constant k B J/K. critical point The point on the phase diagram where liquid and gas coexist at the same density and are indistinguishable. vapor A gaseous state, especially when diffused in the atmosphere and at a temperature below boiling point. Unlike gases, vapors can be condensed via compression. equilibrium vapor pressure The pressure of the vapor at which the rates of condensation 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 and solid) coexist in equilibrium. sublimation Raising temperature of a solid at a fixed pressure below that of the triple point 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 of the 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 a function of energy for a system in thermal equilibrium at a given temperature Here 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 speed up its response the walls of the bulb are made thin and that thinness makes it vulnerable to pressure 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 the different heights of the column. (e) There is a difference in internal pressure if it s held vertically as opposed to horizontally. (f) There are errors associated with the softness of the glass. If the thermometer is raised to a high temperature and then cooled rapidly, it might take weeks for the glass to return to its original volume. Try measuring the freezing point of water before and immediately after reading its boiling point the difference can be as great as 1 C. (g) The mere 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 read warmer if the bath temperature is falling and vice versa The glass will expand first, dropping the mercury level until it too becomes heated and expands.

3 418 CHAPTER 12 THERMAL PROPERTIES OF MATTER 12.4 We know that the antimony expands on solidifying, as does water. Since that would be a very helpful trait for a casting material to have (it would fill all the fine details in the mold), it s reasonable to expect that s the answer to this question The water inside the bomb freezes and expands and the bomb explodes. When I was an undergraduate I foolishly put one of these in a mix of dry ice and alcohol. The resulting violent explosion blew the thing to bits. It s common for water collecting in cracks in rocks to split them when it freezes Besides being more malleable when hot they contracted on cooling, pulling the plates together No. It wouldn t work at all. The thermometer relies on a difference between the two β values to produce a net excess expansion which sends the fluid out of the storage bulb Being a poor conductor, the center of a boulder so treated would remain quite hot while the outside was cooled and contracted rapidly. Pressure would build up, there would be considerable internal stress, and the thing would rupture at any flaw or weak spot Chemists do a lot of heating with open flames and so the coefficient of expansion must be as small as possible. The walls are kept thin to insure that they will heat up almost uniformly and therefore not suffer appreciably different amounts of expansion and/or contraction and so shatter Bulbs are cheaper thinner and can tolerate the changes in temperature associated with ordinary operation, which are fairly gradual. They are made of inexpensive glass with a relatively large β, soadrop of water (or latex paint) can cause enough contraction and stress to shatter a hot light bulb. Clearly, outdoor lamps have to be protected from rain and snow. The heating in a flashbulb is so rapid even the thin walls tend to burst and they therefore are usually enclosed in a tough plastic film to keep them from shattering.

4 CHAPTER 12 THERMAL PROPERTIES OF MATTER The increased pressure at the points of contact melts the ice which, as the pressure is subsequently removed, refreezes. This process called regelation, was named by the English physicist John Tyndall who used it to explain the making of snowballs At atemperature of even 1 Cittakes about 140 atm of pressure ( 2000 lb/in 2 )tomelt ice, so the problem was probably that the snow was just too cold If the curves intersect as in Figure Q13 the region PQR is below the fusion curve and so must represent a solid; it s above the vaporization curve and so must represent a liquid; it s below the sublimation curve and so must represent a vapor. These conclusions are contradictory hence the configuration cannot exist The liquid will expand rapidly: its density decreasing as the density of the vapor increases. The surface meniscus will flatten out and then disappear altogether when the density of liquid and vapor are equal at a pressure of 7.38 MPa At the triple point ( K, 0.61 kpa) water can boil and freeze simultaneously. Place some water in a dish positioned above a bowl of strong sulfuric acid and seal the whole thing in a vacuum chamber. If you have a good high capacity pump the acid isn t necessary. On evacuating the chamber the water will evaporate rapidly and the vapor will be absorbed by the acid. At a low enough pressure the water will begin to boil and becoming cooled by evaporation will soon freeze Yes. They were called permanent because they could not at first be liquified, which suggests a weak intermolecular cohesive force and that, in turn, suggests ideal behavior Equation (12.15), KE av 3 2 k BT, describing the average translational KE of each molecule is what Jeans is talking about but he overlooks the quantum mechanical zero-point energy at 0K the energy of motion is not nil PV nrt,soboth pressures must be equal since everything else is the same. The speeds of the hydrogen molecules must be greater than those of the nitrogen because the average KE is

5 420 CHAPTER 12 THERMAL PROPERTIES OF MATTER the same from Eq. (12.15). The pressures can be equal because the lighter hydrogens hit the walls at greater speeds and they do it more frequently because they traverse the chamber more quickly. The pressure is proportional to the average KE via Eq. (12.14) It can be expected that they have discovered the same absolute zero we know and would be equally familiar with low-temperature physics. But it s quite unlikely that they too happened to use the freezing and boiling points of water at the surface pressure of planet Earth as reference levels. It s even more unlikely that they might use human body temperature as a reference as did Fahrenheit. So they ll probably not have thermometers graduated in Fahrenheit or Celsius degrees or even in kelvins, since the latter matches up with the Celsius scale. Still, it wouldn t take much effort to understand their temperature system and they will surely be as concerned with the energy concentration of their environment as we are Remember that the pressure in the room is more or less constant. Increasing T increases the average KE, which increases the net KE and the P but leads to an over-pressure and an outward current of air. The room leaks warm air to the outside. The temperature goes up because it s dependent on the average KE of each molecule, which is higher. The pressure remains the same because it depends on both the number of molecules per unit volume and their average KE. Compare Eqs. (12.14) and (12.15) According 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. c 8. 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. a 22. b 23. b 24. c 25. e 26. a 27. a

6 CHAPTER 12 THERMAL PROPERTIES OF MATTER 421 Solutions to Problems 12.1 Use Eq. (12.1), with T F 98.6 F: T C 5 9 (T F 32 C) 5 9 ( ) C 37.0 C. (Note that we needn t be concerned with the significant number in the figure 32 Cabove it is defined to be exact.) 12.2 Use Eq. (12.1), with T F 70.0 F: T C 5 9 (T F 32 ) 5 9 ( ) C 21.1 C Fig. P3 is a plot of the Celsius temperature T C versus the Fahrenheit temperature T F, which are related by Eq. (12.1): T C 5 9 (T F 32 ). Since T C is a linear function of T F the plot is a straight line. In Problem (12.1) T F 98.6 F and in Problem (12.2) T F 70.0 F. You can verify that the corresponding temperature readings in Celsius should be 37.0 and 21.1 C, respectively, by examining the plot The slope of a straight line in a plot of y vs x is k, ify kx+ b. Inour case the equation reads T C 5 9 T F 17.8,soinaplot of T C vs T F the slope is 5 9,asyou should verify by applying slope rise/run directly over the plot Read the value of T C1 directly from the plot. It is about 18 C. For more precision, note that the slope of the line is 5/9, so T F1 /T C1 5/9 in the small triangle formed by the line with the two axes. This gives T C1 5 9 T F1 5 9 (32.0 ) 17.8 ; and, since T C1 < 0, T C1 T C C. Alternatively, you can also plug T F1 0 Finto Eq. (12.1) to obtain T C1 : T C1 5 9 (T F1 32 ) 5 9 (0 32.0) C 17.8 C.

7 422 CHAPTER 12 THERMAL PROPERTIES OF MATTER 12.6 The 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 (T F and T C ) give identical readings From Eq. (12.2), with T K, T C T C Use Eq. (12.1), with T C 39.0 C: T F 9 5 T C F From Eq. (12.2), with T 58K, T C T C Since T T C , T (T C ) T C. This means that if a certain change in temperature measures n units in kelvin it should also measure n degrees in Celsius. so the ratio in question is First, convert the temperature from kelvin to Celsius degrees: T C T Now convert it into Fahrenheit: T F 9 5 T C F Use Eq. (12.2). For T C C, T T C K; and for 272 C, T K. T C

8 CHAPTER 12 THERMAL PROPERTIES OF MATTER First convert the temperatures to Celsius scale. For T F 0 F, T C 5 9 (T F 32 C) 5 9 (0 32) C 17.8 C; and for T F 100 F, T C 5 9 (T F 32) 5 9 (100 32) C 37.8 C. Thus the temperature change, in Celsius, is T C 37.8 C ( 17.8 C)55.6 C. The corresponding temperature change in Kelvins is the same as that in Celsius: T 55.6 K The melting temperature in Celsius is T C T C According tot he result of Problem (12.10) T (in kelvin units) T C (in Celsius degrees). Since T C 42C, T 42K Since T C T , doubling T to 2T would result in a new reading in Celsius degree of T C 2T T +(T ) T + T C, which is not the same as 2T C. Similarly, from T F 9 5 T +32,ifT is doubled to 2T the C C C resulting new reading in Fahrenheit degree would be T 9 ( ) 9 F 5 (2T C)+32T C + 5 T C +32 T C + T F, which is generally not the same as 2T F Use Eq. (12.2) to convert T to T C : T C use Eq. (12.1) to convert T to T F : T C. Now T F T C ( ) F The temperature in Celsius is T C 5 9 (T F 32 C) 5 9 ( ) C 2204 C, which in Kelvins is T T C K The temperature in Celsius is T C T C, which is equivalent to T F T C (497) 927 F.

9 424 CHAPTER 12 THERMAL PROPERTIES OF MATTER From Table (12.1), the ratio is 1893 K/600 K 3.16 when the temperatures are measured in Kelvins; and 1620 C/327 C 4.95 when they are measured in Celsius Since the T C vs T F curve is a straight line, the equation it represents can be written as T C k(t F C), where k is the slope of the line and C is a constant. To determine C, note that T C 0when T F C. ThusC 32degrees. Next, use the two points (T F2,T C2 )(212, 100) and (32, 0) to determine the slope k: Thus k T C2 0 T F ( ) 100 T C k(t F C) (T F 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 ( ) 100 T C (T F 32) (T F 32) With the previous problem in mind, write T C kt F + b and determine k and b. For b, let T F T F1 0to obtain T C T C1 kt F1 + b k 0+b b, orb T C1. The slope k can be obtained from the two points, (T F2,T C2 ) and (T F1,T C1 ): k (T C2 T C1 )/(T F2 T F1 ).Thus ( TC2 T T C kt F + b C1 T F2 ) T F + T C1, where in the last step we noted T F1 0. Now plug in T C C [see Problem (12.5)], T C2 100 C, T F2 212 F, and T F 68 Ftoobtain [ ] 100 ( 17.8) T C (68) C Use Eq. (12.3). For Pyrex glass α K 1, and the temperature change is T 80.0 C 80.0 K; so the length of the rod will increase by L αl 0 T ( K 1 )(1.00 m)(80.0 K) m 0.2 mm.

10 CHAPTER 12 THERMAL PROPERTIES OF MATTER 425 (A note in notation: in many books, Cisused for the actual temperature in degrees Celsius whereas C denotes a change in temperature, measured in Celsius degrees. A similar distinction exists between F and F. For example, although 9 F 5 C, we do have 9 F 5C,as achange of 9 Fahrenheit degrees is equivalent to a change of 5 Celsius degrees. We will be following this convention in later problems.) Start from Eq.(12.3), L αl 0 T, with L 2.00 mm m, α /K, and L 10.0 m. Solve for T : T L αl m ( K 1 )(10.0 m) 17K 17C Use Eq. (12.3). For aluminum α K 1, and the temperature change is T 1.00 C 1.00 K; so the length of the rod will increase by L αl 0 T ( K 1 )(1.00 m)(1.00 K) m 25µm Again, use Eq. (12.3). For brass α K 1, and the temperature change is T 1.00 C 1.00 K, so the length of the rod will change by L αl 0 T ( K 1 )(1.00 m)( 1.00 K) m 18.9 µm, where the minus sign indicates that the length of the rod has decreased Apply Eq. (12.3), L αl 0 T, which gives the change in length of an object of length L 0 as a result of T,achange in its temperature. Here α is the coefficient of linear expansion of the material of which the object is made. In this case we are dealing with an aluminum bar, for which α K 1, T 50 C 30 C 20C 20K, and L 0 10m. So the length of the bar will increase by L αl 0 T ( K 1 )(10 m)(20 K) m 5.0 mm Similar to the previous problem, apply L αl 0 T. Here α K 1 for steel, T 35 C 5 C 30C 30K, and L 0 10m. So the length of the rod will increase by L αl 0 T ( K 1 )(10 m)(30 K) m 3.6 mm.

11 426 CHAPTER 12 THERMAL PROPERTIES OF MATTER Apply L αl 0 T. Here α K 1 for steel, the temperature change is T 35 C 0 C 35C 35K, and L m. So the length of the roadway will increase by L αl 0 T ( K 1 )(1280 m)(35 K) 0.54 m We are looking for the percentage change in length for the rod, i.e., L/L 0. Convert T to kelvin: T 168 F 68 F 100 F 5 9 (100) C C K. Thus from Eq. (12.3) L L 0 α T ( K 1 )(55.56 K) % (a) The relationship in question is given by Eq. (12.3), L αl 0 T. (b) Each slab expands by 1 2 L on either side, so the spacing between two adjacent slabs are shortened by 1 2 L L L. The minimum initial spacing d must then be equal to min L, whereupon the gap between the slabs would completely close as temperature rises to the highest value. (c) T T f T i 46 C 10 C 36C 36K. (d) Take α K 1 to obtain d min L αl 0 T ( K 1 )(2.4 m)(36 C ) m 1.0 mm Use Eq.(12.3) to find the new length and width of the sheet. If the original length and width are denoted as a 0 and b 0, respectively, then their new values will be a a 0 + a a 0 + a 0 α T and b b 0 + b b 0 + b 0 α T, respectively. Plug in a cm, b cm, α K 1, and T C C C Ktoobtain the new area as a result of the temperature change: A ab (a 0 + a 0 α T )(b 0 + b 0 α T )a 0 b 0 (1 + α T ) 2 (50.00 cm)(20.00 cm) [ 1+( K 1 )(30.00 K) ] cm 2.

12 CHAPTER 12 THERMAL PROPERTIES OF MATTER According to Problem (12.28) the change in area for the sheet, whose original area is A 0 a 0 b 0, is A 2αa 0 b 0 T 2( K 1 )(50.00 cm)(20.00 cm)(30.00 K) cm 2, which gives the new area to be A a 0 b 0 + A (50.00 cm)(20.00 cm) cm cm 2,ingood agreement with the result of Problem (12.32) Use Eq. (12.3), L αl 0 T,tofind L, the change in height for the steel tower. Here α K 1 for steel, L m, and T T C 5 9 T F 5 9 [95 ( 10)] 58.3 K. Thus the new height is L L 0 + L L 0 (1 + α T )( m) [ 1+( K 1 )(58.3 K) ] m (a) For the pipe of diameter d 0 (2.00 cm) the area is A 1 4 πd π(2.00 cm) cm 2. (b) The new diameter d is found from Eq. (12.3) (with L replaced by d and L 0 by d 0 ): d d 0 (1 + α T )(2.00 cm) [ 1+( K 1 )( )K ] cm. (c) The new area is A 1 4 πd2,sothe change in area is A A A πd2 1 4 πd2 0 1 [ 4 πd2 0 (1 + α T ) 2 1 ] [ A 0 2α T +(α T ) 2 ] 2αA 0 T 2( K 1 )( )K cm 2, where we neglected the term (α T ) 2 since it is much less than α T (a) Ice and water: T C, boiling water: T C. (b) The assumption is that the thermal expansion of the rod is linear, i.e., the length of the rod increases in proportion to the temperature change: L T.

13 428 CHAPTER 12 THERMAL PROPERTIES OF MATTER (c) The slope is L/ T, which by definition equals αl, with α the coefficient of thermal expansion and L the length of the rod. (d) Find the slope of the line using rise-over-run: L T cm cm C 0.00 C cm/c. Thus α ( L/ T )/L ( cm/c )/4.00 cm /C. (e) Checking Table 12.2 for α, wefind the closest match to be brass. (f) As the temperature changes from 0.0 Cto60.0 Cwehave T 60.0 C 0.0 C 60.0C, and so L αl T ( /C )( cm)(60.0 C ) cm, and so its length at 60.0 Cis cm cm cm (a) We may heat up the brass tube to increase its diameter. (b) d αd 0 T, where d 0 (3.995 cm) is the initial diameter of the tube. (c) The diameter of the rod is greater than that of the tube by cm cm cm, so we need to increase d 0 by d cm. The corresponding change in temperature of the tube is then T d αd cm ( K 1 )(3.995 cm) 66.2 K 66.2 C. (d) The new temperature of the tube should be T i + T 20 C C 86 C (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 thermal expansion. (c) ρ 0 V 0 is the initial amount of mass of the sample before its temperature changes, while ρv is its mass afterwards. Since the mass m of the sample is independent of the temperature, ρ 0 V 0 ρv m. (d) If the initial volume is V 0 before the temperature change, then afterwards V V 0 (1+β T ). Thus ρ m V ρ 0 V 0 V 0 (1 + β T ) ρ 0 1+β T ρ (1 β T ), 0 where in the last step we made use of the approximation (1+x) 1 1 x for x β T 1.

14 CHAPTER 12 THERMAL PROPERTIES OF MATTER 429 (e) In this case T 120 C 20 C 100 C 100 K and β K 1,so ρ ρ 0 (1 β T ) ( kg/m 3 ) [ 1 ( K 1 )(100 K) ] kg/m Similar to the previous problem, we use Eq. (12.3), L αl 0 T,tofind L, the change in length for the steel tape. Here α K 1 for steel, L m, and T T f T i, where T i 20 C and T f 5 9 ( ) 36.0 C. Thus the new length of the tape is L L 0 (1 + α T )( m) [ 1+( K 1 )( )K ] m, which is longer than the correct length by m, or 2 mm. This is not likely to trouble the carpenter. (Note that we used 1 K 1C in finding T.) For simplicity, imagine that the sheet of area A 0 assumes the shape of a rectangle of side lengths a 0 and b 0. Then A 0 a 0 b 0. Asaresult of a temperature change T the side lengths increase by a αa 0 T and b αb 0 T, respectively; and so the new area of the sheet is A (a 0 + a)(b 0 + b) (a 0 + αa 0 T )(b 0 + αb 0 T )a 0 b 0 (1 + α T ) 2 A 0 [1 + 2α T + α 2 ( T) 2 ] A 0 (1+2α T), where in the last step we neglected the term proportional to (α T ) 2 since α T 1. The change in area for the sheet is then A A A 0 A 0 (1+2α T ) A 0 2αA 0 T Apply Eq. (12.4) for volume expansion: V βv 0 T. Here β K 1 is the coefficient of volume expansion for mercury, V cm 2 is its initial volume at 32 F, and T T C 5 9 T F 5 9 (T T Ff Fi) 5 9 (212 32) 100 K. (Note that 1 K 9 5 F for temperature change measured in K and in F.) Thus the new volume at 212 Fisgiven by V V 0 + V V 0 (1 + β T ) (0.50 cm 3 ) [ 1+( K 1 )(100 K) ] 0.51 cm Suppose that the initial volume of the mercury in the Pyrex glass is V 0. Asaresult of a temperature change T the mercury (M) expands by V M β M V 0 T, causing the mercury

15 430 CHAPTER 12 THERMAL PROPERTIES OF MATTER level to rise. Here β M is the coefficient of volume expansion for mercury. Meanwhile the Pyrex glass (P) holding the mercury also expands by V P β P V 0 T, causing the mercury level to fall. Here β P is the coefficient of volume expansion for Pyrex glass. Since these two expansions have opposite effects, the apparent change in volume for the mercury is V V M V P (β M β P )V 0 T β eff V 0 T, where β eff β M β P K K K 1 is the apparent coefficient of volume expansion If the inside diameter of the beaker is D and the height of the mercury column is H, then the original volume of the mercury inside the beaker is V πd2 H. The apparent change in volume for the mercury as a result of the temperature change T is V (β M β P )V 0 T, following the result of the previous problem. Thus V 1 4 πd2 H (β M β P )V 0 T, and so the change in height for the mercury column is H (β M β P )V 0 T πd 2 /4 ( K K 1 )(800.0 cm 3 )(0 95)K π(10.00 cm) 2 / cm 1.5 mm Let the original volume of the mercury inside the thermometer be V 0. Then when the temperature increases by T the change in volume for the mercury is V β M V 0 T, where β M is the coefficient of volume expansion for mercury. Similar to the previous problem, if the bore diameter of hole in the stem containing the mercury is D, then V 1 4 πd2 H, where H is the change in height for the mercury column. Equate the two expressions for V : β M V 0 T 1 4 πd2 H, which gives H β MV 0 T πd 2 /4 ( K 1 )(0.400 cm 3 )(90 10)K π(0.010 cm) 2 /4 74cm Let the initial volume of the water (W) in the bowl be V 0.Asaresult of a temperature change T the water expands by V W β W V 0 T, where β W is the coefficient of volume expansion for water. Meanwhile the Pyrex bowl (P) also expands by V P β P V 0 T, where β P is the coefficient of volume expansion for Pyrex. The new volume of water is therefore V 0 + V W,

16 CHAPTER 12 THERMAL PROPERTIES OF MATTER 431 which is greater than that of the bowl by V (V 0 + V W ) (V 0 + V P ) V W V P (β W β P )V 0 T, which in turn is equal to the amount of water that will overflow: V (β W β P )V 0 T ( K K 1 )(100 cm 3 )(50 10)K 0.79 cm The diameter D of the hole expands in accordance with Eq. (12.3): D αd 0 T. Here D cm is the original diameter, α K 1, and T T C K. Thus D αd 0 T ( K 1 )(2.000 cm)(80 K) cm, and the new diameter is D D 0 + D cm cm cm The initial density of the iron block of mass m and volume V i is ρ i m/v i. When the temperature of the iron block increases by T its final volume is V f V i + βv i T,soits new density is ρ f m V f m V i (1 + β T ) ρ i 1+β T 7.85 g/cm 3 1+( K 1 )(0 20)K 7.86 g/cm kg/m At 273 K (0 C) the density of water (W) is ρ W kg/m 3, while that of ice (I) is ρ I 917 kg/m 3. Since the mass m is unchanged as water freezes into ice, m ρ W V W ρ I V I, which gives the volume of the resulting ice to be V I ρ WV W ρ I (999.8 kg/m3 )(1.00 m 3 ) 917 kg/m m The period τ 0 of the pendulum of length L 0 at Cisgiven by L τ 0 2π 0 g 2π m 9.81 m/s s.

17 432 CHAPTER 12 THERMAL PROPERTIES OF MATTER When the temperature drops by T C C C K, the length of the pendulum changes to L L 0 + L L 0 + αl 0 T (where α is the coefficient of linear expansion for aluminum); and so the new period is L τ 2π g 2π 2π s <τ 0, meaning that the clock runs slightly faster. L 0 (1 + α T ) g m [1 + ( K 1 )( 20 K)] 9.81 m/s Similar to the previous problem, at C the period τ 0 of the pendulum of length L 0 is given by τ 0 2π L 0 /g; and after a temperature change T the new period is ( ) L τ 2π g 2π L 0 (1 + α T ) L 2π 0 1+α T τ0 1+α T g g (1.000 s) 1+( K 1 )( )K s >τ 0, meaning that the clock will run slightly slower The initial density ρ 0 of the specimen of mass m and volume V 0 is given by ρ 0 m/v 0. After a temperature change T, V 0 becomes V V 0 + V V 0 + βv 0 T, and the new density is ρ m ( ) V m m V 0 (1 + β T ) 1 1+β T ρ 0 1+β T. V 0 Since β is small, β T 1; sowemay apply the Binomial Theorem, which states that (1+x) n 1+nx for x 1. With x β T and n 1, wehave (1+β T ) 1 1 β T. So finally ρ ρ 0 (1 + β T ) 1 ρ 0 (1 β T ) Consider an object with initial length L 0 undergoing a length change L as a result of a temperature change T : L αl 0 T. The Young s modulus Y of the object is defined as Y stress/strain, so the Thermal stress in question is given by ( ) ( ) L αl0 T Thermal stress Y strain Y Y Yα T. L 0 L 0

18 CHAPTER 12 THERMAL PROPERTIES OF MATTER Use the result of the previous problem to find the stress in the concrete slabs due to thermal expansion: stress Yα T ( Pa)( K 1 )(42 5)K 9.3 MPa. Concrete is strong in compression, with an ultimate compressional strength of MPa (see Table 10.1), so it is more likely to buckle than to break (a) The relationship between P and V of an ideal gas in an isothermal process (during which T constant) is Boyle s Law: PV constant. (b) Since the product of P and V is unchanged, P increases as V decreases. (c) From P 1 V 1 P 2 V 2 and P 2 3P 1 we get V 2 P 1 V 1 P 2 P 1 V 1 3P V The 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 the gas must increase. So the piston moves out. (d) From V 1 /T 1 V 2 /T 2 and T 2 2T 1 we get V 2 V 1T 2 T 1 V 1(2T 1 ) T 1 2V The relationship between T and P of an ideal gas in an constant-volume process (during which V 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 P 1 /T 1 P 2 /T 2 and T T 1 we get P 2 P 1T 2 T 1 P 1(T 1 /2) T 1 P 1 2.

19 434 CHAPTER 12 THERMAL PROPERTIES OF MATTER The 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 unchanged and the piston will not move. (d) From the analysis in part (c) above we know that V 2 V Since the container is sealed V 0, soeq. (12.7) applies: P/T constant. Initially, P i 1.00 atm MPa and T i 273 K, and finally T f 730 K. The final pressure P f can then be obtained from P i /T i P f /T f,as P f P i T f T i (0.101 MPa)(730 K) 273 K MPa According to the hint given in the problem statement we may use Eq. (12.5), PV constant. Initially, P P i MPa and V V i, and finally V f 1 10 V i.thus from P iv i P f V f P f P i V i V f (0.101 MPa)V i 1 10 V i 1.0 MPa Apply Boyle s Law, Eq. (12.5), for constant temperature: PV constant. Let the absolute pressure at the bottom of the lake be P and the volume of the bubble down there be V. As the bubble rises to the surface of the lake its pressure becomes P s ( 1.0 atm) and its volume becomes 3V.ThusPV P s (3V ), or P P s(3v ) V 3P s 3(1.0 atm) 3.0 atm The initial absolute pressure of the gas is P i P A +20.0atm 21.0atm, and the initial volume is V i 1.00 m 3. After the transfer the absolute pressure becomes P f, and the volume

20 CHAPTER 12 THERMAL PROPERTIES OF MATTER 435 becomes V f V i m m 3 (since the gas is uniformly present in both the tank and the chamber). For T constant we have P i V i P f V f,so P f P i V i V f (21.0 atm)(1.00 m3 ) m atm. The final reading of the gauge pressure is then P G 2.10 atm 1.0 atm 1.1 atm Since the pressure is a constant, Eq. (12.6) applies: V i /T i V f /T f constant. With V i m 3, T i 273 K, and T f 300 K, the final volume is V f V it f ( m 3 )(300 K) T i 273 K m Similar to the previous problem, the pressure of the gas is kept constant so V i /T i V f /T f constant, where V i 300 cm 3, T i K, and V f 200 cm 3. Thus the final temperature is T f T iv f ( K)(200 cm3 ) V i 300 cm K Apply Eq. (12.8) to two states of the gas, labeled i and f, respectively: P i V i /T i P f V f /T f constant, where P i 81.3 kpa, V i 5.50 liters m 3, T i K, P f 1.00 atm 101 kpa, and T f K. Thus the final volume of the gas is V f or 4.0 liters. ( )( ) Pi Tf V i ( m 3 ) P f T i ( 81.3 kpa 101 kpa )( ) ( K m 3, K Apply Eq.(12.8) to the initial state of the helium gas (labeled with subscript i) and the final state at STP (labeled with f): P i V i /T i P f V f /T f constant. Here P i 99kPa, V i 1200 cm 3, T i K, P f 1.00 atm 101 kpa, and T f K. Solve for the final volume of the gas: V f ( )( ) ( )( ) Pi Tf 99 kpa ( K V i (1200 cm 3 ) cm 3. P f T i 101 kpa K

21 436 CHAPTER 12 THERMAL PROPERTIES OF MATTER One mole of oxygen gas occupies a volume of V 22.4liters m 3 at STP. Since the molecular mass of an O 2 molecule is u 32.0u, the mass of one mole of oxygen gas is m 32.0g kg. Thus its density is ρ m V kg m kg/m Awater molecule, H 2 O, is made of two hydrogen atoms and one oxygen atom. Thus its mass is m [2(1.008 u) u]( kg/u) kg The volume occupied by one mole of ammonia gas at STP is V 22.4 liters m 3. Since the mass of this much ammonia gas is m g kg, its density is ρ m V kg m kg/m Each oxygen molecule, O 2,ismade of two oxygen atoms, each of mass u [see Problem (12.67)]. Thus its molecular mass is m [2( u/molecule)] ( kg/u) kg/molecule, and the number of oxygen molecules with a total mass of M 16.0 kg is then N M m which corresponds to 16.0 kg kg/molecule molecules, n N N A molecules molecules/mol 500 mol The density of oxygen gas at STP is found in Problem (12.66) to be ρ 1.43 kg/m 3. The volume 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/m m3.

22 CHAPTER 12 THERMAL PROPERTIES OF MATTER The pressure of the balloon is always equal to the atmospheric pressure, which is assumed to be aconstant. Thus Eq. (12.6) applies: V i /T i V f /T f, where in our case V i m 3, T i K, and T f K. Thus the final volume of the balloon at 20 Cis V f V T i f (1.000 m3 )( K) 1.1m 3. T i K Apply Eq. (12.8) to the initial and final states of the hydrogen gas: P i V i /T i P f V f /T f constant. Here P i 2.00 atm, V i 30.0liters, T i 273 K, P f 3.00 atm, and V f 15.0liters. Solve for the final temperature of the gas: ) T f T i ( Pf P i )( Vf V i (273 K) ( 3.00 atm 2.00 atm )( ) 15.0 liters 205 K liters Apply Eq. (12.8) to states 1 and 2 of a given amount of ideal gas: P 1 V 1 /T 1 P 2 V 2 /T 2. Let the mass of the gas in question be m, then m ρ 1 V 1 ρ 2 V 2,orV 1 m/ρ 1, V 2 m/ρ 2. Substitute these expressions for V 1 and V 2 into the first equation above to yield P 1 (m/ρ 1 ) T 1 P 2(m/ρ 2 ) T 2. Cancel m from both sides and the identity in the problem statement follows One mole of hydrogen gas occupies a volume of V 22.4liters m 3 at STP. Since the molecular mass of an H 2 molecule is u 2.0u, the mass of one mole of hydrogen gas is m 2.0g kg. Thus its density is ρ m V kg m kg/m3. As the pressure remains at 1.00 atm while the temperature increases to T 273 C ( )K 546 K, i.e., double the value at STP, the volume of the gas doubles while its density is halved: ρ 1 2 (0.089 kg/m3 )0.045 kg/m (a) For a given amount of ideal gas at constant pressure, its volume V is proportional to its absolute temperature: V/T constant.

23 438 CHAPTER 12 THERMAL PROPERTIES OF MATTER (b) The device measures the temperature of the gas (which equals that of the environment upon thermal equilibrium) by varying its volume V while keeping its pressure fixed (as the mercury drop remains in the same position). First, calibrate the thermometer by submerging it in a thermal bath of known absolute temperature, say T 0,atwhich the volume of the gas reads V 0. Then we may use it to measure the absolute temperature T of another thermal bath by finding the new volume V of the gas which keeps the mercury drop in the same position (which indicates that the pressure remains the same): V 0 /T 0 V/T,or ( ) V T T 0. V 0 (c) Since the capillary tube is open and the mercury drop does not move, the pressure of the gas remains the same. (d) Since the mercury drop does not move, the pressure P exerted on it from the gas sealed by it must equal to that form the outside air. So P P A 1.0atm. (e) Apply the formula in part (b) above, with the reference volume at V cm 3, the reference temperature at T 0 0 C ( ) K K, and the new volume at V cm 3 : ( ) ( V cm 3 ) T T 0 ( K) V cm K, or ( ) C 55 C (a) The relationship between P and V of an ideal gas in an isothermal process (during which T 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 P 1 V 1 P 2 V 2 to find P 2. Note that P 1 1.9atm +1.0atm 2.9atm is the absolute pressure of the gas before the compression. Thus P 2 P 1 V 1 V 2 (2.9 atm)(590 cm3 ) 181 cm atm, and so the final gauge pressure is 9.5 atm 1.0 atm 8.5 atm. (Note that in the Ideal Gas Law, PV/T constant, P is the absolute pressure, not the gauge pressure.) The absolute pressure at a depth h below the surface of the liquid is P i P s + ρgh, where P s 1.0atm. As the bubble rises to the surface of the liquid the final absolute pressure is P f P A. Apply Eq. (12.5) for constant temperature: P i V i P f V f. This gives V f P iv i (P ( A + ρgh)v i 1+ ρgh ) V P f P A P i. A

24 CHAPTER 12 THERMAL PROPERTIES OF MATTER The volume of the gas inside the tire is unchanged, so Eq. (12.7) applies: P/T constant. Initially, P i (33 lb/in. 2 ) [ Pa/(lb/in. 2 ) ] Pa and T i ( ) K, and finally T f ( ) K. Thus the final pressure P f follows from P i /T i P f /T f to be P f P T i f ( Pa)(321.9 K) T i K Pa (a) At a depth h below the mercury level the gauge pressure caused by the weight of the mercury column is P G ρ Hg gh. (b) The barometer reading gives the air pressure P A in the room: P A ρ Hg gh The absolute pressure P at a depth h below the mercury level is the sum of the air pressure and the gauge pressure due to the mercury column: P P A + P G ρ Hg gh + ρ Hg gh ρ Hg g(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 on the right is constant. Thus its pressure P and its volume V are related by P V C, where C is a constant. Since V AL, P C/V C/AL. Toachieve mechanical equilibrium in the two arms of the U-tube we must require P P, i.e., ρ Hg g(h + h) C/AL, which we rewrite as ( ) C 1 H + h ρ Hg ga L, which indicates that H + h varies linearly as a function of 1/L. Fig. P79(b). The slope of the straight line should be C/ρ Hg ga. This explains the plot in (a) The relationship is the Ideal Gas Law: PV/T constant. (b) and (c) Use P 1 V 1 /T 1 P 2 V 2 /T 2 to find the new volume V 2 of the gas as its pressure drops from P kPa to P kPa while its temperature drops from T 1 to 1 2 T : 1 Since V 2 <V 1 the piston moves in. V 2 P 1V 1 T 2 P 2 T 1 (49.8 kpa)(0.58 m3 )( 1 2 T 1) (39.8 kpa)t m 3.

25 440 CHAPTER 12 THERMAL PROPERTIES OF MATTER The situation described in the problem statement is depicted in the diagram to the right. To achieve mechanical balance the gas pressure P i in the graduated cylinder before cooling must satisfy P i + ρ M gh P A, where ρ M is the density of mercury, h is the height of the mercury column ( 3.90 cm), and P A kpa is the air pressure outside. The temperature of the gas then is T i K (room temperature), and its volume is V i 214 cm 3. After cooling the new pressure is P f kpa, the new temperature is T f K, and the new volume of the gas V f must satisfy Eq. (12.8): P i V i /T i P f V f /T f,or P A mercury P i h V f ( )( ) ( )( ) Pi Tf PA ρ V i V M gh Tf P f T i i P f T i [ kpa ( (214 cm 3 3 kg/m 3 )(9.81 m/s 2 ]( ) )( m) K ) kpa 293 K 189 cm With the previous problem in mind, the total pressure P of the water vapor-hydrogen gas mixture inside the beaker satisfies P + ρ W gh P A, where ρ W is the density of water ( kg/m 3 ), h 5.00 cm, and P A 759 mmhg (0.759 m)( kg/m 3 )(9.81 m/s 2 ) Pa. Since the total pressure P is the sum of P H, the pressure due to hydrogen alone; and P V, the pressure of the water vapor ( 1.92 kpa), P H P P V P A ρ W gh P V Pa ( kg/m 3 )(9.81 m/s 2 )( m) Pa Pa 98.8 kpa Label the oxygen gas with subscript O and the nitrogen with N, respectively, and consider the two gases separately. The expansion process for either gas is isothermal (i.e., T constant.),

26 CHAPTER 12 THERMAL PROPERTIES OF MATTER 441 so for the oxygen gas Eq. (12.5) gives P io V io the chamber due to the oxygen gas alone: P fo V fo, which yields P fo, the final pressure in P fo P iov io V fo (5.0 atm)(3.0 liters) 4.0 liters 3.75 atm. Similarly for the nitrogen gas P in V in the final pressure in the chamber, is P fn V fn,sop fn, the contribution from the nitrogen to P fn P inv in V fn (2.0 atm)(1.0 liters) 4.0 liters 0.50 atm. The total pressure P f in the chamber is the sum of the contributions from both gases: P f P fo + P fn 3.75 atm atm 4.3atm Apply Eq. (12.10), PV nrt (the Ideal Gas Law), to solve for V. Here P Pa, n 3.0 mol, and T K. Thus V nrt P (3.0 mol)( J/mol K)(333 K) Pa m Use Eq. (12.10), PV nrt,tosolve for T. Here P Pa, V m 3, and n 1.40 mol. Thus T PV nr ( Pa)( m 3 ) (1.40 mol)( J/mol K) 174 K Use Eq.(12.11), PV Nk B T,tosolve for N, the number of molecules. Here P (10 15 atm) ( Pa/atm) Pa, V 1.0 cm m 3, and T 273 K. Thus N PV k B T ( Pa)( m 3 ) ( J/K)(273 K) That s still about 30,000 molecules left per cubic centimeters Similar to the previous problem, use PV NK B T to solve for N. Here P 1.00 atm Pa, V 1.00 cm m 3, and T 273 K. Thus N PV k B T ( Pa)( m 3 ) ( J/K)(273 K)

27 442 CHAPTER 12 THERMAL PROPERTIES OF MATTER Apply Eq. (12.11), PV Nk B T,tosolve for the number of air molecules, with P Pa, V 0.10 m 3, and T K. Thus N PV k B T ( Pa)(0.10 m 3 ) ( J/K)(293 K) Apply the Ideal Gas law, Eq. (12.10): PV nrt. Here P Pa, V m 3, and T T C K. Thus n PV RT ( Pa)( m 3 ) (8.314 J/mol k)(293 K) 8.21 mol Use Eq. (12.11), PV Nk B T, to solve for N, the number of air molecules. Here P (1.00 atm)( Pa/atm) Pa, V 1.00 m 3, and T 300 K. Thus N PV k B T ( atm)(1.00 m 3 ) ( J/K)(300 K) Apply Eq. (12.10), PV nrt, with V m 3, n mol, and T K. Solve for P : P nrt V which is about 3.1 atm. (0.050 mol)(8.314 J/mol K)(373 K) m Pa, (a) It is the Ideal Gas Law: PV Nk B T, where N is the number of gas molecules and k B is the Boltzmann Constant. This is equivalent to PV nrt, with n N/N A the number of moles of the gas and R k B N A the Universal Gas Constant. (b) T ( ) K 293 K. (c) V 2.0liters cm 3 ( cm 3 )(10 2 m/cm) m 3. (d) Solve for P from the Ideal Gas Law: P nrt V (4.0 mol)( J/mol K)(293 K) m Pa.

28 CHAPTER 12 THERMAL PROPERTIES OF MATTER 443 Thus the gauge pressure is P P Pa Pa Pa (a) It is the Ideal Gas Law: PV Nk B T, where N is the number of gas molecules and k B is the Boltzmann Constant. This is equivalent to PV nrt, with n N/N A the number of moles of the gas and R k B N A the Universal Gas Constant. (b) T ( ) K 347 K. (c) V 4.0liters cm 3 ( cm 3 )(10 2 m/cm) m 3. (d) P P G + P atm +1.0atm 3.0 atm (3.0 atm)( Pa/atm) Pa. (d) Solve for n, the number of moles, from the Ideal Gas Law: n PV RT ( Pa)( m 3 ) ( J/mol K)(347 K) 0.42 mol. Multiply this by the Avogadro s Number to obtain the number of gas molecules: N nn A (0.42 mol)( /mol) According to the result of Problem (12.90) there are N air molecules in a volume of V 1.00 m 3. Thus the average volume occupied by each molecule is v V/N. If this volume is cubic, then the side length l of the cube, which is also the center-to-center distance between air molecules, satisfies v l 3. Solve for l: l v 1/3 ( ) 1/3 ( ) V 1.00 m 3 1/3 N m 3.45 nm, which is about an order of magnitude greater than the size of individual air molecules Apply Eq. (12.11), PV Nk B T,toestimate N, the number of air molecules per breath. Here P 761 mmhg (0.761 m)( kg/m 3 )(9.81 m/s 2 ) Pa, V m 3, and T K. Thus N PV k B T ( Pa)( m 3 ) ( J/K)(310 K) , which is about the same as the number of stars there are in the entire Universe From Eq. (12.10), PV nrt,wefind the number of moles per unit volume to be n/v P/RT. Let the molar mass of nitrogen be m N (28.0g/mol), then the mass of n moles of

29 444 CHAPTER 12 THERMAL PROPERTIES OF MATTER nitrogen is nm N, and the density of nitrogen is given by ρ N nm N /V m N P/RT. Solve for T : T m NP ρ N R ( kg/mol)( Pa) (1.245 kg/m K. )( J/mol K) First find n, the number of moles of helium gas, from Eq. (12.10): n PV RT ( Pa)( m 3 ) (8.314 J/mol K)( )K mol. Since the molar mass of helium gas is 4.0 g/mol, this amounts to (4.0 g/mol)( mol) 0.33 gofhelium For constant pressure, apply Eq. (12.6): V i /T i V f /T f. Here V i 600 cm 3, T i K, and V f 750 cm 3. Solve for T f : T f T i V f V i ( K)(750 cm3 ) 600 cm K First find n, the number of moles of hydrogen gas needed, from PV nrt. Here P Pa, V 2000 m 3, and T ( 55) 218 K; so n PV RT ( Pa)(2000 m 3 ) ( J/mol K)(218 K) mol. Since the molar mass of hydrogen is kg/mol, the total mass of hydrogen needed is m ( kg/mol)( mol) 43kg The volume coefficient β for a substance of volume V at temperature T is defined in the equation V βv T.Inthis case the substance we are dealing with is the ideal gas, which satisfies PV nrt,orv 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 1 T, where in the last step we again used PV nrt.

30 CHAPTER 12 THERMAL PROPERTIES OF MATTER The average molecular speed is v av 1 (1.0 m/s +2.0m/s +3.0m/s +4.0m/s +5.0m/s) 3.0m/s The rms -speed of the molecules is 1 v rms 5 [(1.0 m/s)2 +(2.0m/s) 2 +(3.0m/s) 2 +(4.0m/s) 2 +(5.0m/s) 2 ]3.3m/s, which is greater than v av computed in the previous problem. In fact this is an example of the general result v rms v av, with the equality valid only when all of the individual speeds are identical to each other Use Eq. (12.16), v rms 3k B T/m,tofind v rms.for oxygen molecules with a molecular mass of m kg at T K, 3kB T v rms m 3( J/K)( K) m/s. kg Use Eq. (12.15), KE av 3 2 k T, with T K: B KE av 3 2 k BT 3 2 ( J/K)( K) J Apply Eq. (12.15), with T K: KE av 3 2 k BT 3 2 ( J/K)(293 K) J Use Eq. (12.16), with T K and m (44.0 g/mol)/n A ( kg/mol)/( mol) kg (note that the molar mass of CO 2 is 44.0 g/mol). Thus 3kB T v rms m 3( J/K)( K) m/s. kg

31 446 CHAPTER 12 THERMAL PROPERTIES OF MATTER (a) The relationship is given by Eq. (12.14): KE av 3 2 k T. B (b) Since KE av T,asT is doubled so is KE av. (c) Solve for T from Eq. (12.14): T 2KE av 3k B 2( J) 3( J/K) 34K. (d) T C T C. (e) The average molecular kinetic energy is directly proportional to the absolute temperature of an ideal gas: KE av 3 2 k BT.Aslong as KE av is fixed, so is T, regardless of whether the gas is nitrogen or hydrogen (a) The relationship is given by Eq. (12.14): KE av 3 2 k BT. (b) KE esc 1 2 mv2 1 esc 2 ( kg)( m/s) J. (c) Let KE av 3 2 k BT KE esc and solve for the absolute temperature T : T 2KE esc 3k B 2( J) 3( J/K) K, (d) T C (e) ( ) C 9384 C, or Cto2significant figures. T F 9 5 T C (9384) F. (f) At a given temperature the average translational KE of different kinds of molecules is the same. Hydrogen has the smallest value of m so for the same average KE it has the greatest value of average speed, which makes it more likely for hydrogen to reach the escape speed than any other type of molecules The average translational KE per oxygen molecule at T Kisgiven by Eq. (12.14): KE av 3 2 k BT. The number of such molecules, N, can be found from Eq.(12.11), PV Nk B T, where P 1.00 atm Pa and V 0.50 m 3. Combine these two equations to find the total translational KE: ( )( ) PV 3 KE total NKE av k B T 2 k BT 3 2 PV 3 2 ( Pa)(0.50 m 3 ) J 76kJ.