Chapter 10. Answers to Even Numbered Problems. 2. (a) 251 C. (b) 1.36 atm C, C. 6. (a) 273 C (b) 1.27 atm, 1.74 atm

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1 hapter Answers to Even Numbered Problems. (a) 5 (b).6 atm , (a) 7 (b).7 atm,.74 atm 8. (a) 8 F (b) 45 K. (a) 6 (b) 6. (a) L. m.49 mm (b) fast m..5 km, accordion-like expansion joints at periodic intervals. (a). mm (b) 96 N 4. (a) 6. (a) 6.5 Pa (b) It will not fracture cm (b).94 cm 8. (a). mol (b) 4.8 molecules. 884 balloons.. kg m m kn 8. (a).74 kj m ol (b).9 km s.7 km s 4. (a) ( v rm s) (b) H ( rm s) O v.69 km s (c) Hydrogen escapes; carbon dioxide does not Pa 7

2 74 HAPTER kpa 46. (a).4 cm (b) cm (c). cm m 5..4 m (a) 4 K (b).5% of the original mass 56. L steel 4. cm, L copper 9. cm 58. (a) 6.9 cm (b) 5.5 Pa 6..5 atm 6. (a) 6. cm (b) The stress on the span would be Pa, so it will not crumble

3 Thermal Physics 75. (a) onverting from elsius to Fahrenheit, 9 9 TF T + ( 5.87) + 4 F 5 5 and converting to Kelvin, T T K F 5 5 (b) T T F and T T K.6 Since we have a linear graph, we know that the pressure is related to the temperature as P A + BT, where A and B are constants. To find A and B, we use the given data:.9 atm A + B( -8. ) () and.65 atm A + B 78. () Solving equations () and () simultaneously, we find: A.7 atm, and B 4.65 atm P.7 atm atm T Therefore, (a) At absolute zero the gas exerts zero pressure ( P ), so T.7 atm atm (b) At the freezing point of water, T and P.7 atm +.7 atm At the boiling point of water, T, so P.7 atm atm.74 atm.8 [Note that some rules concerning significant figures are deliberately violated in this solution to better illustrate the method of solution.] Let L be the final length of the aluminum column. This will also be the final length of the quantity of tape now stretching from one end of the column to the other. In order to determine what the scale reading now is, we need to find the initial length this quantity of tape had at. (when the scale markings were presumably painted on the tape). L and require that Thus, we let this initial length of tape be ( ) tape L ( L ) + α ( T ) ( L ) + α ( T ) ( L ), which gives tape steel colum n al tape ( L ) colum n + α al T + α ( T) steel

4 76 HAPTER or ( L ) 6 ( 8.7 m ) + ( 4 ( ) )( 9.4. ) 6 + ( ( ) )( 9.4. ) tape 8.7 m. (a) As the temperature of the pipe increases, the original 5.-m length between the water heater and the floor above will expand by 6 L αl ( T) ( 7 )( 5. m )( 46 ). m If this expansion occurs in a series of 8 ticks, the expansion per tick is ( ) 4 movem entper tick L 8. m 8. m. mm (b) When the pipe is stuck in the hole, the floor exerts a friction force on the pipe preventing it from expanding. Just before a tick occurs, the pipe is compressed a distance of. mm. The force required to produce this compression is given by Y F A L L, as the equation defining Young s modulus, 4 L F YA L 5. m Pa.55 m 96 N 5. m.5 (a) The gap width is a linear dimension, so it increases in thermal enlargement as the temperature goes up. (b) At 9, the length of the piece of steel that is missing, or has been removed to create the gap, is L L + L L α + ( T ). This gives 6 L.6 cm cm.7 (a) From the ideal gas law, nrt, we find P nr. Thus, if both n and V are T V constant as the gas is heated, the ratio PT is constant giving P T P P P i T T K 9 K 67 P P f i f or f i f Ti i (b) If both pressure and volume double as n is held constant, the ideal gas law gives: f f ( Pi)( V ) i Tf Ti Ti 4Ti 4( K) K 97 i i i i i.9 (a) 5 6 (. Pa atm )(. m ) n RT ( 8. J m ol K )( 9 K ) 5 4. mol

5 Thermal Physics 77 Thus, N n N A molecules ( ) mol mol 6..5 molecules (b) Since both V and T are constant, n n RT P, or n RT P P. Pa -5 5 ( 4. mol ) 4. n mol P. Pa. From the ideal gas law, nrt, with n n, we have 5.8 Pa.7 m 5. Pa.5 m T T K 56 K 87. With n held constant, the ideal gas law gives V P T. atm K 4.5 V P T. atm K Since the volume of a sphere is V ( 4 ) r, V V ( r r) V 4.5 m 7. m Thus, r r V π.4 The pressure at a depth of m in the ocean is P P +ρ gh atm Pa 5 kg m 9.8 m s m. Pa 5 At pressure P atm. Pa, the air in the bell occupies a volume V ( πr ) h π(.5 m ) ( 4. m ) 8. m At the ocean bottom, the volume of this air will be 5 P T. Pa 78 K V V 6 ( 8. m ).6 m P T. Pa 98 K V.6 m The height of this cylindrical volume is h.64 m πr π (.5 m ) so the height the water will rise inside the bell as it sinks to the bottom is h h h 4. m.64 m.84 m.7 The average kinetic energy of the molecules of any gas at K is

6 78 HAPTER J K KE mv kb T.8 K 6. J.4 onsider a time interval of. min 6 s, during which 5 bullets bounce off Superman s chest. From the impulse-momentum theorem, the magnitude of the average force exerted on Superman is F av I 5 p 5 m v t t t ( v ) bullet ( ) ( ) 5 8. kg 4 m s 4 m s 6 s 6 N.45 As the pipe undergoes a temperature change T , the expansion of the horizontal segment is L αl T x x The expansion of the vertical section is 6 Ly αly( T) 7 ( ) ( 4 cm )( 8.5 ).649 mm cm cm.6 mm The total displacement of the pipe elbow is at or L Lx + Ly.6 mm mm.66 mm L y.649 mm θ tan tan 78. L x.6 mm L r.66 mm at78. below the horizontal.5 When gas the supports the piston in equilibrium, the gauge pressure of the gas is F mg ( 5. kg)( 9.8 m s ) P gauge 9.8 Pa A A.5 m, and the absolute pressure is 5 P P + P Pa atm gauge The ideal gas law gives the volume as V nrt P, so the height of the cylindrical space is V nrt (. mol)( 8. J m ol K )( 5 K ) h.4 m A P A. 5 Pa+ 9.8 Pa.5 m.5 (a) The volume of the liquid expands by V βv ( T) and the volume of the glass liquid

7 Thermal Physics 79. The amount of liquid that must overflow flask expands by V ( α ) V ( T) flask into the capillary is V V V V ( β α)( T). The distance the overflow liquid flask liquid will rise into the capillary is then Voverflow V h ( β α)( T) A A 4 (b) For a mercury thermometer, 6 6 α glass (. ( ) ) 9.6 ( ) β Hg.8 and (assuming Pyrex glass),. Thus, the expansion of the mercury is almost tim es the expansion ofthe flask, making it a rather good approximation to neglect the expansion of the flask..55 After expansion, the increase in the length of one span is. L α L T 6 5 m.. m giving a final length of L L + L 5 m +. m L L 5 m y From the Pythagorean theorem, y L L 5 +. m 5 m.74 m P V T.58 (a) From the ideal gas law,, or T T P V T The initial conditions are: P atm, V 5. L 5. m,and T. 9 K The final conditions are: F k h P atm + atm +, V V+ A h,and T 5 5 K A A Thus, kh A h 5 K + + A atm 9 K V. N m h. m h 5 or (. m )(. N m ) ( 5. m ) 9 Simplifying and using the quadratic formula yields

8 8 HAPTER h.69 m 6.9 cm (b) kh P atm + A (. N m )(.69 m ) 5 5. Pa+.5 Pa. m.6 Let container be maintained at T T 7 K, while the temperature of container is raised to T 7 K. Both containers have the same constant volume, V, and the same initial pressures, P P P. As the temperature of container is raised, gas flows from one container to the other until the final pressures are again equal, P P P. The total mass of gas is constant, n + n n + n () so From the ideal gas law, n, so equation () becomes RT P + +, or P RT RT RT RT + T T T Thus, P TT. ( atm ) 7 7 P T T + T atm.6 Assume that you fill a -gallon container with gasoline when the temperature is. When the temperature decreases to, your container will not be full because your gasoline has undergone a decrease in volume of 4 - V βv T 9.6 gallon.9 gallon Had you purchased the gasoline when the temperature was, you would have gotten a full gallons, or.9 gallons more than you now have. The increased mass of the gasoline in your container would have been kg.786 L m m ρ( V) 7 (.9 gal ).5 kg m gal L