= = mol J K K mol N m. N/m N/m ( )( ) is the universal gas constant.

Transcription

1 THE THERMAL BEHAVIOR OF MATTER 7 EXERCISES Section 7. Gases 7. INTERPRET This problem involves the ideal-gas law, which we can use to ind the volume o mol o Martian atmosphere given its temperature and pressure. DEVELOP Apply Equation 7. PV = nrt with P = 0.0P E, T = 5 K, n =.0 mol, and R = 8.4 J/K. EVALUATE Solving the ideal-gas law or the volume and inserting the given quantities gives nrt (.0 mol)( 8.4 J/K)( 5 K) V = = = 5.8 m P Pa ( )( ) The dimensions o this expression are ( ) ( ) mol J K K mol N m = = mol m N/m N/m but a mole is a dimensionless number, so the inal dimensions are m, as expected or a volume. 8. INTERPRET We are dealing with an ideal gas. Given the pressure, temperature, and volume we are to ind the number o gas molecules. DEVELOP We shall use the ideal-gas law, pv = NkT (Equation 7.), to ind the number o molecules N. EVALUATE Inserting the given quantities gives One mole has 5 (.8 0 Pa)( m ) PV N = = =. 0 kt (.8 0 J/K)( 50 K) N A = molecules. Thus, we have about 0.5 mole o molecules in the system. 9. INTERPRET This problem involves an ideal gas, so we can apply the ideal-gas law to ind the pressure o the gas at the given temperature and volume. DEVELOP In terms o moles, the ideal-gas law is given by Equation 7., PV = nrt. The volume is V =.0 L =.0 0 m, and T = 50 C = K. EVALUATE Solving or the pressure and inserting the given quantities gives nrt (.5 mol)( 8.4 J/K)( K) 6 P = = =.8 0 Pa V.0 0 m This is about 0 times standard atmospheric pressure. 0. INTERPRET You want to veriy that the tank you purchased contains the amount o argon that it is supposed to. You can treat it as an ideal gas. DEVELOP You know the volume, the temperature and the pressure o the gas. Using the ideal-gas law (Equation 7.), you can ind the number o moles in the tank: n= pv / RT, where R = 8.4 J/K mol is the universal gas constant. 7-

2 7- Chapter 7 EVALUATE Remembering that the temperature needs to be in Kelvin and the volume needs to be in the number o moles is: pv ( 4 MPa)( 6.88 L) m n = = 40 mol o = RT 8.4 J/K mol 0 C+7 0 L ( )( ) m, This is 5 mol less than the vendor claimed, so you did not get your money s worth. I the right amount o gas were in the tank, then you would have measured a pressure o nearly 6 MPa. Notice that at no point did we have to consider the speciic properties o argon. The ideal-gas law applies to any gas. The units work out since Pa m = N m = J.. INTERPRET This problem involves an ideal gas, so we can apply the ideal-gas law. We are to ind the volume o an ideal gas given its temperature and pressure, then ind its new temperature i the pressure is increased and the volume is cut in hal. DEVELOP In terms o moles, the ideal-gas law takes the orm o Equation 7., pv = nrt. For part (a), p =.5 atm = Pa = Pa, T = 50 K, and n =.0. For part (b), we can take the ratio o the ideal-gas law applied to part (a) an that applied to part (b) to get pv a a nrta Ta = = pv nrt T b b b b which we can solve or T b given that V b = V a / and P b = 4.0 atm. Note that we have used the act that the number o moles does not change. EVALUATE (a) The volume V a o the gas is nrt (.0 mol) 8.4 J/ ( mol K) ( 50 K) V = = 5 P.55 0 Pa =.7 0 m = 7 L (b) Upon compressing the gas and increasing its pressure, the new temperature is 4.0 atm 0.5V Tb = ( 50 K) 0 K.5 atm = V to two signiicant igures. As expected, the temperature increases i we compress the gas.. INTERPRET We treat air molecules as ideal gas. Given the pressure, temperature, and volume, we want to ind the number o air molecules. DEVELOP We shall use the ideal-gas law, pv = NkT, given in Equation 7., to ind the number o molecules. EVALUATE The number o air molecules is 0 PV (0 Pa)(0 m ) 7 N = = = 0 kt (.8 0 J/K)(7 K) system. One mole has N = molecules. Thus, we have about mole o molecules in the A. INTERPRET This problem in an exercise in calculating the thermal speed o ideal-gas molecules at a given temperature. DEVELOP The thermal speed (also called the rms, or root-mean-square speed) is, rom Equation 7.4, vth = kt m, where m is the mass o a molecule. From Appendix C we estimate the mass o a H molecule to be kg.

3 The Thermal Behavior o Matter 7- EVALUATE Inserting the given quantities into Equation 7.4 gives v ( )( ).8 0 J/K 800 K = =.6 km/s.66 0 kg ( ) th 7 This is about 00 times aster than the standard speed o sound (4 m/s). 4. INTERPRET In this problem we want to compare the thermal speeds o two dierent molecules that are at dierent temperatures. DEVELOP The thermal speed o a molecule is given by Equation 7.4: kt vth = m where T is the temperature and m is the mass o the gas molecule. We can use this to compare the thermal speeds o the two molecules in question. EVALUATE Comparing the thermal speeds or H ( mh = u) and SO ( mso = 64 u) at the given temperatures, we ind vth(h ) TH m SO 75 K 64u = = =.6 v (SO ) T m 50 K u th SO H So hydrogen is aster. The thermal speed o a gas molecule is proportional to T and inversely proportional to m. Section 7. Phase Changes 5. INTERPRET This problem involves the latent heat o usion, which is the energy it takes to liberate the molecules that compose the ice to orm water. We are asked to ind the energy required to melt a 65-g ice cube. DEVELOP The energy required or a solid-liquid phase transition at the normal melting point o water (0 C) is (Equation 7.5) Q = ml, where m = kg and L = 4 kj/kg (rom Table 7.). EVALUATE The heat required to melt the ice cube is This is equivalent to 5. kcal. (See Table 7. or the heats o transormation.) Q= ml = ( kg)( 4 kj/kg) = kj 6. INTERPRET This problem is about melting, and it involves the heat o usion. We want to identiy the substance in question given its mass and the energy required to melt it. DEVELOP Using Equation 7.5, we ind the heat o transormation rom solid to liquid to be Q 00 J L = = = 5 kj/kg m kg Compare this result with the data given in Table 7. to determine the substance. EVALUATE The closest igure in Table 7. is the heat o usion or lead L = 5 kj/kg, so the substance is lead. The heat o usion L is a chemical property o a material. Thus, knowing L allows us to identiy the material. 7. INTERPRET This problem involves the latent heat o vaporization, which is the energy required to pass rom the liquid to the gas phase. Given the latent heat o vaporization (Table 7.) and the energy required to vaporize the substance, we are to calculate the mass o the substance. DEVELOP Assuming the vaporization takes place at the normal boiling point or oxygen at atmospheric pressure, we may use Equation 7.5, Q = L v m, where L v = kj/kg (rom Table 7.) and Q = 840 kj.

4 7-4 Chapter 7 EVALUATE The mass o oxygen in the sample is Q 840 kj m = = =.9 kg L kj/kg v Given that O is approximately g/mol, this corresponds (at standard temperature and pressure) to a volume o ( 900 g)( 8.4 J/K)( 7 K) nrt V = = =.7 m p 5 ( g/mol)( 0 Pa) 8. INTERPRET This problem is about phase change o CO rom gas to solid. We are asked to ind the heat (i.e., thermal energy) that must be extracted rom the given amount o CO to solidiy it, which is the same as the heat required to sublime it (turn it rom solid into gas). DEVELOP The relevant latent heat o transormation in this process is the latent heat o sublimation, which is L s = 57 kj/kg. Use this in Equation 7.5 to ind the heat needed to solidiy CO. EVALUATE Equation 7.5 gives Q= ml s = ( 0.5 kg)( 57 kj/kg) = 40 kj to two signiicant igures. This is the heat that must be extracted to turn CO gas into rozen CO, or dry ice. To revert to the gaseous state, the same amount o heat must be absorbed. 9. INTERPRET This problems involves the phase transormation o a liquid to a gas, so the latent heat o vaporization comes into play. Because the liquid is at its boiling point, any heat added to the liquid will cause vaporization instead o a temperature rise. DEVELOP From Table 7., the latent heat o vaporization or O is kj/kg. Use this in Equation 7.5, Q = L v m, to ind the heat needed to vaporize 8 kg o O. EVALUATE Inserting the given quantities into Equation 7.5 gives Q= ml v = ( 8 kg)( kj/kg) = 6.0 MJ to two signiicant igures. At standard temperature and pressure (STP, T = 7.5 K, p = 0 5 Pa), 8 kg o O would occupy a volume o nrt ( 8 0 g)( 8.4 J/K)( 7 K) V = = = 5 9 m p g/mol 0 Pa Section 7. Thermal Expansion ( )( ) 0. INTERPRET This is a thermal expansion problem: we are given the length o a wire and the temperature, and we wish to know the change in length at a higher temperature. DEVELOP Thermal expansion is given by Equation 7.7, ΔL α = L Δ T 6 The wire is made o copper, and α = 7 0 K or copper (see Table 7.). The temperature change is ΔT = 8 K and the initial length is L = 0 cm. EVALUATE Inserting the given quantities gives 6 ( )( )( ) Δ L= αlδ T = 7 0 K 0 cm 8 K =. cm A small change, but enough that power and phone lines sag noticeably on hot days.. INTERPRET This is a problem in thermal expansion. We know the initial volume, the material, and the change in temperature, and we need to ind the new volume. DEVELOP The change in volume is given by Equation 7.6, ΔV β = V Δ T

5 The Thermal Behavior o Matter 7-5 The initial volume is V =.00 L. The material is ethyl alcohol, which has a volume coeicient o expansion o 5 β = 75 0 K, and the change in temperature is ΔT = 8 K. The new volume V o the liquid is EVALUATE Inserting the given quantities gives ( β ) V = V +Δ V = V + Δ T 5 ( β ) ( ) ( )( ) V = V + Δ T =.00 L K = L This is enough to observe in your own rerigerator: Seal a volume o liquid at room temperature, put it in the rerigerator, and observe the deormation o the container when it cools.. INTERPRET This problem involves thermal expansion (volume expansion in this case). We are given the diameter o a Pyrex sphere and are to ind its new diameter due to a given temperature change. DEVELOP The volume V o the marble is related to its radius r by V = 4πr /. The change in volume is given by Equation 7.6, ΔV β = V Δ T where β = α = (. 0 6 K ) = K (see Table 7. and Problem 7) and ΔT = 65 C = 65 K. The new volume is V = V +Δ V = V + βδ T so the new diameter will be ( ) 4 4 = = + Δ = + Δ ( ) V πr πr β T d d β T where we have used d = r. EVALUATE Inserting the given quantities into the expression above gives 6 ( ) ( )( ) = + Δ = + = d d β T cm K 65 K.000cm Because we are dealing with a sphere, we can also consider the linear expansion o the diameter using α.. INTERPRET This problem deals with thermal expansion o a steel washer. The quantity o interest is the diameter o the washer, so the relevant quantity is the coeicient o linear expansion, α. DEVELOP The coeicient o linear expansion is deined as (see Equation 7.7): L/ L α = Δ Δ T 6 For steel, its value is (see Table 7.) α = 0 K. Solve this equation or ΔT. EVALUATE From the equation above, we get ΔL 9.55 mm 9.5 mm Δ T = = = 6 K αl 0 6 K 9.5 mm ( )( ) Since the initial temperature is 0 C we must heat the washer to 6 C. Since α is very small, a large increase in temperature results in a small increase in the washer s diameter. 4. INTERPRET This problem involves the thermal expansion o a one-dimensional object, so we can use the linear expansion ormula to ind the total distance a 5000-km rail would expand or a 60 C temperature change. DEVELOP Equation 7.7, L/ L α = Δ Δ T gives the relationship between temperature change and linear expansion. For steel, α = 0 6 K and the temperature change is ΔT = 40 C ( 5 C) = 65 C = 65 K.

6 7-6 Chapter 7 EVALUATE Solving the expression or ΔL and inserting the given quantities gives 6 ( )( )( ) Δ L= αlδ T = 0 K 5000 km 65 K =.9 km PROBLEMS This is a signiicant expansion, but it is spread over the 5000-km distance. 5. INTERPRET The system o interest is the solar corona, which we treat as an ideal gas. The quantity o interest is the number density o air molecules. DEVELOP The number density implied by the ideal-gas law (Equation 7.) is N p pv = NkT = V kt EVALUATE Applying the above equation to the solar corona, we obtain N V corona p 0 Pa = = = 0 m 6 kt (.8 0 J/K)( 0 K) 5 I we assume the Earth s atmosphere has standard temperature and pressure, the particle density is 5 N p.0 0 Pa m V = = = kt (.8 0 J/K)(7 K) STP So the corona is over 0 billion times less dense than on Earth. Scientists are still not entirely certain how the corona ends up being so hot. 6. INTERPRET We re asked to igure out how much a balloon expands as it rises to an altitude o lower pressure and temperature. DEVELOP We ll neglect the tension in the balloon material, so the helium gas will expand until its outward pressure on the balloon matches the inward pressure rom the surrounding air. The balloon s temperature will also come into equilibrium with its surroundings. The helium obeys the ideal-gas law at both the initial and inal altitudes, and since the number o molecules is constant, we have: pv i i pv = T T i EVALUATE Solving or the inal volume, V p T ( ) ( ) ( ) ( ) ( ) 0.65 atm 9 K.0 atm 6 K 8.0 L L i = Vi = = p Ti I the balloon s pressure and temperature are in equilibrium with its surroundings, you might wonder why it rises. By Archimedes s principle rom Chapter 5, the balloon will rise until its weight is equal to the weight o air it displaces. I we neglect the contribution rom the balloon material, the inal volume will be V = mhen/ ρair, where mhe is the mass o a single helium atom, N is the total number o helium atoms and ρ air is the air density, which depends on altitude. 7. INTERPRET The object o interest is the cylinder compressed with air. We are given the pressure, temperature, and volume, and want to ind the number o moles (i.e., the number o air molecules) in the cylinder. DEVELOP We shall treat the air as an ideal gas (although this is somewhat risky at80 atm) and use the ideal-gas law PV = nrt given in Equation 7., to ind the number n o moles. The volume o the cylinder is d V = π h= π( 0.0 m) (.0 m) = 0.0 π m 4

7 EVALUATE (a) Applying the ideal-gas law gives 5 ( 80 atm)(.0 0 Pa/atm)( 0.0 π m ) pv n = = = 5 mol RT ( 8.4 J/K mol)( 9 K) where we have used T = 0 C = 9 K as the room temperature. (b) I the pressure is p = atm, then the volume would be The Thermal Behavior o Matter 7-7 nrt p 80 atm V = = V = ( 0.0 π m ) = 5.65 m p p atm When temperature is held constant, PV = constant or an ideal gas. Thereore, decreasing the pressure increases the volume in a proportional amount. 8. INTERPRET You re arguing that a whipped cream can exploded because o an error on the part o the manuacturer. DEVELOP It s true that the pressure is the relevant parameter, not the temperature. You can assume that the can is in danger o exploding when the pressure is above the value p. max You can estimate this value rom the manuacturer s claim that a ull can should not exceed a temperature o T max = 50 C. By the ideal-gas law, the maximum pressure is related to this temperature, as well as the initial volume and initial number o moles o the propellant: nrt i max pmax = V i You can t calculate the maximum pressure, but you can compare it to pressure, p, o the can that exploded, which had T = 60 C, V > Vi and n< n. i EVALUATE The ratio o the pressure to the maximum pressure is p n Vi T ( K) = < () 0.5 p n V T = K max i max ( ) The can exploded when its pressure was about hal the maximum pressure, so the manuacturer appears to be at ault. By the above reasoning, a hal-ull can could supposedly withstand temperatures as high as about 70 C. However, at such a high temperature the whipped cream will vaporize, thus adding its pressure to that o the propellant. 9. INTERPRET The object o interest is the lask illed with air, which we treat as an ideal gas. We explore the eect o changing temperature and pressure. The maximum pressure in the lask will occur when the gas inside the lask, which is initially at STP, is heated to the boiling point o water (00 C). To ind the number o moles that escape when the lask is opened, we consider that the gas escapes so ast that the temperature o the gas can be considered to be constant on this timescale. DEVELOP When the lask is immersed in boiling water, its volume remains ixed. Thereore, the ideal-gas law pv = nrt (Equation 7.) applied at each temperature gives pv = nrt p T = pv = nrt p T The initial conditions o the gas are p = atm, V =.00 L =.00 0 m, and T = 9 K. The maximum pressure in the lask occurs when T = 00 C = 7 K. For part (b), we irst calculate the number o moles o gas initially in the lask. Again applying the ideal-gas law, we ind the number o molecules to be pv 5 ( atm)(.0 0 Pa/atm)( 0 m ) = = = RT 8.4 J/ K mol 9 K n ( ) ( ) 0.5 mol

8 7-8 Chapter 7 When the lask is opened at T = 7 K the pressure rapidly decreases to p = atm, so the quantity o gas remaining in the lask is pv T n = = n RT T so the number o moles that escaped rom the lask is Δn = n n. Ater the lask is closed and cooled back down to T = 0 C = 9 K, we again apply the ideal-gas law using n to ind the new pressure. This gives nrt n p = = p V n EVALUATE (a) From the equation above, we ind the maximum pressure reached in the lask to be T 7 K p = p = ( atm ) =.7 atm T 9 K (b) Ater opening the lask the quantity o gas let in the lask is T 9 K n = n = ( 0.5 mol ) = mol T 7 K Thereore, the amount that escaped is Δ n= n n = mol. (c) Ater sealing the lask and cooling it to 9 K, the pressure o the gas in the lask is nrt n mol p = = p = ( atm) atm V n = 0.5 mol As expected, the pressure in the lask is greatest or part (a), when the temperature is highest and there are the most moles o gas in the lask, and the pressure is the lowest or part (c), when the reverse is true. Pressure is proportional to the number o molecules in the volume, so ater some gas molecules escape rom the lask, the pressure decreases. 40. INTERPRET We want to know how long it will take to warm up a rostbitten hand. We can calculate the total energy needed and divide by the rate that energy is absorbed rom the water bath. DEVELOP We model the rostbitten hand as pure ice with mass m. A certain amount o heat will be needed to irst melt the ice: Q = L m, which is Equation 7.5 with L = 4 kj/kg or water. Aterward, the water will have T = 7 C. This will require Q = mcδt, where c = 4.84 kj/kg K rom Table 6.. The time to supply the ull heat will be: to be warmed rom the reezing temperature ( T = 0C ) to the normal body temperature ( ) ( ) t = Q+ Q / P. EVALUATE The melting and warming steps will require energies o: Q = Lm= ( 4 kj/kg)( 0. kg) = 40. kj Q = mcδ T = 0. kg 4.84 kj/kg K 7 K = 8.6 kj ( )( )( ) Using the rate that heat is conducted into the hand, Q+ Q 40. kj kj t = = = 7 s P 800 W This seems reasonable, assuming the bath supplies heat to the hand at a constant rate. In act, the outside o the hand will come into near temperature equilibrium with the surrounding water, thus slowing the rate o heat low to the inner part o the hand. 4. INTERPRET This problem involves inding the time required to transorm the given amount o 00 C water to gas with the given rate o heating.

9 The Thermal Behavior o Matter 7-9 DEVELOP From Equation 7.5, the energy absorbed during the vaporization o water at its boiling point is Q = L v m. I this energy is supplied at in a time t, then the power must be P = Q/t, so t = Q/P. Given that P = 500 W, we can solve or the time t. EVALUATE The time it takes to boil away the water is ( )( ) Q L 57 kj/kg.kg vm t =.66 0 s 7.6 min P = P =.5 kw = = This seems like a reasonable time, and it explains why it is a good idea to add water occasionally to steamers when cooking with steam. 4. INTERPRET This problem is about melting, and it involves heat o usion. We want to know how much ice would melt by exploding a -megaton nuclear bomb in the Greenland ice cap. 5 DEVELOP A -megaton nuclear device releases about Q = J o energy. The heat o usion or water is L = 4 kj/kg, so we can use Equation 7.5 Q = L m to ind the mass m o ice that is melted. EVALUATE The amount o ice at the normal melting point o 0 C that melts is 5 Q J 0 m = = =.5 0 kg L 4 kj/kg The amount o ice melted by this explosion corresponds to a lake approximately 00-m deep, -km long, and 00-m wide. 4. INTERPRET You want to know how long it will take your camping stove to melt snow. Note that the stove here is the same as the one in Problem 6.56 that was used to boil water. DEVELOP You can integrate the given power, P, to ind the total heat that the snow has absorbed. You can then equate that to the amount o energy needed to melt the snow. Since the snow starts o around 0 C, it doesn t need to be warmed up to the reezing point. From Equation 7.5: Q= L m, where L = 4 kj/kg rom Table 7.. EVALUATE The heat absorbed by the snow over a given time is: t 0 ( ) (.kw) (.5W/s) Q= P t dt = at+ bt = t+ t You want to know how long until this absorbed heat melts the snow ( )( ) Q= L m= 4 kj/kg 5.0 kg = 670 kj This requires solving a quadratic equation with the quadratic ormula rom Appendix A: ( 00W) ( 00W) 4(.5W/s)( 670 kj).5w/s ( ) + + t = = 88 s 4 min In Problem 6.56, it took the same stove 9 minutes to boil water.5 kg o water. I we double this result, we see that it takes about 0% less time to melt snow than to boil the same amount o water. 44. INTERPRET This problem is about melting, so heat o usion is involved. We want to know how much energy is needed to melt a given quantity o ice. DEVELOP We can look up the surace area o Lake Superior: A = 8,000 km. The total mass o ice is then m= ρv = ρah= = 0 (97 kg/m )(8. 0 m )(. m) kg I we can assume the ice is initially at 0 C, the energy needed to melt it will be Q= L m, where L = 4 kj/kg rom Table 7..

10 7-0 Chapter 7 EVALUATE (a) Substituting the values, the heat required to thaw Lake Superior is: ( )( ) Q= L m= = 9 4 kj/kg kg.7 0 J EJ Where we have used the preix exa (0 8 ). (b) I the ice melts in three weeks, the average power must be 9 Q.7 0 J P = = =.8 0 W = 8 TW t s The power per unit area (or intensity) in this case is 9 W/m. In Chapter 6, we learned that the Earth absorbs energy rom the Sun at a rate o 960 W/m averaged over its cross-sectional area. Considering that the Sun is shining on Lake Superior only a raction o the three-week period, the calculated power is reasonable. 45. INTERPRET This problem involves the latent heat o usion, with which we can calculate the time it takes to reeze water that is at 0 C i we remove energy a the rate given. DEVELOP From Equation 7.5, the rerigerator must extract the energy Q = L m, where m = 0.75 kg and L = 4 kj/kg (rom Table 7.). At the rate P = Q/t, where P is the power applied by the rerigerator, it will take a time t = Q/P to reeze the water. EVALUATE Inserting the given quantities into the expression above gives ( )( ) Q L 4 kj/kg 0.75 kg m t =.64 0 s 4.9 min P = P = kw = = This is a good igure to keep in mind i you need to make ice cubes or a party. 46. INTERPRET You re asked to determine i a highly absorbing black carbon can cause a layer o Arctic ice to completely melt. DEVELOP The ice has a thickness o d =.5 m. You assume that the only energy input is rom absorbed sunlight: E = a SAt ( ) where a is the raction o light relected away, S = 00 W/m is the solar intensity, A is the exposed surace area o ice, and t is the time, which we will assume is a quarter o a year or the summer season. The ice will melt i E Q= L m, where L = 4 kj/kg and m= ρ Ad. EVALUATE Combining the above relations, the criteria or the ice to melt can be written as: L ( )( )( ) ρd 4 kj/kg 97 kg/m.5 m a = = 68% St 00 W/m s ( )( 4 ) So the ice shouldn t melt completely under normal conditions when a = 90%, but the presence o black carbon could cause complete melting, assuming a = 50% in this case. The parameter a is called the albedo. An object with zero albedo would absorb all incoming radiation, like a perect blackbody. 47. INTERPRET This problem involves a change in temperature and a phase change (solid to liquid). We will apply the concepts o speciic heat and heat o usion to ind temperature o the system in equilibrium. DEVELOP From Example 7.4, we expect that the 50 g o ice at 0 C will completely melt in the.0 kg o water at 5 C. To ind the equilibrium temperature, consider that the water must irst warm the ice rom 0 C to 0 C, which will require an energy Q = m ice c ice ΔT, with ΔT = 0 K, m ice = kg, and c ice = 050 J/(kg K) (see Table 6.). Next, the water will melt the ice, which costs an energy Q = L m ice, with L = 4 kj/kg (see Table 7.), and inally, the water will warm up the newly melted water rom 0 C to the equilibrium

11 The Thermal Behavior o Matter 7- temperature T, which will cost an energy Q = m ice c water T. The sum o these energies must equate to the energy lost by the water, or ( ) = + + = ( Δ + + ) m c T T Q Q Q m c T L c T water water 0,water ice ice water where T 0,water = 5 C and c water = 484 J/(kg K) (see Table 6.). EVALUATE Solving the expression above or the equilibrium temperature T gives T = mwatercwatert0,water mice ( ciceδ T + L ) ( mice + mwater ) cwater ( ) ( ) ( ) ( ) ( ) ( ) ( kg+.0 kg) 484 J/ ( kg K) { } =.0 kg 484 J/ kg K 5 C kg 050 J/ kg K 0 C + 4 kj/kg = 0 C As expected, the water has not reached 0 C, so all the ice melts and our initial assumption is conirmed. 48. INTERPRET This problem involves raising the temperature o ice to the melting point and then changing the phase, so both speciic heat and heat o usion are involved. DEVELOP The energy needed to raise the temperature is given by Equation 6., Q = mciceδ T, with ΔT = T inal T initial = 0 C ( 0 C) = 0 C = 0 K. Equation 7.5, Q = L m, gives the heat o usion required to change the phase rom solid to liquid. Use Table 6. to ind the speciic heat c ice o ice and Table 7. to ind the latent heat o usion L or water. The total energy required to go rom ice at 0 C to water at 0 C is the sum o Q and Q. EVALUATE Adding up the two energies, we obtain Q = Q + Q = mc Δ T + ml tot ice { } ( ) ( ) ( ) warm water melt ice = 0 kg.05 kj/ kg K 0 K + 4 kj/kg = 0.05 MJ +.4 MJ =.545 MJ.5 MJ About 94% o the energy is actually used to melt the ice, only 6% or raising the temperature. 49. INTERPRET This problem involves raising the temperature o water, which involves the speciic heat o water, then changing its phase rom liquid to gas, which involves the latent heat o vaporization. Using these two concepts, we are to ind the initial temperature o the water given that the 90% o the energy needed to boil the water away is used to change the phase, with only 0% being used to raise its temperature. DEVELOP Equation 6. Q = mcδt gives the energy needed to raise the water s temperature, where ΔT = 00 C T where T is the initial temperature o the water. The energy needed to boil the water (i.e., change it rom the liquid phase to the gas phase) is Q = ml v. The quantities L v and c can be ound in Tables 7. and 6., respectively. The problem statement says that Q = Q /0, so ml mcδ T = mc( 00 C T ) = 0 Lv T = ( 00 C) 0 c we can solve or the initial water temperature T. EVALUATE Inserting the speciic heat and latent heat o vaporization into the expression above gives T = 57 kj/kg ( 00 C) 46. C kj/ ( kg K) = Much more heat is required to boil the water away (i.e., change its phase rom liquid to gas) than to raise its temperature rom 46 C to 00 C. 50. INTERPRET The problem deals with both raising the temperature o water and vaporizing it, so both the speciic heat and the latent heat o vaporization are involved. v

12 7- Chapter 7 DEVELOP The energy needed to raise the temperature o water is given by Equation 6., Q = mcδ T. Equation 7.5, Q = Lm, gives the energy or phase change. The total energy required is Q tot = Q + Q. The mass o water involved can be calculated rom the density o water; m = ρv, with ρ =.00 0 kg/m. Given that the reactor provides energy at the rate P = 00 MW, the time t it takes to boil away the water is t = Q/P. EVALUATE Thus, the amount o time t required is t ( Δ + ) ρ ( Δ + ) Q m c T L V c T L P P P v = = = v (.0 0 kg/m )( 40 m ){ 4.84 kj/ ( kg K) ( 00 C 0 C) + 57 kj/kg} = 00 MW = = 9 min =.5 h Failsae cooling systems are crucial or preventing nuclear meltdown s 5. INTERPRET This problem involves mixing ice with water and letting the mixture come to equilibrium. We are to calculate the minimum amount o ice needed so that the inal equilibrium mixture is at 0 C. This calculation will involve the speciic heat o water to calculate the energy required to raise the temperature o the ice to 0 C and the latent heat o usion to calculate the energy required to melt the ice. DEVELOP For the inal equilibrium temperature in Example 7.4 to be 0 C, the original.0 kg o water must lose at least Q = 6.8 kj o heat energy (see Example 7.4). It could lose more energy, i some or all o it roze, but this would clearly require a greater amount o ice. The amount o ice needed to absorb this thermal energy and just melt, without exceeding 0 C, is given by summing the energy needed to raise its temperature to zero (Q = mcδt, where ΔT = 0 C) and the energy needed to melt the ice (Q = L v m). Equating Q with Q + Q gives ( ) Q = Q + Q = m cδ T + L v which we can solve or m. EVALUATE Solving or m gives cδ T + L.050 kj/ ( kg K) ( 0 C) + 57 kj/kg v m = = = 77 g Q 6.8 kj The amount o original ice that could produce a inal temperature o 0 C and reeze all the original water is m ice ( Δ + W ) ( )( + ) mw cw T L.0 kg kj/kg = = = 9 kg c ΔT 0.5 kj/kg ice ice The amount o ice that would produce a inal temperature o 0 C with none o the ice melted and none o the water rozen is m m c (.0 kg) 4.84 kj/ ( kg K) ( 5 C).050 kj/ ( kg K) ( 0 C) ΔT W W W ice = = = ciceδtice. kg For 77 g < m ice <. kg, some o the original ice melts, and or. kg < mice < 9 kg, some o the original water reezes. 5. INTERPRET Our system consists o both ice and warm water. We want to know how much ice at 0 C is needed to bring the water to 0 C. This process involves cooling the water (i.e., using the speciic heat o water) and melting the ice (i.e., using the latent heat o usion o ice). DEVELOP Assume that the only heat transer is between the water and the ice. To cool the water to 0 C, the amount o heat that must be extracted is (using Equation 6.) ( )( )( ) Q = mwcwδ T = 6 kg 4.84 kj/kg K 5 C 0 C =.67 MJ The amount o heat used to melt ice at 0 C is, rom Equation 7., Q = L m. ice For the ice to just melt and not increase in temperature, we must have Q = Q.

13 The Thermal Behavior o Matter 7- EVALUATE To bring the water temperature down to 0 C, the minimum mass o ice needed is thus ΔQ.67 MJ m = = = 5.0 kg L 4 kj/kg Note that the punch would be diluted with 5.0 kg o melt-water. To avoid this dilution, you could use less ice at a temperature below 0 C. 5. INTERPRET This problem involves mixing an equal mass o ice and water and letting the mixture reach equilibrium. We are to calculate at what temperature the water must be i the inal mixture is to contain equal amounts o ice and water. DEVELOP Assume that all the heat gained by the ice was lost by the water, with no heat transer to the container or the surroundings. An equilibrium mixture o ice and water (at atmospheric pressure) must be at 0 C, and i the masses o ice and water start out and remain equal, there is no net melting or reezing. Thus any energy spent raising the temperature o the ice and melting it must be balanced by energy lost by the water as its temperature lowers and it reezes. These energies are given by Equations 6. (or the temperature change), Q = mc and (or the melting or reezing) Q = L m. We can thereore sum these energies or both ice and water and equate the result, which gives ice w Q ice w Q Q Q Q = m c (0 C T ) + L m = Q = m c ( T 0 C) + L m ice ice ice ice ice w w w w w where T ice = 0 C. Given that m ice = m w, we can solve or the water temperature T w. EVALUATE Solving or T W and inserting the given quantities gives cice ( T ice ).050 kj/ ( kg K) Tw = = 0 C = 4.9 C c 4.84 kj/ kg K w ( ) ( ) Because any thermal energy spent melting the ice is balanced by reezing the water, only the speciic heats o water and ice come into play. 54. INTERPRET The evaporation o sweat removes heat rom the body. We re asked to ind how much heat is removed rom a marathon runner during a race. DEVELOP It takes 4. MJ to evaporate a kg o water, and we re told that a marathoner typically sweats L per hour. We assume that all o this sweat is evaporated and that all o the needed heat comes rom the marathoner s body. EVALUATE During a -hour marathon, the amount o heat expelled due to sweating is: kg Q = (.4 MJ/kg) ( L/h)( h) = MJ L This is equivalent to 500 kilocalories. However, running or hours should only burn about 500 kilocalories. That would mean the marathoner is losing more than twice the amount o energy he/she is consuming. That wouldn t make sense, so we might assume that some o the sweat alls o beore evaporating. 55. INTERPRET This problem involves the latent heat o usion o water, which is the energy required per unit mass to change ice to water (or vice versa, but with the opposite sign). We can use this concept to ind the energy required to melt 0 kg o ice in 6 min, and rom there ind the power. DEVELOP I the melting occurs at atmospheric pressure and i the ice is at 0 C, the energy required to melt the ice is given by Equation 7., Q = L m, where L is the latent heat o usion, which is given in Table 7.. To melt the ice in a time t = 6 min would require a power P = Q/t.

14 7-4 Chapter 7 EVALUATE The power required is ( )( ) Q L 0 kg 4 kj/kg m P = 9 kw. t = t = 6 60 s = This is equivalent to the power needed to light W bulbs. 56. INTERPRET This problem involves both a temperature rise and a phase change, so both the speciic heat and the latent heat (o vaporization o water) will enter into the calculation. The object o interest is the water in the microwave oven. DEVELOP The energy needed to raise the temperature o the water is given by Equation 6., Q = mcδt. Equation 7.5, Q = L v m, gives the heat needed to vaporize water. In 0 minutes, the total heat energy is transerred to the water (i we ignore energy absorbed by a container or lost to the surroundings) is Qtot = Pt = (0.5 kw)(0 60 s) = 600 kj Out o this total energy, the energy consumed in raising the water s temperature to the normal boiling point is Q = mc Δ T = ( 0. kg) 4.84 kj/ ( kg K) ( 00 C 0 C) = 00 kj and the dierence Q = Qtot Q = 500 kj is let to vaporize some o the water. The amount o water vaporized can be ound by using Equation 7.5, Q = L v m, where L v is the latent heat o vaporization rom Table 7.. EVALUATE Using Equation 7.5, the mass m v o water vaporized is Q 500 kj mv = = = g Lv 57 kj/kg Thereore, only Δ m= m mv = 00 g g = 79 g o boiling water (or less than oz) is all that remains (to two signiicant igures). The excess heat rom the microwave oven vaporizes the water. This is precisely what causes your ood to dry out when you heat it in the microwave oven or too long. 57. INTERPRET This problem involves a change in temperature and phase (liquid to gas) or water, so both the speciic heat and the latent heat o vaporization come into play. We use these to calculate how long it takes, given a constant input power o 00 MW, to boil away hal o the water. DEVELOP The reactor must irst raise the temperature o the water to the boiling point, which requires an energy Q = mcδ T (Equation 6.), where ΔT = 00 C 0 C = 90 K. Next, hal the water must be boiled o, which requires an energy Lvm Q = (Equation 7.5), where L v is the latent heat o vaporization (see Table 7.). The total energy required is the sum o these two, so at the given power P, it will take a time t = (Q + Q )/P to boil o hal the water. EVALUATE Inserting the given quantities, we ind 5 ( Δ + L ) ( kg) { 4.84 kj/ ( kg K) ( 90 K) kj/kg } mc T v t = = =.4 0 s = 56 min 4 P 0 0 kw For this solution, we have assumed that the water is heated uniormly, which may or may not be true depending on the geometry o the situation. I the water does not circulate well, it is possible that hal the water boils o beore the entire mass o water is heated to 00 C. In this case, the time would be less than we ound above, which is thereore the maximum time it can take or hal the water to boil o. 58. INTERPRET This problem involves heat transer between ice and water. Some o the ice may melt and some o the water may reeze, which will involve the latent heat o usion o water. In addition, the ice and the water will change temperature, so the speciic heats o ice and water are involved.

15 The Thermal Behavior o Matter 7-5 DEVELOP Assume that all the heat lost by the water is gained by the ice. The temperature o the water drops and that o the ice rises. I either reaches 0 C, a change o phase occurs, reezing or melting, depending on which reaches 0 C irst. To cool to 0 C, the amount o heat the water would lose is ( ) ( ) ( ) Qw,= mwcwδ Tw =.0 kg 4.84 kj/ kg K 5.0 K =0.9 kj On the other hand, to warm the ice to 0 C, it would gain an energy Qi= mc i iδ Ti = (.0 kg).050 kj/ ( kg K) ( 40 K) = 8.0 kj Evidently, the water reaches 0 C irst, so we will assume it reezes and the ice simply warms up. Because all the energy lost by the water in cooling and then changing phase to ice is assumed to be gained by the ice, we must have Q i = Q w, + Q w,, where Q w, = L m (Equation 7.) is the energy is takes to reeze the water. We can thereore solve or the mass m o water that is changed to ice. EVALUATE Thereore, the mass o water m that reezes is Qi Qw, 6. kj m = = = 0.8 kg L 4 kj/kg The inal mixture is thus at 0 C and contains.8 kg o ice and.0 kg 0.8 kg = 0.8 kg o water. The equilibrium temperature in this case is 0 C where water and ice coexist. 59. INTERPRET This problem involves the linear thermal expansion o steel. We are to ind the temperature at which the hole in the steel plate will become large enough to allow the marble to pass through it. DEVELOP The diameter o the hole expands with the coeicient o linear expansion o steel. Using Equation 7.7 ΔL αδ T = L with L =.000, L 0 = 0.997, we can solve or ΔT. EVALUATE Solving or T and inserting the given quantities gives ΔL 0.00 cm Δ T = = = 5 K Lα cm 0 6 K ( )( ) so the plate must be heated to 5 K above room temperature. Note that we have treated the problem as i we were considering a steel disk o initial diameter cm, instead o a hole in a steel plate. The treatment is valid because such a hole must expand at the same rate as the disk because i we put the disk in the hole, it must it at all temperatures. 60. INTERPRET This problem is about thermal expansion. Since it involves volume, the relevant quantity is the coeicient o volume expansion β, whose value can be used to identiy the substance in question. DEVELOP As in Example 7.5, we assume that the thermal expansion o the cylinder is negligible compared to that o the liquid. Then the entire change in volume is due to the liquid. The amount o volume change can be calculated rom Equation 7.6, ΔVV β = Δ T EVALUATE Substituting the values given in the problem statement, we have ΔVV ( 75 ml) ( 000 ml) 5 β = = = 75 0 K ΔT 50 K From Table 7., we see that this matches the coeicient or ethyl alcohol. The thermal expansion coeicient or ethyl alcohol is much greater than the material with which the cylinder is made, so our assumption is justiied. 6. INTERPRET This problem involves the linear expansion o Pyrex and steel. We are to ind the temperature at which the diameter o the Pyrex tube is μm greater than the diameter o the steel ball.

16 7-6 Chapter 7 DEVELOP Since the coeicient o linear expansion o steel is greater than that o Pyrex glass, the unit must be cooled to provide clearance. The dierence in the contraction o steel and Pyrex must be twice the given clearance on one side, so ( ) ΔL Δ L = μm = α α L Δ T steel pyrex steel pyrex which we can solve or ΔT (ind the values or α steel and α pyrex in Table 7.). EVALUATE Solving or ΔT and inserting the known quantities gives cm.0 0 cm Δ T = = =.7 K 6 ( αsteel αpyrex ) L (.) 0 K (.0 cm) Since we must cool the system by this amount, the inal temperature o the system will be T = 0 K K = 07 K. Reversing this process is a good technique to create tightly itted parts. 6. INTERPRET You re in charge o speciying the size o an expansion tank or a car s uel system. This will involve calculating the thermal expansion o gasoline. DEVELOP The amount that a material expands due to an increase in temperature is given in Equation 7. 6: 5 Δ V / V = βδt. For gasoline, the coeicient o volume expansion is β = 95 0 K rom Table 7.. EVALUATE The car you are working on has a gas tank with a volume o V = 75 L. You assume the tank is illed initially with gasoline at 0 C. Once the gasoline comes into equilibrium with the outside temperature o 5 C, the volume will have expanded by 5 Δ V = VβΔ T = 75 L 95 0 K 5 K =.8 L ( )( )( ) Your expansion tank needs to be at least this big. The expansion tank size is a little over % o the main tank s size, which seems reasonable. Notice that gasoline has the highest coeicient o volume expansion o all the liquids in Table INTERPRET This problem involves the linear expansion o a rod as it is heated, so the coeicient o linear expansion will come into play. We are asked to calculate the height d o the apex o the triangle ormed by the rod that cracks upon expanding because it is ixed between two immovable walls. DEVELOP I the two straight pieces in Fig. 7. are o equal length, the Pythagorean Theorem gives where L L ( α T ) ( ) ( ) 0 d = L L = 0 + Δ is the total expanded length o the rod. EVALUATE Inserting the expression or L gives L 0 d = αδ T + α Δ T Checking the limits, we see that or ΔT 0K, d 0, as expected. 64. INTERPRET You have to quickly calculate how much ice to buy to keep your parent s rerigerator rom warming up during a temporary black out. DEVELOP For simplicity, let s imagine that you will be placing bags o ice along the interior walls o the rerigerator, as shown in the igure below. In this way, the heat rom the outside room will low directly into the ice o o at a rate o H =Δ T/ R, where R = 0. K/W is the thermal resistance o the ridge walls, and Δ T = 0 C 0 C is the temperature between the room and the ice.

17 The Thermal Behavior o Matter 7-7 The heat will melt the ice in t = L m/ H, so you want to buy enough ice that this time equals the 5 hours it supposedly will take to return the electric power to your home. EVALUATE Solving or the required mass o ice gives th tδt ( s)( 0 K) m = = = = 7 kg L L R 4 kj/kg 0. K/W ( )( ) You might have some diiculty stuing 7 -kg bags o ice in the ridge. Note that we haven t taken into account the act that the ridge interior will not start out at 0 C, so some extra heat will low between your ood and the ice. 65. INTERPRET This problem involves the temperature change and phase change (solid-liquid) o Glauber salt. Given its thermal parameters, we are to calculate how long the salt takes to cool to 60 F, and how long the salt takes to solidiy at 90 F. DEVELOP In cooling rom 95 F to 90 F, the liquid expels a heat Q = mc liq ΔT liq (Equation 6.). Changing phase at 90 F rom liquid to solid expels a urther amount Q = L m (Equation 7.5). Finally, cooling the solid salt rom 90 F to 60 F expels the heat Q = mc sol ΔT sol. Thus, the total heat expelled is Q= Q+ Q + Q = mcliqδ Tliq + Lm+ mcsolδ Tsol Given that the house loses heat at a rate o P = 0,000 Btu/h, the time t it takes to cool to 60 F is t = Q/P. EVALUATE (a) Inserting the given quantities gives mcliqδ Tliq + L m + mcsolδtsol t = P lbs = Btu/h + + = 6h {( 5 F) 0.68 Btu/ ( lb F) 04 Btu/lb 0.46 Btu/ ( lb F) ( 0 F) } (b) The time spent during just the solidiication at 90 F is ml ( lb )( 04 Btu/lb ) = = 5 h 4 P.0 0 Btu/h Most o the time the salt is at 90 because the latent heat o usion is much greater than the heat liberated by temperature change. 66. INTERPRET For this problem, we are to prove that the coeicient o volume expansion o an ideal gas at constant pressure is the inverse temperature in the Kelvin scale. DEVELOP As mentioned in the text ollowing Equation 7.6, the coeicient o volume expansion β is deined in general as β = dv/ V dv dt = V dt The ideal-gas law pv = NkT is given by Equation 7.. The two equations can be combined to give the proo.

18 7-8 Chapter 7 For an ideal gas at constant pressure, V = NkT/p. This gives dv/dt = nk/p. Substituting the equation EVALUATE into the above expression or β gives dv Nk Nk β = = = = VdT V p NkT T The unit o β are K, which is the inverse temperature in the Kelvin scale. 67. INTERPRET For this problem, we are to show at what temperature between 0 C and 0 C water has its greatest density. We are given the expression or the coeicient o volume expansion as a unction o temperature. DEVELOP We do not actually need to dierentiate the density or the volume [ρ(t) = constant mass/v(t)] because Equation 7.6 shows that dv/dt = βv = 0 when β(t) = 0. Thus, the maximum density (or minimum volume) occurs or a temperature satisying a+ bt + ct = 0, which allows us to solve or T. EVALUATE The quadratic ormula gives ± T = c b b 4ac or, since both a and c are negative, T = b b 4 a c c Canceling a actor o 0 5 rom the given coeicients, we ind ( ) ( )( ) C T = =.97 C The second root, 80. C, can be discarded because it is outside the range o validity, 0 T 0 C, o the original unction β (T). Thus, the maximum density o water occurs at a temperature close to 4 C. That this represents a minimum volume can be veriied by plotting V(T), or rom the second derivative, dv dβ dv dβ = V + β = V dt dt dt β + dt = V β + b+ ct > 0 or T =.97 C. ( ) 68. INTERPRET The problem involves inding the volume o water at a given temperature, given its coeicient o volume expansion as a unction o temperature. DEVELOP In Problem 68, we learned that water s coeicient o volume expansion in the temperature range rom 0 C to about 0 C is given approximately by a = temperature is C, b 5 =.70 0 C, and dv V c β = a+ bt + ct, where T is in Celsius and 7 =.0 0 C. Thus, the volume as a unction o = β = + + dt ( a bt ct ) dt Integrate this expression and impose the boundary condition that V(0 C) = L to ind the volume o water at C. EVALUATE Integrating the expression above gives V dv V T T = ln = βdt ( a bt ct ) dt a ( T T) b( T T ) c( T T ) V V V = + + = + + T T 4 For T = 0C, T = C, and coeicients a, b and c given above, the right hand side is.6 0. With V = L, exponentiation gives V = Ve =.0004 L

19 The Thermal Behavior o Matter 7-9 The raction o volume change is ( V V)/ V = INTERPRET This problem involves the conversion o gravitational potential energy into thermal energy. We are to ind the gravitational potential energy equivalent to the thermal energy needed to melt a alling ice cube on impact. DEVELOP The assumptions stated in the problem (no air resistance or heat exchange with the environment) imply that the change in the gravitational potential energy o the ice cube, per unit mass, must equal the heat o transormation o the ice cube. Expressed mathematically, this gives mgh = ml which we can solve or the height h. EVALUATE Solving or h, we ind L h = = ( 4 kj/kg )( 9.8m/s ) = 4. km g O course, the expression or potential energy dierence, mgh, is not valid over such a large range, but mgy RE/( RE + y) only changes this result to 4. km. The thermal energies o ordinary macroscopic objects are very large compared to their mechanical energies. 70. INTERPRET This problem deals with simple harmonic motion and the thermal expansion o a brass pendulum. The quantity o interest is the pendulum s length, so the relevant quantity is the coeicient o linear expansion, α. DEVELOP N swings o a pendulum clock produce a time reading o t = Nτ where τ = π L/ g (Equation.5) is the period. I the clock is accurate at T = 0 C, at some other temperature T, the length o the pendulum becomes L = L ( + α Δ T ), where ΔT = 7 C 0 C = K. Thereore, the ratio o the periods is τ L α τ = L = + Δ The error in the clock is the dierence in its time-readings or N swings at T, relative to T : τ Δ t = t t = Nτ Nτ = Nτ τ Solve this equation to ind the time it takes or the clock to err by minute. EVALUATE Substituting the ratio o the periods into the expression or Δt and solving or t gives ( α ) Δ t = t + ΔT Δt.0 min t = = = 4 d + Δ K K 6 ( α T ) ( )( ) where we used α = K or brass rom Table 7.. Because L < L, τ < τ, Δ t = t t < 0, and the clock at 7 C is ast. The period o a pendulum, τ = π L/ g, increases with its length L. Due to thermal expansion, the pendulum length at 0 C is greater than that at 7 C, and consequently, its period is longer. 7. INTERPRET We are asked to derive the given equation or the volume expansion coeicient β. DEVELOP Equation 7.6 gives the volume expansion coeicient as ΔV β = V ΔT For ininitesimally small changes, this becomes dv β = V dt T

20 7-0 Chapter 7 Using the chain rule in the expression or β gives dv dv dl β = = V dt L dl dt Given that dv = d ( L ) = L dl dl and, rom Equation 7.7, dl = αl dt we can evaluate the chain-rule expression or β in terms o α. EVALUATE Inserting the expressions above or the derivatives gives volume expansion coeicient as β dv dl = = ( L )( L) L dl dt L α = α Alternatively, use the binomial approximation or ( ) Δ V = L+ΔL L = L Δ L, keeping only the lowest order term in ΔL. Since Δ V = βv Δ T and L αl T, L αl Δ T = βl Δ T, or β = α. Δ = Δ one inds ( ) 7. INTERPRET This problem involves pressure, temperature, and volume. We will assume that the ideal-gas law applies and use it to ind the volume o one mole under the given conditions. DEVELOP In terms o moles, the ideal-gas law (Equation 7.) is pv = nrt where R = 8.4 J ( K mol). The pressure is p = 90 atm, the temperature is T = 70 K, and we are interested in the volume o mole, so n =.0. We want to ind the volume o one mole o gas, and see i this volume is less than L. EVALUATE Inserting the given quantities into the ideal-gas law gives pv = nrt (.0 mol) 8.4 J ( K mol) ( 70 K) nrt V = = = = p 5 ( 90 atm)(.0 0 Pa/atm) m 0.67 L This design will work, as one liter contains more than a mole at this pressure and temperature. 7. INTERPRET This problem involves the determination o the linear expansion coeicient o a metal wire as it is heated. We analyze the data to see how the vertical distance between the hanging mass and the top varies with temperature. DEVELOP Pythagorean Theorem gives ( ) ( ) L d + y y = ( L d ) 4 where L = L 0 ( + α Δ T ) is the total expanded length o the wire. EVALUATE (a) Inserting the expression or L gives 0( ) d d y = L + αδt L0( + αδt) = ( L0 d ) + L0αΔ T where we drop terms involving α. The expression is o the orm y = a+ b( ΔT), so plotting y as a unction o ΔT will allow us to determine the expansion coeicient α.