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- 1. Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 3 • The specific heat of aluminum is more than twice the specific heat of copper. A block of copper and a block of aluminum have the same mass and temperature (20ºC). The blocks are simultaneously dropped into a single calorimeter containing water at 40ºC. Which statement is true when thermal equilibrium is reached? (a) The aluminum block is at a higher temperature than the copper block. (b) The aluminum block has absorbed less energy than the copper block. (c) The aluminum block has absorbed more energy than the copper block. (d) Both (a) and (c) are correct statements. Picture the Problem We can use the relationship TmcQ Δ= to relate the amount of energy absorbed by the aluminum and copper blocks to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum block: TcmQ Δ= AlAlAl Express the energy absorbed by the copper block: TcmQ Δ= CuCuCu Divide the second of these equations by the first to obtain: Tcm Tcm Q Q Δ Δ = AlAl CuCu Al Cu Because the block’s masses are the same and they experience the same change in temperature: 1 Al Cu Al Cu <= c c Q Q or AlCu QQ < and )(c is correct. 11 • A real gas cools during a free expansion, while an ideal gas does not cool during a free expansion. Explain the reason for this difference. Determine the Concept Particles that attract each other have more potential energy the farther apart they are. In a real gas the molecules exert weak attractive forces on each other. These forces increase the internal potential energy during an expansion. An increase in potential energy means a decrease in kinetic energy, and a decrease in kinetic energy means a decrease in translational kinetic energy. Thus, there is a decrease in temperature. 363
- 2. Chapter 18364 21 •• An ideal gas in a cylinder is at pressure P and volume V. During a quasi-static adiabatic process, the gas is compressed until its volume has decreased to V/2. Then, in a quasi-static isothermal process, the gas is allowed to expand until its volume again has a value of V. What kind of process will return the system to its original state? Sketch the cycle on a graph. Determine the Concept During a reversible adiabatic process, is constant, where γ < 1 and during an isothermal processes is constant. Thus, the pressure rise during the compression is greater than the pressure drop during the expansion. The final process could be a constant volume process during which heat is absorbed from the system. A constant-volume cooling will decrease the pressure and return the gas to its original state. γ PV PV V P adiabatic isothermal iVi2 1 f VV = constant volume Estimation and Approximation 25 •• A ″typical″ microwave oven has a power consumption of about 1200 W. Estimate how long it should take to boil a cup of water in the microwave assuming that 50% of the electrical power consumption goes into heating the water. How does this estimate correspond to everyday experience? Picture the Problem Assume that the water is initially at 30°C and that the cup contains 200 g of water. We can use the definition of power to express the required time to bring the water to a boil in terms of its mass, heat capacity, change in temperature, and the rate at which energy is supplied to the water by the microwave oven. Use the definition of power to relate the energy needed to warm the water to the elapsed time: t Tmc t W P Δ Δ = Δ Δ = ⇒ P Tmc t Δ =Δ Substitute numerical values and evaluate Δt: ( ) ( ) min1.6s63.97 W600 C30C100 Kkg kJ 184.4kg200.0 Δ ≈= °−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =t , an elapsed time that seems to be consistent with experience.
- 3. Heat and the First Law of Thermodynamics 365 Heat Capacity, Specific Heat, Latent Heat 29 • How much heat must be absorbed by 60.0 g of ice at –10.0ºC to transform it into 60.0 g of water at 40.0ºC? Picture the Problem We can find the amount of heat that must be absorbed by adding the heat required to warm the ice from −10.0°C to 0°C, the heat required to melt the ice, and the heat required to warm the water formed from the ice to 40.0°C. Express the total heat required: waterwarmicemelticewarm QQQQ ++= Substitute for each term to obtain: ( )waterwaterficeice waterwaterficeice TcLTcm TmcmLTmcQ Δ++Δ= Δ++Δ= Substitute numerical values (See Tables 18-1 and 18-2) and evaluate Q: ( ) ( )( ) ( ) kJ3.31 C0C0.04 Kkg kJ 184.4 kg kJ 5.333C0.01C0 Kkg kJ 05.2kg0.0600 = ⎥ ⎦ ⎤ °−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ + +⎢ ⎣ ⎡ °−−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =Q Calorimetry 33 • While spending the summer on your uncle’s horse farm, you spend a week apprenticing with his farrier (a person who makes and fits horseshoes). You observe the way he cools a shoe after pounding the hot, pliable shoe into the correct size and shape. Suppose a 750-g iron horseshoe is taken from the farrier’s fire, shaped, and at a temperature of 650°C, dropped into a 25.0-L bucket of water at 10.0°C. What is the final temperature of the water after the horseshoe and water arrive at equilibrium? Neglect any heating of the bucket and assume the specific heat of iron is K)J/(kg460 ⋅ . Picture the Problem During this process the water will gain energy at the expense of the horseshoe. We can use conservation of energy to find the equilibrium temperature. See Table 18-1 for the specific heat of water.
- 4. Chapter 18366 Apply conservation of energy to obtain: 0horseshoethecoolwaterthewarm i i =+=∑ QQQ or ( ) ( ) 0C650C0.10 fFeFefwaterwater =°−+°− tcmtcm Solve for tf to obtain: ( ) ( ) FeFewaterwater FeFewaterwater f C650C0.10 cmcm cmcm t + °+° = Substitute numerical values and evaluate tf: ( ) ( ) ( ) ( ) ( ) ( ) C1.12 Kkg kJ 460.0kg750.0 Kkg kJ 184.4kg0.25 C650 Kkg kJ 460.0kg750.0C0.10 Kkg kJ 184.4kg0.25 f °= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ °⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =t 37 •• A 200-g piece of ice at 0ºC is placed in 500 g of water at 20ºC. This system is in a container of negligible heat capacity and is insulated from its surroundings. (a) What is the final equilibrium temperature of the system? (b) How much of the ice melts? Picture the Problem Because we can not tell, without performing a couple of calculations, whether there is enough heat available in the 500 g of water to melt all of the ice, we’ll need to resolve this question first. See Tables 18-1 and 18-2 for specific heats and the latent heat of fusion of water. (a) Determine the energy required to melt 200 g of ice: ( ) kJ70.66 kg kJ 5.333kg0.200ficeicemelt = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == LmQ The energy available from 500 g of water at 20ºC is: ( ) ( ) kJ84.14 C02C0 Kkg kJ 184.4kg0.500Δ waterwaterwatermaxavailable, −= °−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ == TcmQ Because maxavailable,Q < :icemeltQ C.0isretemperatufinalThe °
- 5. Heat and the First Law of Thermodynamics 367 (b) Equate the energy available from the water maxavailable,Q to miceLf and solve for mice to obtain: f maxavailable, ice L Q m = Substitute numerical values and evaluate mice: g125 kg kJ 5.333 kJ84.14 ice ==m 43 •• A 100-g piece of copper is heated in a furnace to a temperature tC. The copper is then inserted into a 150-g copper calorimeter containing 200 g of water. The initial temperature of the water and calorimeter is 16.0ºC, and the temperature after equilibrium is established is 38.0ºC. When the calorimeter and its contents are weighed, 1.20 g of water are found to have evaporated. What was the temperature tC? Picture the Problem We can find the temperature t by applying conservation of energy to this calorimetry problem. See Tables 18-1 and 18-2 for specific heats and the heat of vaporization of water. Use conservation of energy to obtain: 0 sampleCu thecool rcalorimete thewarm waterthe warm water vaporize i i =+++=∑ QQQQQ or 0ΔΔΔ CuCuCuwcalcalOHOHOHwf,vaporizedO,H 2222 =+++ TcmTcmTcmLm Substituting numerical values yields: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0C0.38 Kkg kJ 386.0g100C0.16C0.38 Kkg kJ 386.0g150 C0.16C0.38 Kkg kJ 184.4g200 Kkg kJ 2257g1.20 C =−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +°−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ + °−°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ t Solving for tC yields: C618C °=t Work and the PV Diagram for a Gas 51 • The gas is first cooled at constant volume until it reaches its final pressure. It is then allowed to expand at constant pressure until it reaches its final volume. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process.
- 6. Chapter 18368 Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat absorbed by the gas during this process (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram. 2 0 1 0 P, atm 3.00 2.00 1.00 1.00 2.00 3.00 V, L The work done by the gas equals the area under the curve: ( )( ) J054 L m10 L2.00 atm kPa101.325 atm2.00L2.00atm2.00Δ 33 gasby = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×=== − VPW (b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain: ( ) ( ) ( ) gasbyint,1int,2 gasbyint,1int,2 onintin WEE WEE WEQ +−= −−−= −Δ= Substitute numerical values and evaluate Qin: ( ) J618J054J456J912in =+−=Q 57 •• An ideal gas initially at 20ºC and 200 kPa has a volume of 4.00 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 100 kPa. Find (a) the work done by the gas, and (b) the heat absorbed by the gas during the expansion.
- 7. Heat and the First Law of Thermodynamics 369 Picture the Problem The PV diagram shows the isothermal expansion of the ideal gas from its initial state 1 to its final state 2. We can use the ideal-gas law for a fixed amount of gas to find V2 and then evaluate for an isothermal process to find the work done by the gas. In Part (b) of the problem we can apply the first law of thermodynamics to find the heat added to the gas during the expansion. ∫PdV 200 100 2 1 V2 P, kPa 293 K 4.00 V, L (a) Express the work done by a gas during an isothermal process: ∫∫∫ === 2 1 2 1 2 1 11gasby V V V V V V V dV VP V dV nRTdVPW Apply the ideal-gas law for a fixed amount of gas undergoing an isothermal process: 2211 VPVP = ⇒ 1 2 1 2 V P P V = Substitute numerical values and evaluate V2: ( ) L00.8L4.00 kPa100 kPa200 2 ==V Substitute numerical values and evaluate W: ( )( ) ( )[ ] ( ) J555 L m10 LkPa5.554 L4.00 L00.8 lnLkPa800 lnLkPa800L00.4kPa200 33 L8.00 L00.4 L8.00 L00.4 gasby = ×⋅=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅= ⋅== − ∫ V V dV W (b) Apply the first law of thermodynamics to the system to obtain: onintin WEQ −Δ= or, because ΔEint = 0 for an isothermal process, onin WQ −= Because the work done by the gas is the negative of the work done on the gas: ( ) J555gasbygasbyin ==−−= WWQ
- 8. Chapter 18370 Remarks: in an isothermal expansion the heat added to the gas is always equal to the work done by the gas (ΔEint = 0). Heat Capacities of Gases and the Equipartition Theorem 59 •• The heat capacity at constant pressure of a certain amount of a diatomic gas is 14.4 J/K. (a) Find the number of moles of the gas. (b) What is the internal energy of the gas at T = 300 K? (c) What is the molar heat capacity of this gas at constant volume? (d) What is the heat capacity of this gas at constant volume? Picture the Problem (a) The number of moles of the gas is related to its heat capacity at constant pressure and its molar heat capacity at constant pressure according to . For a diatomic gas, the molar heat capacity at constant pressure is PP nc'C = Rc' 2 7 P = . (b) The internal energy of a gas depends on its number of degrees of freedom and, for a diatomic gas, is given by nRTE 2 5 int = . (c) The molar heat capacity of this gas at constant volume is related to its molar heat capacity at constant pressure according to Rc'c' −= PV . (d) The heat capacity of this gas at constant volume is the product of the number of moles in the gas and its molar heat capacity at constant volume. (a) The number of moles of the gas is the ratio of its heat capacity at constant pressure to its molar heat capacity at constant pressure: P P c' C n = For a diatomic gas, the molar heat capacity is given by: Kmol J 1.292 7 P ⋅ == Rc' Substitute numerical values and evaluate n: mol495.0 mol4948.0 Kmol J 1.29 K J 4.14 = = ⋅ =n (b) With 5 degrees of freedom at this temperature: nRTE 2 5 int =
- 9. Heat and the First Law of Thermodynamics 371 Substitute numerical values and evaluate Eint: ( ) ( ) kJ09.3K300 Kmol J 314.8mol4948.02 5 int =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =E (c) The molar heat capacity of this gas at constant volume is the difference between the molar heat capacity at constant pressure and the gas constant R: Rc'c' −= PV Because Rc' 2 7 P = for a diatomic gas: RRRc' 2 5 2 7 V =−= Substitute the numerical value of R to obtain: Kmol J 8.20 Kmol J 79.20 Kmol J 314.82 5 V ⋅ = ⋅ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =c' (d) The heat capacity of this gas at constant volume is given by: VV nc'C' = Substitute numerical values and evaluate :VC' ( ) K J 3.10 Kmol J 79.20mol4948.0V = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =C' 65 •• Carbon dioxide (CO2) at a pressure of 1.00 atm and a temperature of –78.5ºC sublimates directly from a solid to a gaseous state without going through a liquid phase. What is the change in the heat capacity at constant pressure per mole of CO2 when it undergoes sublimation? (Assume that the gas molecules can rotate but do not vibrate.) Is the change in the heat capacity positive or negative during sublimation? The CO2 molecule is pictured in Figure 18-22. Picture the Problem We can find the change in the heat capacity at constant pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in the solid and gaseous states. Express the change in the heat capacity (at constant pressure) per mole as the CO2 undergoes sublimation: solidP,gasP,P CCC −=Δ (1)
- 10. Chapter 18372 Express Cp,gas in terms of the number of degrees of freedom per molecule: ( ) NkNkfC 2 5 2 1 gasP, == because each molecule has three translational and two rotational degrees of freedom in the gaseous state. We know, from the Dulong-Petit Law, that the molar specific heat of most solids is 3R = 3Nk. This result is essentially a per-atom result as it was obtained for a monatomic solid with six degrees of freedom. Use this result and the fact CO2 is triatomic to express CP,solid: Nk Nk C 9atoms3 atom 3 solidP, =×= Substitute in equation (1) to obtain: NkNkNkC 2 13 2 18 2 5 PΔ −=−= Quasi-Static Adiabatic Expansion of a Gas 69 •• A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is (a) isothermal and (b) adiabatic. Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In Part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In Part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas: i i i P nRT V = Substitute numerical values and evaluate Vi: ( ) ( ) 33 i m103.118 kPa400 K300 Kmol J 8.314mol0.500 − ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =V (a) Because the process is isothermal: K300if == TT
- 11. Heat and the First Law of Thermodynamics 373 Use the ideal-gas law for a fixed amount of gas with T constant to express Vf: ffi VPVP = ⇒ f i if P P VV = Substitute numerical values and evaluate Vf: ( ) L7.80 L795.7 kPa160 kPa400 L3.118f = =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V Express the work done by the gas during the isothermal expansion: i f gasby ln V V nRTW = Substitute numerical values and evaluate Wby gas: ( ) ( ) kJ14.1 L3.118 L7.795 lnK300 Kmol J 8.314mol0.500gasby = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =W Noting that the work done by the gas during the process equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas: ( ) kJ1.14 kJ1.140onintin = −−=−Δ= WEQ (b) Using γ = 5/3 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf: γγ ffii VPVP = ⇒ γ1 f i if ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = P P VV Substitute numerical values and evaluate Vf: ( ) L5.40 L403.5 kPa160 kPa400 L118.3 53 f = =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V Apply the ideal-gas law to find the final temperature of the gas: nR VP T ff f =
- 12. Chapter 18374 Substitute numerical values and evaluate Tf: ( )( ) ( ) K208 Kmol J 8.314mol0.500 m105.403kPa160 33 f = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × = − T For an adiabatic process: 0in =Q Apply the first law of thermodynamics to express the work done on the gas during the adiabatic process: TnRTCQEW Δ=−Δ=−Δ= 2 3 Vininton 0 Substitute numerical values and evaluate Won: ( )( ) ( ) J745 K300K208 KJ/mol8.314mol0.5002 3 on −= −× ⋅=W Because the work done by the gas equals the negative of the work done on the gas: ( ) J745J574gasby =−−=W Cyclic Processes 73 •• A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This expansion is represented by the straight line from 1 to 2 in the PV diagram (Figure 18-23). The gas is then compressed isothermally. This compression is represented by the straight line from 2 to 1 in the PV diagram. Calculate the work per cycle done by the gas. Picture the Problem The total work done as the gas is taken through this cycle is the area bounded by the two processes. Because the process from 1→2 is linear, we can use the formula for the area of a trapezoid to find the work done during this expansion. We can use ( )ifprocessisothermal ln VVnRTW = to find the work done on the gas during the process 2→1. The net work done during this cycle is then the sum of these two terms. Express the net work done per cycle: 1221 gason thegasby thenet →→ += += WW WWW (1)
- 13. Heat and the First Law of Thermodynamics 375 Work is done by the gas during its expansion from 1 to 2 and hence is equal to the negative of the area of the trapezoid defined by this path and the vertical lines at V1 = 11.5 L and V2 = 23 L. Use the formula for the area of a trapezoid to express W1→2: ( ) ( ) atmL3.17 atm1.0atm2.0 L11.5L232 1 trap21 ⋅−= +× −−= −=→ AW Work is done on the gas during the isothermal compression from V2 to V1 and hence is equal to the area under the curve representing this process. Use the expression for the work done during an isothermal process to express W2→1: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =→ i f 12 ln V V nRTW Apply the ideal-gas law at point 1 to find the temperature along the isotherm 2→1: ( )( ) ( )( ) K280 Katm/molL10206.8mol1.00 L5.11atm0.2 2 = ⋅⋅× == − nR PV T Substitute numerical values and evaluate W2→1: ( )( )( ) atmL9.15 L23 L5.11 lnK280Katm/molL10206.8mol00.1 2 12 ⋅=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅⋅×= − →W Substitute in equation (1) and evaluate Wnet: kJ14.0 atmL J101.325 atmL40.1 atmL15.9atmL3.17net −= ⋅ ×⋅−= ⋅+⋅−=W 75 ••• At point D in Figure 18-24 the pressure and temperature of 2.00 mol of an ideal monatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point C. Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle.
- 14. Chapter 18376 Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle: DCCBBAADtotgas,by →→→→ +++= WWWWW 1. Use the ideal-gas law to find the volume of the gas at point D: ( ) ( ) ( ) L29.54 atm kPa 101.325atm2.00 K360 Kmol J 8.314mol2.00 D D D = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = = P nRT V 2. We’re given that the volume of the gas at point B is three times that at point D: ( ) L62.88 L54.2933 DCB = === VVV Use the ideal-gas law to find the pressure of the gas at point C: ( ) ( ) atm6667.0 L62.88 K360 Kmol atmL 10206.8mol2.00 2 C C C = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × == − V nRT P We’re given that the pressure at point B is twice that at point C: ( ) atm333.1atm6667.022 CB === PP 3. Because path DC represents an isothermal process: K360CD == TT Use the ideal-gas law to find the temperatures at points B and A: ( )( ) ( ) K719.8 Kmol atmL 108.206mol2.00 L88.62atm1.333 2 BB BA = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × = == − nR VP TT
- 15. Heat and the First Law of Thermodynamics 377 Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that: atm00.42 DA == PP The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right. Point P V T (atm) (L) (K) A 4.00 29.5 720 B 1.33 88.6 720 C 0.667 88.6 360 D 2.00 29.5 360 4. For the path D→A, 0AD =→W and: ( )DA2 3 AD2 3 ADint,AD TTnR TnREQ −= Δ=Δ= →→→ Substitute numerical values and evaluate QD→A: ( ) ( ) kJ979.8K360K720 Kmol J 8.314mol2.002 3 AD =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =→Q For the path A→B: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == →→ A B BA,BABA ln V V nRTQW Substitute numerical values and evaluate WA→B: ( ) ( ) kJ15.13 L29.54 L88.62 lnK720 Kmol J 8.314mol2.00BA =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =→W and, because process A→B is isothermal, 0BAint, =Δ →E For the path B→C, , and:0CB =→W ( )BC2 3 VCBCB ΔΔ TTnR TCUQ −= == →→ Substitute numerical values and evaluate QB→C: ( )( )( ) kJ979.8K720K360KJ/mol8.314mol2.002 3 CB −=−⋅=→Q For the path C→D: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =→ C D DC,DC ln V V nRTW
- 16. Chapter 18378 Substitute numerical values and evaluate WC→D: ( ) ( ) kJ576.6 L62.88 L54.92 lnK603 Kmol J 8.314mol2.00DC −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ =→W Also, because process A→B is isothermal, 0BAint, =Δ →E , and kJ58.6DCDC −== →→ WQ Qin, Won, and ΔEint are summarized for each of the processes in the table to the right. Process Qin Won ΔEint (kJ) (kJ) (kJ) D→A 98.8 0 8.98 A→B 2.13 −13.2 0 B→C 98.8− 0 −8.98 C→D 58.6− 6.58 0 Referring to the table, find the total work done by the gas per cycle: kJ6.6 kJ6.580kJ13.20 DCCBBAADtotgas,by = −++= +++= →→→→ WWWWW Remarks: Note that, as it should be, ΔEint is zero for the complete cycle. General Problems 79 • The PV diagram in Figure 18-25 represents 3.00 mol of an ideal monatomic gas. The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path AEC, find (a) the initial and final temperatures of the gas, (b) the work done by the gas, and (c) the heat absorbed by the gas. Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process EDC is isobaric, we can find the area under this line geometrically and then use the 1st law of thermodynamics to find QAEC.
- 17. Heat and the First Law of Thermodynamics 379 (a) Using the ideal-gas law, find the temperature at point A: ( )( ) ( ) K65K65.2 Kmol atmL 108.206mol3.00 L4.01atm4.0 2 AA A == ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × = = − nR VP T Using the ideal-gas law, find the temperature at point C: ( )( ) ( ) K81K81.2 Kmol atmL 108.206mol3.00 L0.02atm1.0 2 CC C == ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × = = − nR VP T (b) Express the work done by the gas along the path AEC: ( )( ) kJ1.6kJ1.62 atmL J101.325 atmL15.99 L4.01L20.0atm1.0 Δ0 ECECECAEAEC == ⋅ ×⋅= −= +=+= VPWWW (c) Apply the first law of thermodynamics to express QAEC: ( )ATTnRW TnRW TCWEWQ −+= Δ+= Δ+=Δ+= C2 3 AEC 2 3 AEC VAECintAECAEC Substitute numerical values and evaluate QAEC: ( )( )( ) kJ2.2K65.2K81.2KJ/mol8.314mol3.00kJ1.62 2 3 AEC =−⋅+=Q Remarks The difference between WAEC and QAEC is the change in the internal energy ΔEint,AEC during this process. 83 •• As part of a laboratory experiment, you test the calorie content of various foods. Assume that when you eat these foods, 100% of the energy released by the foods is absorbed by your body. Suppose you burn a 2.50-g potato chip, and the resulting flame warms a small aluminum can of water. After burning the potato chip, you measure its mass to be 2.20 g. The mass of the can is 25.0 g, and the volume of water contained in the can is 15.0 ml. If the temperature increase in the water is 12.5°C, how many kilocalories
- 18. Chapter 18380 (1 kcal = 1 dietary calorie) per 150-g serving of these potato chips would you estimate there are? Assume the can of water captures 50.0 percent of the heat released during the burning of the potato chip. Note: Although the joule is the SI unit of choice in most thermodynamic situations, the food industry in the United States currently expresses the energy released during metabolism in terms of the ″dietary calorie,″ which is our kilocalorie. Picture the Problem The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it. The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it: 500 g0.30 g150 g0.30 servingg-150 == Q Q or g0.30servingg150 500QQ = Letting f represent the fraction of the heat captured by the can of water, express the energy transferred to the aluminum can and the water in it during the burning of the potato chip: ( ) Tcmcm TcmTcm QQfQ Δ ΔΔ OHOHAlAl OHOHAlAl OHAlg0.30 22 22 2 += += += where ΔT is the common temperature change of the aluminum cup and the water it contains. Substituting for yields and solving for yields: g0.30Q servingg-150Q ( ) f Tcmcm Q Δ500 OHOHAlAl servingg-150 22 + = Substitute numerical values and evaluate :servingg-150Q ( ) ( ) ( ) kcal652cal10256 J4.184 cal1 J1007.1 500.0 C5.12 Kkg kJ 184.4kg0150.0 Kkg kJ 900.0kg0250.0500 36 servingg-150 ≈×=××= °⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ =Q 89 •• A thermally insulated system consists of 1.00 mol of a diatomic gas at 100 K and 2.00 mol of a solid at 200 K that are separated by a rigid insulating
- 19. Heat and the First Law of Thermodynamics 381 wall. Find the equilibrium temperature of the system after the insulating wall is removed, assuming that the gas obeys the ideal-gas law and that the solid obeys the Dulong–Petit law. Picture the Problem We can use conservation of energy to relate the equilibrium temperature to the heat capacities of the gas and the solid. We can apply the Dulong-Petit law to find the heat capacity of the solid at constant volume and use the fact that the gas is diatomic to find its heat capacity at constant volume. Apply conservation of energy to this process: 0Δ solidgas =+= QQQ Use to substitute for QTCQ ΔV= gas and Qsolid: ( ) ( ) 0K200K100 equilsolidV,equilgasV, =−+− TCTC Solving for Tequil yields: ( )( ) ( )( ) solidV,gasV, solidV,gasV, equil K200K100 CC CC T + + = Using the Dulong-Petit law, determine the heat capacity of the solid at constant volume: RnC solidsolidV, 3= The heat capacity of the gas at constant volume is given by: RnC gas2 5 gasV, = Substitute for CV,solid and CV,gas and simplify to obtain: ( )( ) ( )( ) ( )( ) ( )( ) solidgas2 5 solidgas2 5 solidgas2 5 solidgas2 5 equil 3 3K200K100 3 3K200K100 nn nn RnRn RnRn T + + = + + = Substitute numerical values for ngas and nsolid and evaluate Tequil: ( )( )( ) ( )( )( ) ( ) ( ) K171 mol00.23mol00.1 mol00.23K200mol00.1K100 2 5 2 5 equil = + + =T 95 ••• (a) Use the results of Problem 94 to show that in the limit that ETT >> , the Einstein model gives the same expression for specific heat that the Dulong–Petit law does. (b) For diamond, TE is approximately 1060 K. Integrate numerically dEint = dT to find the increase in the internal energy if 1.00 mol of diamond is heated from 300 to 600 K. v ′c
- 20. Chapter 18382 Picture the Problem (a) We can rewrite our expression for by dividing its numerator and denominator by ' cV TT e E and then using the power series for ex to show that, for T > TE, . In Part (b), we can use the result of Problem 94 to obtain values for every 100 K between 300 K and 600 K and use this data to find ΔU numerically. Rc' 3V ≈ ' cV (a) From Problem 94 we have: ( )2 2 E V 1 3 E E − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = TT TT e e T T Rc' Divide the numerator and denominator by TT e E to obtain: TTTT TT TTTT eeT T R e eeT T Rc' EE E EE 2 1 3 12 1 3 2 E 2 2 E V − +− ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = +− ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Express the exponential terms in their power series to obtain: E 2 E 2 EE 2 EE for ... 2 1 12... 2 1 12 EE TT T T T T T T T T T T ee TTTT >>⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−+−+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=+− − Substitute for TTTT ee EE 2 − +− to obtain: R T TT T Rc' 3 1 3 2 E 2 E V = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈ (b) Use the result of Problem 94 to verify the following table: T (K) 300 400 500 600 cV (J/mol⋅K) 9.65 14.33 17.38 19.35
- 21. Heat and the First Law of Thermodynamics 383 The following graph of specific heat as a function of temperature was plotted using a spreadsheet program: 5 7 9 11 13 15 17 19 21 300 350 400 450 500 550 600 T (K) CV(J/mol-K) Integrate numerically, using the formula for the area of a trapezoid, to obtain: ( )( )( ) ( )( )( ) ( )( )( ) kJ62.4 Kmol J 35.1938.17K100mol00.1 Kmol J 38.1733.14K100mol00.1 Kmol J 33.1465.9K100mol00.1Δ 2 1 2 1 2 1 = ⋅ ++ ⋅ ++ ⋅ +=U
- 22. Chapter 18384