PV/T = k or PV = kt Describe the difference between an ideal gas and a real gas.

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1 10.1 Thermodynamics 10.2 Processes The second law of thermodynamics and entroy 10.1 Thermodynamics From the combined gas laws, we determined that: P/T = k or P = kt State the equation of state for an ideal gas Describe the difference between an ideal gas and a real gas Describe the concet of the absolute zero of temerature and the Kelvin scale of temerature Solve roblems using the equation of state of an ideal gas. IBO The ideal gas equation n ideal gas is a theoretical gas that obeys the equation of state of an ideal gas exactly. They obey the equation P = nrt when there are no forces between molecules at all ressures, volumes and temeratures. Remember from vogadro s hyothesis that one mole of any gas contains the vogadro number of articles N equal to articles. It also occuies 22.4 dm 3 at 0 C and kpa ressure (STP). The internal energy of an ideal gas would be entirely kinetic energy as there would be no intermolecular forces between the gaseous atoms. s temerature is related to the average kinetic energy of the atoms, the kinetic energy of the atoms would deend only on the temerature of the ideal gas. If the value of the universal gas constant is comared for different masses of different gases, it can be demonstrated that the constant deends not on the size of the atoms but rather on the number of articles resent (the number of moles). Thus for n moles of any ideal gas: P / nt = R P = nrt This is called the equation of state of an ideal gas, where R is the universal gas constant and is equal to 8.31 J mol 1 K -1. The equation of state of an ideal gas is determined from the gas laws and vogadro s law Real and ideal gases n ideal gas is a theoretical gas that obeys the ideal gas equation exactly. Real gases conform to the gas laws under certain limited conditions but they can condense to liquids, then solidify if the temerature is lowered. Furthermore, there are relatively small forces of attraction between articles of a real gas, and even this is not allowable for an ideal gas. real gas obeys the gas laws at low ressure and at high temerature (above the temerature at which they liquefy), and we refer to such a gas as an ideal gas Physics Ch 10 final.indd 273 3/12/2007 1:27:08 PM

2 Chater 10 Most gases, at temeratures well above their boiling oints and ressures that are not too high, behave like an ideal gas. In other words, real gases vary from ideal gas behaviour at high ressures and low temeratures bsolute zero of temerature When the variation in volume as a function temerature is lotted for an ideal gas, a grah similar to Figure 1001 is obtained. Pressure / P kpa T C K Figure 1002 ariation of ressure with temerature olume / m 3 The Pressure (dmonton) Law of Gases states that: The ressure of a fixed mass of gas at constant volume is directly roortional to its its temerature. P T P kt P 1 = ---- = T 1 k T C K Figure 1001 ariation of volume with temerature Therefore, P T 1 = P T 2 Note that from the extraolation of the straight line that the volume of gases would be theoretically zero at C called absolute zero. The scale chosen is called the Kelvin scale K and this is the fundamental unit of thermodynamic temerature. When a ressure versus volume grah is drawn for the collected data, a hyerbola shae is obtained, and when ressure is lotted against the recirocal of volume a straight line is obtained. See Figure The Charles (Gay-Lussac) Law of gases states that: The volume of a fixed mass of gas at constant ressure is directly roortional to its absolute (Kelvin) temerature. ressure, P / mm Hg ressure, P / mm Hg volume, / cm 3 1 /cm 3 P P This can also be stated as: Figure 1003 Pressure-volume grahs The volume of a fixed mass of gas increases by 1 / of its volume at 0 C for every degree Celsius rise in temerature rovided the ressure is constant. When the variation in ressure as a function temerature is lotted for an ideal gas, a grah similar to Figure 1002 is obtained. Boyle s Law for gases states that the ressure of a fixed mass of gas is inversely roortional to its volume at constant temerature. P 1 -- P constant When the conditions are changed, with the temerature still constant = Physics Ch 10 final.indd 274 3/12/2007 1:27:09 PM

3 P 1 1 = P 2 2 lthough the ressure and the recirocal of volume have a directly roortional linear lot, it is the first volumetemerature grah that is used to define absolute zero. lthough different samles of an ideal gas have different straight-line variations, they still extraolate back to absolute zero Using the equation of state of an ideal gas ( Nm -1 ) (molar mass / 1.25 kg m -3 ) = 1 (8.31 J mol -1 K -1 ) ( K) So that the molar mass = J mol -1 K K 1.25 kg m Nm -1 n = mol = kg mol -1 The ideal gas is helium with a molar mas of kg mol -1. Exercise 10.1 Examle 1 weather balloon of volume 1.0 m 3 contains helium at a ressure of N m -2 and a temerature of 35 C. What is the mass of the helium in the balloon if one mole of helium has a mass of kg? Solution 1. If the average translational kinetic energy E K at a temerature T of helium (molar mass 4 g mol 1 ), then the average translational kinetic energy of neon (molar mass 20 g mol -1 ) at the same temerature would be:. 1/5 E K B. 5 E K C. 5 E K D. E K Use the equation, P = nrt, we have ( Nm -1 ) (1.0 m 3 ) = n (8.31 J mol -1 K -1 ) ( K) So that, n = mol Then, the mass of helium = (39.46 mol) ( kg mol -1 ) = kg. The mass of helium in the balloon is 0.16 kg. Examle 2 n ideal gas has a density of 1.25 kg m -3 at STP. Determine the molar mass of the ideal gas. Solution Use the equation, P = nrt, with = molar mass/density. For I mole 2. samle of gas is contained in a vessel at 20 C at a ressure P. If the ressure of the gas is to be doubled and the volume remain constant, the gas has to be heated to:. 40 C B. 293 C C. 586 C D. 313 C 3. Real gases behave most like ideal gases at. low temeratures and high ressures B. high temeratures and low ressures C. low temeratures and low ressures D. high temeratures and high ressures 4. The Kelvin temerature of an ideal gas is a measure of:. the average otential energy of the gas molecules B. the average seed of the gas molecules C. the average ressure of the gas molecules D. the average kinetic energy of the gas molecules Physics Ch 10 final.indd 275 3/12/2007 1:27:10 PM

4 Chater Two identical containers and B contain an ideal gas under the different conditions as shown below PROCESSES CONTINER N moles Temerature T Pressure P The ratio P : P B would be: CONTINER B 3N moles Temerature T / 3 Pressure P B Deduce an exression for the work involved in a volume change of a gas at constant ressure State the first law of thermodynamics Identify the first law of thermodynamics as a statement of the rincile of energy conservation.. 3 B. 2 C. 3 2 D The internal volume of a gas cylinder is m3. n ideal gas is umed into the cylinder until the ressure is 15 MPa at a temerature of 25 C Describe the isochoric (isovolumetric), isobaric, isothermal and adiabatic changes of state of an ideal gas Draw and annotate thermodynamic rocesses and cycles on P diagrams Calculate from a P diagram the work done in a thermodynamic cycle. (a) (b) (c) (d) Determine the number of moles of the gas in the cylinder Determine the number of gas atoms in the cylinder Determine the average volume occuied by one atom of the gas Estimate the average searation of the gas atoms 7. cylinder of an ideal gas with a volume of 0.2 m 3 and a temerature of 25 C contains molecules. Determine the ressure in the cylinder. 8. (a) State what is meant by the term ideal gas. (b) In terms of the kinetic theory of gases, state what is meant by an ideal gas. (c) Exlain why the internal energy of an ideal gas is kinetic energy only Solve roblems involving state changes of a gas. IBO Work done in volume change Consider a mass of gas with ressure enclosed in a cylinder by a iston of cross-sectional area as in Figure ressure, F surface area, cylinder Δl Figure 1005 Exansion of a gas at constant ressure Physics Ch 10 final.indd 276 3/12/2007 1:27:10 PM

5 The ressure,, on the iston = force er unit area So that, = F - Therefore, the force on the iston, F, is given by F = volume, temerature and change in internal energy in determining the state of a system. Heat can be transferred between a system and its environment because of a temerature difference. nother way of transferring energy between a system and its environment is to do work on the system or allow work to be done by the system on the surroundings. Suose the iston is moved a distance l when the gas exands. Normally, if the gas exands, the volume increases and the ressure decreases, as was determined from Boyle s Law for ideal gases in the revious section. However, if the distance l is a small Δ l, then the ressure can be considered constant. If the ressure is constant then the force F will be constant. The work done by the gas is: ΔW = F Δ l = Δ l since ressure = Force F / rea = Δ since volume Δ = l That is, (work done / J) = (ressure / Nm -2 ) (volume change / m 3 ) So that, W = = ( 2 1 ) The sign of the work done by the gas deends on whether volume change is ositive or negative. When a gas exands, as is the case for Figure 1005, then work is done by the gas, and the volume increases. s is ositive, then W is ositive. This equation is also valid if the gas is comressed. In the comression, work is done on the gas and the volume is decreased. Therefore, Δ is negative which means that W will be negative. From the first law of thermodynamics this means that ositive work is done on the gas and First law of In order to distinguish between thermal energy (heat) and work in thermodynamic rocesses If a system and its surroundings are at different temeratures and the system undergoes a rocess, the energy transferred by non-mchanical means is referred to as thermal energy (heat). Work is defined as the rocess in which thermal energy is transferred by means that are indeendent of a temerature difference. In thermodynamics the word system is used often. system is any object or set of objects that is being investigated. The surroundings will then be everything in the Universe aart from the system. For examle, when a volume of gas in a cylinder is comressed with a iston, then the system is the cylinder-gas-iston aaratus and the surroundings is everything else in the Universe. closed system is one in which no mass enters or leaves the system. It is an isolated system if no energy of any kind enters or leaves the system. Most systems are oen systems because of the natural dynamic rocesses that occur in the Universe. In Chater 3, internal energy U was defined as the sum total of the otential energy and kinetic energy of the articles making u the system. From a microscoic viewoint, the internal energy of an ideal gas is due to the kinetic energy of the thermal motion of its molecules. There are no intermolecular forces and thus there cannot be any increase in otential energy. Therefore a change in the temerature of the gas will change the internal energy of the gas. thermodynamics Thermodynamics is the name given to the study of rocesses in which thermal energy is transferred as heat and as work. It had its foundations with engineers in the 19th century who wanted to know what were the limitations of the Laws of Physics with regard to the oeration of steam engines and other machines that generate mechanical energy. Thermodynamics treats thermal energy from the macroscoic oint of view in that it deals with the thermodynamic variables of ressure, From the macroscoic oint of view of thermodynamics, one would exect that the internal energy of the system would be changed if: work is done on the system work is done by the system thermal energy is added to the system thermal energy is removed from the system Internal energy is a roerty of the system that deends on the state of the system. In thermodynamics, a change of Physics Ch 10 final.indd 277 3/12/2007 1:27:11 PM

6 Chater 10 state of an ideal gas occurs if some macroscoic roerty of the system has changed eg. hase, temerature, ressure, volume, mass, internal energy. const rea = Work done = ( 2 1 ) isobaric (i.e., same ressure rocess) Heat and work can change the state of the system but they are not roerties of the system. They are not characteristic of the state itself but rather they are involved in the thermodynamic rocess that can change the system from one state to another. 1 2 Figure 1006 Work done by a gas exanding at constant ressure The absolute value for internal energy is not known. This does not cause a roblem as one is mainly concerned with changes in internal energy, denoted by ΔU, in thermodynamic rocesses. The first law of thermodynamics is a statement of the Law of Conservation of Energy in which the equivalence of work and thermal energy transfer is taken into account. It can be stated as: the heat added to a closed system equals the change in the internal energy of the system lus the work done by the system. That is, or, Q = U + W = U + U = Q W where +Q is the thermal energy added to the system and +W is the work done by the system. n isobaric transformation requires a volume change at constant ressure, and for this to occur, the temerature needs to change to kee the ressure constant. = constant, or / T = constant. For an isobaric exansion, work is done by the system so ΔW is ositive. Thermal energy is added to cause the exansion so ΔQ is ositive. This means that ΔU must be ositive. For an isobaric comression, all terms would be negative. Isochoric (isovolumetric) rocesses grah of ressure as a function of volume change when the volume is ket constant is shown in Figure Such a rocess is said to be isochoric. When the volume is ket fixed, the curve of the transformation is said to be an isochore. If thermal energy leaves the system, then Q is negative. If work is done on the system, then W is negative. For an isolated system, then W = Q = 0 and ΔU = Changes of state of an 278 ideal gas Isobaric rocesses grah of ressure as a function of volume change when the ressure is ket constant is shown in Figure Such a rocess is said to be isobaric. Note that the work done by the gas is equal to the area under the curve. Figure 1007 isochoric (i.e., same volume rocess) Constant volume transformation Note that the work done by the gas is equal to zero as Δ = 0. There is zero area under the curve on a diagram. However, the temerature and ressure can both change and so such a transformation will be accomanied by a thermal energy change. = constant, or / T = constant. For an isochoric exansion, no work is done by the system so ΔW is zero. Thermal energy leaves the system so ΔQ is negative. This means that ΔU must be negative. For an isobaric exansion, ΔW is zero, and ΔQ and ΔU are ositive Physics Ch 10 final.indd 278 3/12/2007 1:27:11 PM

7 Isothermal rocesses thermodynamic rocess in which the ressure and the volume are varied while the temerature is ket constant is called an isothermal rocess. In other words, when an ideal gas exands or is comressed at constant temerature, then the gas is said to undergo an isothermal exansion or comression. Figure 1008 shows three isotherms for an ideal gas at different temeratures where Consider an ideal gas enclosed in a thin conducting vessel that is in contact with a heat reservoir, and is fitted with a light, frictionless, movable iston. If an amount of heat Q is added to the system which is at oint of Figure 1008, then the system will move to another oint on the grah, B. The heat taken in will cause the gas to exand isothermally and will be equivalent to the mechanical work done by the gas. Because the temerature is constant, there is no change in internal energy of the gas. That is, T 1 < T 2 < T 3. T = 0 and U = 0 Q = W T1 T2 T3 isothermal rocess If the gas exands isothermally from to B and then returns from B to following exactly the same ath during comression, then the isothermal change is said to be reversible. The conditions described above would follow this criterion. B diabatic rocesses Figure 1008 Isotherms for an ideal gas n adiabatic exansion or contraction is one in which no heat Q is allowed to flow into or out of the system. For the entire adiabatic rocess, Q = 0. The curve of an isothermal rocess reresents a Boyle s Law relation T = constant, or = constant = nrt The moles of gas n, the molar gas constant R, and the absolute temerature T are constant. For an isothermal exansion, temerature is constant so ΔU is zero. Work is done by the system so ΔW is ositive. This means that ΔU must be ositive. In order to kee the temerature constant during an isothermal rocess the gas is assumed to be held in a thin container with a high thermal conductivity that is in contact with a heat reservoir an ideal body of large mass whose temerature remains constant when heat is exchanged with it. eg. a constant temerature water bath. the exansion or comression should be done slowly so that no eddies are roduced to create hot sots that would disrut the energy equilibrium of the gas. To ensure that no heat enters or leaves the system during an adiabatic rocess it is imortant to make sure that the system is extremely well insulated. carry out the rocess raidly so that the heat does not have the time to leave the system. The comression stroke of an automobile engine is essentially an adiabatic comression of the air-fuel mixture. The comression occurs too raidly for areciable heat transfer to take lace. In an adiabatic comression the work done on the gas will lead to an increase in the internal energy resulting in an increase in temerature. U = Q W but Q = 0 U = W In an adiabatic exansion the work done by the gas will lead to a decrease in the internal energy resulting in a decrease in temerature. Figure 1009 shows the relationshi that exists between an adiabatic and three isothermals. Note that the adiabatic curve is steeer than the isotherm B because the adiabatic rocess has to occur raidly so that the heat does not have Physics Ch 10 final.indd 279 3/12/2007 1:27:12 PM

8 Chater 10 time to leave the system. The gas exands isothermally from oint to oint B, and then it is comressed adiabatically from B to C. The temerature increases as a result of the adiabatic rocess from T 1 to T 3. If the gas is then comressed at constant ressure from the oint C to, the net amount of work done on the gas will equal the area enclosed by BC. T1 T2 Figure 1009 T3 adiabatic C B isothermal rocess n isothermal and adiabatic rocess For an adiabatic comression, no heat enters or leaves the system so ΔQ is zero. Work is done on the system so ΔW is negative. This means that ΔU must be negative. For an isobaric exansion, ΔQ is zero, and ΔW and ΔU are ositive. In Figure 1010 the area BDE = work done by the gas during isothermal exansion. The area CDE = work done by the gas during an adiabatic exansion. (a) (b) (c) Work done by gas Work done on gas Net work done in exanding as it is comressed by gas Figure 1011 Pressure volume diagrams and Work and the thermodynamic cycle thermodynamic engine is a device that transforms thermal energy to mechanical energy (work) or mechanical energy to thermal energy such as in refrigeration and airconditioning systems. Cars, steam trains, jets and rockets have engines that transform fuel energy (chemical energy) into the kinetic energy of their motion. In all resently manufactured engines, the conversion is accomanied by the emission of exhaust gases that waste some of the thermal energy. Consequently, these engines are not very efficient as only art of the thermal energy is converted to mechanical energy. n engine has two crucial features: It must work in cycles to be useful. The cyclic engine must have more than one heat reservoir. 280 Figure 1010 E adiabatic isothermal Work done during exansion of a gas The work done by a gas and the work done on a gas can be seen using the following grahical reresentation of a ressure volume diagram (Figure 1011). B C D thermodynamic cycle is a rocess in which the system is returned to the same state from which it started. That is, the initial and final states are the same in the cyclic rocess. cycle for a simle engine was shown in Figure 5.4. The net work done in the cycle is equal to the area enclosed by the cycle. Suose a iston was laced on a heat reservoir, such as the hot late of a stove. Thermal energy is sulied by the thermal reservoir, and work is done by the gas inside the iston as it exands. But this is not an engine as it only oerates in one direction. The gas cannot exand indefinitely, because as the volume of the iston increases, the ressure decreases (Boyle s Law). Some oint will be reached when the exanding gas will not be able to move the iston. For this simle engine to function, the iston must eventually be comressed to restore the system to its original osition ready to do work. For a cycle to do net work, thermal contact with the original heat reservoir must be broken, and temeratures other than that of the original heat reservoir must lay a art in Physics Ch 10 final.indd 280 3/12/2007 1:27:13 PM

9 the rocess. In the above examle, if the iston is returned to its original osition while in contact with the hot late, then all the work that the gas did in the exansion will have to be used in the comression. On a diagram, one would draw an isotherm for the exansion and an isotherm for the comression lying on to of the exansion isotherm but in the oosite direction. Therefore, the area enclosed by the cycle would be zero. However, if the gas is comressed at a lower temerature the internal ressure of the system will be lower than during the exansion. Less work will be needed for the comression than was roduced in the exansion, and there will be net work available for transformation to mechanical energy. Heat engines and heat ums heat engine is any device that converts thermal energy into work. Examles include etrol and diesel engines, fossil-fuelled (coal, oil and natural gas) and nuclear ower lants that use heat exchangers to drive the blades in turbines, and jet aircraft engines. lthough we cannot convert all the random motion associated with the internal energy into useful work, we can at least extract some useful work from internal energy using a heat engine. To make a heat engine, we need a source of heat such as coal, etrol (gasoline), diesel fuel, aviation fuel or liquefied etroleum gas (LPG) and a working fluid. working fluid is a substance that undergoes a thermodynamic change of state and in the rocess does work on the surroundings. Common working fluids are water that is heated and converted to steam as used in ower stations and etrol-air mixtures as used in car engines. Most heat engines contain either istons with intake and exhaust valves that work in thermodynamic cycles as in a car engine, or rotating turbines used in heat exchanger systems in ower stations. Fluid enters carrying energy, Q H, at temerature T H. Fluid leaves carrying energy, Q L = Q H W, at temerature T L. Figure 1012 Hot reservoir at T H Q H ENGINE Q L Cold reservoir at T L W Engine does useful work, W. Energy flow diagram of a heat engine The heat inut Q H is reresented as coming from the high temerature reservoir T H which is maintained at a constant temerature. Thermal energy Q L is taken from the hot reservoir. This thermal energy is used to do work in the heat engine. Then thermal energy can be given to the low temerature reservoir T L without increasing its temerature. If a erfect engine comleted a cycle, the change in internal energy U would be zero because all the heat would be converted to work. However, there is no erfect heat engine and the flow diagram in Figure 1012 is more the reality. t this stage, we will assume that the change in internal energy is zero. From the First Law of Thermodynamics That is, U = 0 = Q W, so that W = Q H Q L = W Q Thus for a cycle, the heat added to the system equals the work done by the system lus the heat that flows out at lower temerature. n ideal gas can be used as a heat engine as in the simle cycle in Figure The urose of a heat engine is to convert as much of the heat inut Q H from a high temerature reservoir into work. Figure 1012 shows an energy flow diagram of a tyical heat engine. 6 (kpa) C D 2 B 4 10 (m 3 ) Figure 1013 Behaviour of an ideal gas Physics Ch 10 final.indd 281 3/12/2007 1:27:13 PM

10 Chater 10 From to B, the gas is comressed (volume decreases) while the ressure is ket constant an isobaric comression. The amount of work done by the gas is given by the area under the 2 kpa isobar. Using the fact that W = Δ, we have that W = 2 kpa (4 10) m 3 = J The mixture burns raidly and the hot gases then exand against the iston in the ower stroke. The exhaust valve is oened as the iston moves uwards during the exhaust stroke, and the cycle begins again. Gas vaor and mixture intake valve oen iston intake stroke exhaust valve closed intake valve closed comression stroke exhaust valve closed From B to C, the volume is ket constant as the ressure increases an isochoric increase in ressure. This can be achieved by heating the gas. Since Δ = 0, then no work is done by the gas, W = 0. exhaust sent fuel gases crankshaft intake valve closed crankshaft ignition exhaust valve closed From C to D, the gas exands (volume increases) while the ressure is ket constant an isobaric exansion. The amount of work done by the gas is given by the area under the 6 kpa isobar. Now, we have that W = Δ, so that intake valve closed exhaust valve oen intake valve closed ower stroke exhaust valve closed W = 6 kpa (10 4) m 3 = J From D to, the gas is cooled to kee the volume constant as the ressure is decreased an isochoric decrease in ressure. gain Δ = 0 and no work is done by the gas, W = 0. That is, the net work done by the gas is therefore J J = J. The internal combustion engine Figure 1014 shows a series of schematic diagrams for the cycle of an internal combustion engine as used in most automobiles. With the exhaust valve closed, a mixture of etrol vaour and air from the carburettor is sucked into the combustion chamber through the inlet valve as the iston moves down during the intake stroke. Both valves are closed and the iston moves u to squeeze the mixture of etrol vaour and air to about 1/8 th its original volume during the comression stroke. With both valves closed, the mixture is ignited by a sark from the sark lug. Figure 1014 Four-stroke internal combustion engine Motor cars usually have four or six istons but five and eight cylinders are also common. The istons are connected by a crankshaft to a flywheel which kees the engine turning over during the ower stroke. utomobiles are about 25% efficient. ny device that can um heat from a low-temerature reservoir to a high-temerature reservoir is called a heat um. Examles of heat ums include the refrigerator and reverse cycle air-conditioning devices used for sace heating and cooling. In the summer comonent of Figure 1012, the evaorator heat exchanger on the outside extracts heat from the surroundings. In the winter comonent, the evaorator heat exchanger is inside the room, and it exhausts heat to the inside air. In both cases, thermal energy is umed from a low-temerature reservoir to a high- temerature reservoir. WINTER room condenser T H Figure 1015 evaorator T L SUMMER room evaorator T L reverse cycle heat um condenser T H Physics Ch 10 final.indd 282 3/12/2007 1:27:14 PM

11 Figure 1016 shows the energy flow that occurs in a heat um cycle. By doing work on the system, heat Q L is added from the low temerature T L reservoir, being the inside of the refrigerator. greater amount of heat Q H is exhausted to the high temerature T H reservoir. High temerature reservoir at Q H T H W The refrigerator a heat um refrigerator is a device oerating in a cycle that is designed to extract heat from its interior to achieve or maintain a lower temerature inside. The heat is exhausted to the surroundings normally at a higher temerature. tyical refrigerator is reresented in Figure motor driving a comressor um rovides the means by which a net amount of work can be done on the system for a cycle. Even though refrigerator cabinets are well insulated, heat from the surroundings leaks back inside. The comressor motor can be heard to switch on and off as it ums this heat out again. Q L Low temerature reservoir at Figure 1016 T L Energy flow diagram for a heat um n ideal gas can be used as a heat um as in the simle cycle in Figure (kpa) volatile liquid called Freon is circulated in a closed system of ies by the comressor um, and, by the rocess of evaorative cooling, the vaorised Freon is used to remove heat. Evaorative cooling was discussed in Chater 3. The comressor maintains a high-ressure difference across a throttling valve. Evaoration of the Freon occurs in several loos called the evaorator ies that are usually inside the coldest art of the fridge. s the liquid evaorates on the low-ressure, low-temerature side, heat is added to the system. In order to turn from a liquid to a gas, the Freon requires thermal energy equal to the latent heat of vaorisation. This energy is obtained from the contents of the fridge. B Figure 1017 C (m 3 ) n ideal gas as a heat um insulation evaorative ies freezer comartment throttling valve liquid condenser ies Because the cycle is traced in an anticlockwise direction, the net work done on the surroundings is negative. HET IN heat in cold food HET OUT Freon gas cooling fins comressor um vaour at very high temerature Figure 1018 The tyical small refrigerator On the high-ressure, high-temerature side of the throttling valve, thermal energy is removed from the system. The vaorised Freon in the comressor ies is comressed by the comressor um, and gives u its Physics Ch 10 final.indd 283 3/12/2007 1:27:15 PM

12 Chater 10 latent heat of vaorisation to the air surrounding the comressor ies. The heat fins act as a heat sink to radiate the thermal energy to the surroundings at a faster rate. The fins are ainted black and they have a relatively large surface area for their size. The Carnot engine Before the First Law of Thermodynamics was even established, Nicolas Léonard Sadi Carnot ( ), a young engineer, was able to establish the theoretical maximum efficiency that was ossible for an engine working between two heat reservoirs. In 1824, he formulated that: No engine working between two heat reservoirs can be more efficient than a reversible engine between those reservoirs. Carnot argued that if thermal energy does flow from a cold body to a hot body then work must be done. Therefore, no engine can be more efficient than an ideal reversible one and that all such engines have the same efficiency. This means that if all engines have the same efficiency then only a simle engine was needed to calculate the efficiency of any engine. The net work is the area enclosed by BCD. In the case given, the Carnot engine is working in a clockwise cycle BCD. Thermal energy is absorbed by the system at the high temerature reservoir T H and is exelled at the low temerature reservoir T L. Work is done by the system as it exands along the to isotherm from to B, and along the adiabat from B to C. Work is then done on the system to comress it along the bottom isotherm from C to D and along the left adiabat from D to. The efficiency of the Carnot cycle deends only on the absolute temeratures of the high and low temerature reservoirs. The greater the temerature difference, the greater the efficiency will be. s a result of the Carnot efficiency, many scientists list a Third Law of Thermodynamics which states: It is imossible to reach the absolute zero of temerature, 0 K. The efficiency of the Carnot cycle would be 100% if the low temerature reservoir was at absolute zero. Therefore absolute zero is unattainable Solving roblems on state Consider an ideal erfectly insulated, frictionless engine that can work backwards as well as forwards. The diagram would have the form of that shown in Figure Examle 1 changes of a gas maximum temerature B T H Q H olume exansion If 22 J of work is done on a system and J of heat is added, determine the change internal energy of the system. C D Q = 0 olume comression adiabatic comression D isothermal exansion B isothermal comression adiabatic exansion C B Q = 0 olume exansion C minimum temerature Solution Using the formula, Q = ΔU + W, we have that 340 J = ΔU + (-22) J 340 J = ΔU + ( 22) J 1 2 C D T C Q L olume comression so that ΔU = 340 J + 22 J = 362 J Figure 1019 The Carnot engine That is, the change in internal energy of the system is J Physics Ch 10 final.indd 284 3/12/2007 1:27:16 PM

13 Examle 2 Solution 6.0 dm 3 of an ideal gas is at a ressure of kpa. It is heated so that it exands at constant ressure until its volume is 12 dm 3. Determine the work done by the gas. Solution Using the formula W = Δ, we have that (a) The fuel-air mixture enters the iston at oint. The comression B is carried out raidly with no heat exchange making it an adiabatic comression. The ignition and combustion of the gases introduces a heat inut Q H that raises the temerature at constant volume from B to C. The ower stroke is an adiabatic exansion from C to D. Thermal energy Q L leaves the system during the exhaust stroke, and cooling occurs at constant volume from D to. W = kpa (12 6.0) dm 3 = Pa (12 6.0) 10-3 m 3 = J That is, the work done by the gas in the exansion is J. (b) The Figure Below shows the changes that occur for each rocess in the cycle. Q H C constant ressure maximum temerature adiabatic exansion constant D volume ( 1 ) Examle 3 adiabatic comression Q L minimum temerature thermal system containing a gas is taken around a cycle of a heat engine as shown in the Figure below. (a) (b) Starting at oint, describe the cycle. Label the diagram fully showing the maximum and minimum temerature reservoirs. (c) 1 2 The net work is reresented by the enclosed area BCD. If we assume that the area is aroximately a rectangle with sides of Pa and 200 cm 3, we have: Nm m 3 = 80 J. (c) Estimate the amount of work done in each cycle. Examle x 10 5 Pa Q C D For the comression stroke of an exerimental diesel engine, the air is raidly decreased in volume by a factor of 15, the comression ratio. The work done on the air-fuel mixture for this comression is measured to be 550 J. B Q (a) What tye of thermodynamic rocess is likely to have occurred? (cm 3 ) (b) What is the change in internal energy of the airfuel mixture? (c) Is the temerature likely to increase or decrease? Physics Ch 10 final.indd 285 3/12/2007 1:27:17 PM

14 Chater 10 (a) (b) Solution Because the comression occurs raidly areciable heat transfer does not take lace, and the rocess can be considered to be adiabatic, Q = 0. ΔU = Q W = 0 ( 550) J Therefore, the change in internal energy is 550J. 4. The Figure below shows the variation of ressure with volume during one comlete cycle of a simle heat engine. C Y X B (c) The temerature rise will be very large resulting in the sontaneous ignition of the air-fuel mixture. 0 0 Exercise n ideal gas was slowly comressed at constant temerature to one quarter of its original volume. In the rocess, J of heat was given off. The change in internal energy of the gas was J B. 0 J C J D J The total work done is reresented by the area:. X + Y B. X Y C. X D. Y 5. The Figure below shows the variation of the ressure with volume of a gas during one cycle of the Otto engine. C 2. When an ideal gas in a cylinder is comressed at constant temerature by a iston, the ressure of the gas increases. Which of the following statement(s) best exlain the reason for the ressure increase? 286 I. the mean seed of the molecules increases II. the molecules collide with each other more frequently III. the rate of collision with the sides of the cylinder increases.. II only B. III only C. I and II only D. II and III only 3. n ideal gas in a thermally insulated cylinder is comressed raidly. The change in state would be:. isochoric B. isothermal C. adiabatic D. isobaric 1 2 During which rocess does the gas do external work?. B B. CD C. BC and CD D. B and CD 6. system absorbs 100 J of thermal energy and in the rocess does 40 J of work. The change in internal energy is:. 60 J B. 40 J C. 100 J D. 140 J D Q Physics Ch 10 final.indd 286 3/12/2007 1:27:17 PM

15 7. Work is done when the volume of an ideal gas increases. During which of the following state rocesses would the work done be the greatest?. isochoric B. isothermal C. isobaric D. adiabatic 8. How much heat energy must be added at atmosheric ressure to 0.50 kg of ice at 0 C to convert it to steam at 100 C? 9. If J of heat is added to a gas that exands and does J of work, what is the change in internal energy of the gas m 3 of an ideal gas is cooled at constant normal atmosheric ressure until its volume is 1/6 th its original volume. It is then allowed to exand isothermally back to its original volume. Draw the thermodynamic rocess on a diagram. 11. system consists of 3.0 kg of water at 75 C. Stirring the system with a addlewheel does J of work on it while J of heat is removed. Calculate the change in internal energy of the system, and the final temerature of the system. 12. gas is allowed to exand adiabatically to four times its original volume. In doing so the gas does 1750 J of work. (a) (b) (c) How much heat flowed into the gas? Will the temerature rise or fall? What is the change in internal energy of the gas? 13. For each of the rocesses listed in the following table, suly the symbol +,, or 0 for each missing entry. Process Q W U Isobaric comression of an ideal gas + Isothermal comression of an ideal gas diabatic exansion Isochoric ressure dro Free exansion of a gas 14. Helium gas at 312 K is contained in a cylinder fitted with a movable iston. The gas is initially at 2 atmosheres ressure and occuies a volume of 48.8 L. The gas exands isothermally until the volume is 106 L. Then the gas is comressed isobarically at that final ressure back to the original volume of 48.8L. It then isochorically returns back to its original ressure. ssuming that the helium gas behaves like an ideal gas (a) (b) (c) (d) (e) (f) (g) (h) Calculate the number of moles of helium gas in the system. Determine the ressure after the isothermal exansion. Draw a diagram of the thermodynamic cycle. ssuming that the isotherm is a diagonal line rather than a curve, estimate the work done during the isothermal exansion. Determine the work done during the isobaric comression. Determine the work done during the isochoric art of the cycle. Calculate the net work done by the gas. Calculate the final temerature of the helium. 15. (a) Distinguish between an isothermal rocess and an adiabatic rocess as alied to an ideal gas. (b) fixed mass of an ideal gas is held in a cylinder by a moveable iston and thermal energy is sulied to the gas causing it to exand at a constant ressure of kpa as shown in the Figure below. thermal energy iston The initial volume of the container is m 3 and after exansion the volume is 0.10 m 3. The total energy sulied to the gas during the rocess is 7.0 kj. (i) (ii) (iii) State whether this rocess is isothermal, adiabatic or neither of these rocesses. Determine the work done by the gas. Calculate the change in internal energy of the gas Physics Ch 10 final.indd 287 3/12/2007 1:27:18 PM

16 Chater This question is about a diesel engine cycle as shown in the Figure below. Mark on the diagram each of the state changes that occur at B, BC, CD and D. Identify the maximum and minimum temerature reservoirs and label Q H and Q L. B Q C 1 2 D Q 10.3 THE SECOND LW OF THERMO- DYNMICS ND ENTROPY State that the second law of thermodynamics imlies that thermal energy cannot sontaneously transfer from a region of low temerature to a region of high temerature State that entroy is a system roerty that exresses the degree of disorder in the system State the second law of thermodynamics in terms of entroy changes Discuss examles of natural rocesses in terms of entroy changes. IBO 2007 Introduction We are always told to conserve energy. But according to the First Law of Thermodynamics, in a closed system, energy is conserved, and the total amount of energy in the Universe does not change no matter what we do. lthough the First Law of Thermodynamics is correct, it does not tell the whole story. How often have you seen a videotae layed in reverse sequence. iews of water flowing uhill, demolished buildings rising from the rubble, eole walking backwards. In none of the natural Laws of Physics studied so far have we encountered time reversal. If all of these Laws are obeyed, why then does the time-reversed sequence seem imrobable? To exlain this reversal aradox, scientists in the latter half of the nineteenth century came to formulate a new rincile called the Second Law of Thermodynamics. This Law allows us to determine which rocesses will occur in nature, and which will not. There are many different but equivalent ways of stating the Second Law of Thermodynamics. Much of the language used for the definitions had its origins with the hysicists Physics Ch 10 final.indd 288 3/12/2007 1:27:18 PM

17 who formulated the Law, and their desire to imrove the efficiency of steam engines. These statements of the Second Law of Thermodynamics will be develoed within this section The second law of thermodynamics and temerature The second law of thermodynamics imlies that thermal energy cannot sontaneously transfer from a region of low temerature to a region of high temerature. The Clausius statement of the second law of thermodynamics Just as there is no cyclic device that can convert a given amount of heat comletely into work, it follows that the reverse statement is also not ossible. The Clausius statement of the Second Law of Thermodynamics can be stated as It is imossible to make a cyclic engine whose only effect is to transfer thermal energy from a colder body to a hotter body. The Kelvin Planck statement of the second law of thermodynamics ll attemts to construct a heat engine that is 100% efficient have failed. The Kelvin Planck statement of the Second Law of Thermodynamics is a qualitative statement of the imossibility of certain tyes of rocesses. There is no erfect refrigerator and no eretual motion machine Entroy Recall that in thermodynamics, a system in an equilibrium state is characterised by its state variables (,, T, U, n ). The change in a state variable for a comlete cycle is zero. In contrast, the net thermal energy and net work factors for a cycle are not equal to zero. It is imossible for an engine working in a cycle to transform a given amount of heat from a reservoir comletely into work. or Not all the thermal energy in a thermal system is available to do work. It is ossible to convert heat into work in a non-cyclic rocess. n ideal gas undergoing an isothermal exansion does just that. But after the exansion, the gas is not in its original state. In order to bring the gas back to its original state, an amount of work will have to be done on the gas and some thermal energy will be exhausted. The Kelvin Planck statement formulates that if energy is to be extracted from a reservoir to do work, a colder reservoir must be available in which to exhaust a art of the energy. In the latter half of the nineteenth century, Rudolf Clausius roosed a general statement of the Second Law in terms of a quantity called entroy. Entroy is a thermodynamic function of the state of the system and can be interreted as the amount of order or disorder of a system. s with internal energy, it is the change in entroy that is imortant and not its absolute value Change in entroy The change in entroy ΔS of a system when an amount of thermal energy Q is added to a system by a reversible rocess at constant absolute temerature T is given by S Q -- T The units of the change in entroy are J K 1. = Physics Ch 10 final.indd 289 3/12/2007 1:27:19 PM

18 Chater 10 Examle heat engine removes 100 J each cycle from a heat reservoir at 400 K and exhausts 85 J of thermal energy to a reservoir at 300 K. Comute the change in entroy for each reservoir. or The entroy of the Universe increases. or Natural rocesses tend to move toward a state of greater disorder. Solution Natural rocesses and Since the hot reservoir loses heat, we have that Q 100 J S = --- = = 0.25 J K 1 T 400 K For the cold reservoir we have Q 85 J S = -- = = J K 1 T 300 K change in entroy lthough the local entroy may decrease, any rocess will increase the total entroy of the system and surroundings, that is, the universe. Take for examle a chicken growing inside an egg. The entroy of the egg and its contents decreases because the inside of the egg is becoming more ordered. However, the entroy of surroundings increases by a greater factor because the rocess gives off thermal energy. So the total energy of the Universe is increasing. The change in entroy of the hot reservoir is 0.25 J K -1 and the change in entroy of the cold reservoir is 0.28 J K -1. The change in entroy of the cold reservoir is greater than the decrease for the hot reservoir. The total change in entroy of the whole system equals J K -1. That is, S = S H + S L = = J K 1 So that the net change in entroy is ositive. In this examle and all other cases, it has been found that the total entroy increases. (For an ideal Carnot reversible cycle it can equal zero. The Carnot cycle was discussed earlier). This infers that total entroy increases in all natural systems. The entroy of a given system can increase or decrease but the change in entroy of the system ΔSs lus the change in entroy of the environment ΔSenv must be greater than or equal to zero. i.e., 290 S = S S + S env 0 In terms of entroy, the Second Law of Thermodynamics can be stated as The total entroy of any system lus that of its environment increases as a result of all natural rocesses. In the beginning of Section 10.3, irreversible rocesses were discussed. block of ice can slide down an incline lane if the frictional force is overcome, but the ice cannot sontaneously move u the incline of its own accord. The conversion of mechanical energy to thermal energy by friction as it slides is irreversible. If the thermal energy could be converted comletely to mechanical energy, the Kelvin-Planck statement of the second Law would be violated. In terms of entroy, the system tends to greater disorder, and the entroy increases. In another case, the conduction of thermal energy from a hot body to a cold body is irreversible. Flow of thermal energy comletely from a cold body to a hot body violates the Clausius statement of the Second Law. In terms of entroy, a hot body causes greater disorder of the cold body and the entroy increases. If thermal energy was given by a cold body to a hot body there would be greater order in the hot body, and the entroy would decrease. This is not allowed by the Second Law. Irreversibility can also occur if there is turbulence or an exlosion causing a non-equilibrium state of the gaseous system. The degree of disorder increases and the entroy increases. Entroy indicates the direction in which rocesses occur. Hence entroy is often called the arrow of time. statistical aroach to the definition of entroy was first alied by Ludwig Boltzmann. Boltzmann ( ), Physics Ch 10 final.indd 290 3/12/2007 1:27:19 PM

19 an ustrian hysicist, was also concerned with the heat death of the Universe and irreversibilty. He concluded that the tendency toward dissiation of heat is not an absolute Law of Physics but rather a Statistical Law. Consider air molecules in a container. t any one instant, there would be a large number of ossibilities for the osition and velocity of each molecule its microstate and the molecules would be disordered. Even if there is some momentary order in a grou of molecules due to chance, the order would become less after collision with other molecules. Boltzmann argued that robability is directly related to disorder and hence to entroy. In terms of the Second Law of Thermodynamics, robability does not forbid a decrease in entroy but rather its robability of occurring is extremely low. If a coin is flied 100 times, it is not imossible for the one hundred coins to land heads u, but it is highly imrobable. The robability of rolling 100 sixes from 100 dice is even smaller. small samle of a gas contains billions of molecules and the molecules have many ossible microstates. It is imossible to know the osition and velocity of each molecule at a given oint in time. The robability of these microstates suddenly coming together into some imrobable arrangement is infinitesimal. In reality, the macrostate is the only measurable art of the system. The Second Law in terms of robability does not infer that a decrease in entroy is not allowed but it suggests that the robability of this occurring is low. final consequence of the Second Law is the heat degradation of the Universe. It can be reasoned that in any natural rocess, some energy becomes unavailable to do useful work. n outcome of this suggests that the Universe will eventually reach a state of maximum disorder. n equilibrium temerature will be reached and no work will be able to be done. ll change of state will cease as all the energy in the Universe becomes degraded to thermal energy. This oint in time is often referred to as the heat death of the Universe. Exercise The efficiency of a heat engine is the ratio of. the thermal energy inut to the thermal energy outut B. the thermal energy outut to the thermal energy inut C. the work outut to the thermal energy inut D. the work outut to the thermal energy outut 2. heat engine is most efficient when it works between objects that have a. large volume B. large temerature difference C. large surface area D. small temerature difference 3. The four-stroke engine is often said to consist of the suck, squeeze, bang and blow strokes. Use a series of diagrams to describe what this means. 4. Exlain the difference between internal and external combustion engines, and give an examle of each. 5. car engine oerates with an efficiency of 34% and it does J of work each cycle. Calculate (a) (b) the amount of thermal energy absorbed er cycle at the high-temerature reservoir. the amount of exhaust thermal energy sulied to the surroundings during each cycle. 6. On a hot day, a erson closed all the doors and windows of the kitchen and decided to leave the door of the refrigerator oen to cool the kitchen down. What will haen to the temerature of the room over a eriod of several hours. Give a full qualitative answer. 7. Modern coal-fired ower lants oerate at a temerature of 520 C while nuclear reactors oerate at a temerature of 320 C. If the waste heat of the two lants is delivered to a cooling reservoir at 21 C, calculate the Carnot efficiency of each tye of lant Physics Ch 10 final.indd 291 3/12/2007 1:27:20 PM

20 Chater It takes J of thermal energy to melt a given samle of a solid. In the rocess, the entroy of the system increases by 1740 J K -1. Find the melting oint of the solid in C. 9. If 2.00 kg of ure water at 100 C is oured into 2.00 kg of water at 0 C in a erfectly insulated calorimeter, what is the net change in entroy. (ssume there is 4.00 kg of water at a final temerature of 50 C). 10. Use the concets of entroy and the arrow of time to exlain the biological growth of an organism. 11. You are given six coins which you shake and then throw onto a table. Construct a table showing the number of microstates for each macrostate. 12. Describe the concet of energy degradation in terms of entroy. 13. Using an examle, exlain the meaning of the term reversal aradox. 14. What is meant by the heat death of the universe? Physics Ch 10 final.indd 292 3/12/2007 1:27:20 PM