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# Physics for Engineers and Scientists I PY 205

NCS

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This 5 page Class Notes was uploaded by Asa Wisozk on Thursday October 15, 2015. The Class Notes belongs to PY 205 at North Carolina State University taught by David Aspnes in Fall. Since its upload, it has received 19 views. For similar materials see /class/223911/py-205-north-carolina-state-university in Physics 2 at North Carolina State University.

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Date Created: 10/15/15

On thermodynamics Giancoli Chs 1720 Thermodynamics deals with internal energy ie energy stored or released as heat as a result of an object interacting with another object When you stop your car from 65 mph the kinetic energy 12mv2 of the car does not vanish but is transformed into internal energy of the brake discs and brake pads raising their temperatures Examples of added energy causing effects other than heating include melting solids ice to water for example and converting a liquid to a vapor water to steam for example These are phase changes Much of thermodynamics concerns gases rather than liquids or solids because gases can store energy efficiently via volume pressure and temperature changes and can conveniently turn internal energy in the form of heat into useful work think intemalcombustion engines Chapter 17 This chapter introduces the Fahrenheit Celsius Centigrade and Kelvin temperature scales As you know 32 F 0 C 27316 K The Fahrenheit scale is never used in scientific work so we don t refer to it further We use either Celsius Centigrade C or Kelvin K without the The equation to convert Celsius to Kelvin and vice versa is T c 27316 K TK Thermodynamic expressions such as the ideal gas law PV nRT NkT use only Kelvin A good rule to follow is that all temperatures need to be converted to K be applying any thermodynamic equation Depending on the question you might have to convert temperatures back to C but the calculations themselves must be done in K Chapter 17 also introduces the ideal gas law PV nRTNkT where P pressure V volume 71 number ofmoles in V R ideal gas constant 8314 Jmole K T is the absolute temperature in K N is the number of molecules in V and k is Boltzmann s Constant 138 X 10 23 JK Since 1 mole of gas contains 6023 X 1023 atoms in the case ofmonatomic gas such as He or molecules in the case of a polyatomic gas such as N2 or 02 it follows that R k X Avogadro s Number Chapters 18 and 19 provide the atomicscale basis of thermodynamics Physics takes place in real time in real space and on the atomic scale and thermodynamics is no exception The average kinetic energy of an atom in a monatomic gas such as He is 12mv2 32kT whereas that of a diatomic molecule such as N2 or 02 the stored energy is 52kT In both cases T is in Kelvin Why the difference in prefactors 32 and 52 between monatomic and diatomic gases We didn t cover it in detail but the reason is easy to understand and is covered in Giancoli 198 Each degree of freedom represents an independent way that an atom or molecule can store energy With the enormous number of collisions taking place in any gas on the average energy gets divided equally among the different degrees of freedom with each degree of freedom getting on the average 12kT Thus the 3 in the factor of 32 for the monatomic gas follows because there are 3 independent directions x y and z and v2 vx2 vy2 vz2 so the kinetic energies in the 3 different directions 12kT each simply add For diatomic molecules we not only have these 3 translational degrees of freedom but can also store energy by rotating the molecule and in addition by internal vibrations of the molecule KE plus these additional two degrees of freedom means that diatomic molecules can store an average energy of 52kT If there are N molecules in a volume Vof gas then the total internal energy Eim in Vis 32NkT for a monatomic gas and 5 2NkT for a diatomic gas Taking advantage of the ideal gas law PV NkT we can also write Eim 32PVfor a monatomic gas and Eim 5 2PVfor a diatomic gas Again these internal energies depend ONLY on temperature We ll come back to this later Since the internal energy of each degree of freedom depends only on temperature and since collisions do a great job of spreading energy around it s not surprising that objects initially at different temperatures end up equilibrating to the same temperature Again this process becomes obvious if we consider what happens on the atomic scale Chapter 18 also shows that what we call pressure is simply the average of the enormous number of impulses that strike an object in a gas as a result of its kinetic energy Since only the KB is involved we have for any gas monatomic or otherwise 12mv2 3 2kT This allows the rootmeansquare velocityv Hz to be evaluated For a given concentration of rm molecules P is proportional to T Chapter 19 introduces the First Law of Thermodynamics which is the mathematical statement that energy can neither be created nor destroyed you can t win We write this most efficiently as AEth AQ AW where AEth is the change of internal energy of a system AQ is the heat energy added to the system from another system and AW is the work done by the system on another system If heat ows out of the system then AQ is negative and if work is done on the system AW is negative Heat energy is measured in calories The calorie is de ned as the amount of energy required to raise 1 gm of water from 155 to 165 C and 1 calorie 4186 J The calorie used in connection with food and nutrition is actually a kilocalorie The measure of the amount of heat needed to raise the temperature of a material by an amount AT is the heat capacity at constant pressure Cp or constant volume CV using the conventional notation Thus AQ meAT where AQ is the heat needed or given up m the mass Cp the heat capacity and AT the change of temperature For liquid water Cp 1 cal gm C For liquids and solids the change of volume with pressure or temperature is so negligible that it makes little difference whether we use Cp or CV One can also specify speci c heat which is the ratio of heat required to that of water Giancoli Table 191 gives a table of speci c heats for various materials The amounts of heat required to melt or vaporize a material are represented by the latent heats of fusion and vaporization respectively We have AQ mLf or AQ va for melting or vaporization respectively where m is the mass of material that has been melted or vaporized For phase changes in the reverse direction AQ is negative In Fig 195 Giancoli gives an outstanding summary of the heat input required to heat ice from 40 C to 0 C then to melt the ice to form water then to heat the water to 100 C then to boil the water to form steam then nally to heat the steam Since steam is a triatomic gas the expression for average internal energy is somewhat different than that for a monatomic or diatomic gas the average energy stored per molecule is about 62kT 3kT Now we consider gases These are more complicated but also much more interesting because we can easily change their temperature pressure and volume and hence exchange internal energy for work and vice versa In short they represent a convenient means of storing releasing and transferring energy The freebody diagram of thermodynamics is the PVplot where P is plotted against V Before getting into speci c conditions we note that the following 3 equations must hold independent of the constraints that we ll apply in the next paragraphs Ideal gas law PV nRT NkT De nition of differential work from early inthe course dW F dx FAAdx PdV Internal energy of a mortatomic gas Eim 32NkT 32PVand that of a diatomic gas Eint 52NkT 52PV V2 By integrating dW PdV we obtain W2 W1 IPdV which you recognize as the area under V1 the Pchrve Thus the PVplot gives a direct measure of the work involved in a given process Now let s consider particular cases 1 isothermal By definition here T constant hence by the ideal gas law PV constant and since Eim N T Eim must be constant as well Then by the first law AQ AW so heat gets transferred directly into an equivalent amount of work or vice versa The thermal bath that maintains T constant acts here as a source or sink of energy We can use the ideal gas law to evaluate the work V2 V2 W2 W1 deVnRTjd VnRT1nEPim V1 V1 V V1 V1 Isothermals form part of the cycle of a Carnot cycle which is that of an ideal engine as discussed below The trajectory of an isothermal process in aPVplot is a hyperbola PV constant 2 isobaric By definition here P constant so the ideal gas law shows that in an isobaric process where Vchanges T and hence the internal energy must also change The work done in an isobaric process is easily evaluated V2 WZ VV1iPdVPltVz V1 V1 The trajectory of an isobaric process in aPVplot is obviously a horizontal line 3 isovolumetric By definition here V constant so in an isovolumetric process no work is done Thus by the first law AEim AQ so any heat added or removed involves only the internal energy The ignition and exhaust parts of the Otto cycle of internalcombustion engines are isovolumetric processes The trajectory of an isovolumetric process in aPVplot is obviously a vertical line 4 adiabatic An adiabatic process is one where no heat is added or removed from the system hence AEth AW That is work done on or by a system comes entirely at the expense of the internal energy The compression and power parts of the Otto cycle are adiabatic processes since they happen too fast for any appreciable heat to be removed from the system Adiabatic processes also form part of the Camot cycle We can get an expression for the work done in an adiabatic process and hence the trajectory of P vs Vin aPVplot by considering the definition of differential work and using the differential connection between P V and Eim Using a diatomic gas air as an example we rst have AE PdV VdP Next we have dW PdV dE PdVVdP int int Rearranging terms gives We integrate this to get lnP 2 Zan 2 which is 2 V 2 P P1 5 V1 usually written PW constant where y 75 14 for a diatomic gas For a monatomic gas with no internal degrees of freedom and therefore all supplied energy going into KE 7 53 167 The trajectory of an adiabatic process in aPVplot is given by PM constant One of the most de nitive proofs that the above is actually correct and that we really do know what is going on at the atomic scale is that y for He and Ne is 167 and that for N2 and O is 140 see Giancoli Table 194 These values of y are a manifestation of atomic properties on a macroscopic or average scale Chapter 20 introduces the 2quotd Law of Thermodyamics which states that it is impossible to build a heat engine that operates in a cycle and turns all supplied heat energy into work without having to eliminate some heat as waste you can t even break even Cycle is important here because while the internal energy changes during a cycle at the end of the cycle it must return to the same value that it had at the beginning Hence we can write Emmet QH QL W 0 over the cycle where QH is the heat added during the highertemperature part of the cycle QL is the heat discharged during the lowertemperature part of the cycle and W is the work done Two cycles are important the Carnot cycle for theoretical reasons and the Otto cycle for practical reasons The Carnot cycle consists of an isothermal expansion at a high temperature TH where QH is added to the gas followed by an adiabatic expansion then an isothermal compression at a lower temperature TL where QL is removed from the gas and nally an adiabatic compression back to its original state see Giancoli Fig 207 The work W done is the area enclosed The ef ciency 6 of any cycle is de ned not surprisingly is the ratio of the amount of work W delivered to the amount of heat QH invested For the Carnot cycle Giancoli shows this T 1 g L 1 T L It is not surpr1s1ng that e for the Carnot cycle 6 can be H H W to be e QH described by the temperatures of the two isothermals since TH and TL are constants of the cycle The Otto cycle of a 2cycle engine Giancoli Fig 208 consists of an adiabatic compression from a larger volume VH to a smaller volume VL compression stroke followed by an isovolumetric injection of heat ignition of the airfuel mixture ie the conversion of chemical to thermal energy followed by an adiabatic expansion from VL back to VH power stroke then an isovolumetric removal of heat QL to reach the original state exhaust phase A 4cycle engine adds exhaust and intake strokes that are essentially isobaric at atmospheric pressure and hence contribute nothing to work The ef ciency of the Otto cycle is given by 771 W QL VL e 1 1 aga1n probably not surpr1smg smce VH and VL are constants of Q Q h J the cycle Note that VHVL is the compression ratio which is typically about 11 for a gasoline engine and 22 for a diesel From the above expression the reason for the higher fuel ef ciency of a diesel engine is obvious 7 the parameters of its operating cycle are simply more favorable for ef ciency Considering that to a good approximation air consists of diatomic molecules for engines in general y 14 These compression ratios then yield theoretical efficiencies of about 62 and 71 for gasoline and diesel engines respectively Giancoli also discusses refrigerators and heat pumps where the objective is to extract heat from a volume and discharge it into a surrounding medium usually ambient air Efficiency here is replaced by the coefficient of performance COP QL W QLQH QL For a Carnot cycle QH and QL can be replaced by TH and TL respectively As a practical note heat pumps generally work reasonably well in the South in winter but are impractical in the North where extracting heat by further cooling already cold air is inefficient

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