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Chapter 12: Laws of Thermodynamics Suggested homework assignment: 4,21,33,39,48 Work in Thermodynamics Processes Work done on a gas • Energy can be transferred to a system by heat and by work done on the system. • In this chapter, all the systems we are concerned with are volumes of gas and they are in thermodynamic equilibrium: every part of the gas is at the same temperature and pressure. • Consider a gas contained by a cylinder with a movable piston and in equilibrium. The gas occupies a volume V and exerts a uniform pressure P on the cylinder wall. Work in Thermodynamics Processes Work done on a gas (cont’d) • The gas is compressed slowly enough so that the system remains essentially in thermodynamic equilibrium. • As the piston is pushed downward by an external force F through a distance Dy, the work done on gas is: W FDy PADy DV ADy W PDV under constant pressure • If the gas is compressed, DV is negative and the work done on the gas is positive. If the gas expands, the work done on the gas is negative. Work in Thermodynamics Processes PV diagram • PV diagram W PDV P(V f Vi ) (area under the graph) • In general, the area under the graph in PV diagram is equal in magnitude to the work done on the gas, whether or not the process proceeds at constant pressure. Note that Pa x m3 = J. isobaric process First Law of Thermodynamics First law of thermodynamics • The first law of thermodynamics is another energy conservation law that relates changes in internal energy – the energy associated with the position and movement of all the molecules of a system – to energy transfers due to heat and work. • Two mechanisms to transfer energy between a system and its environment : a macroscopic displacement of an object by a force and by heat which occurs through random molecular collisions. Both mechanisms result in a change in internal energy DU. First law of thermodynamics DU U f U i Q W Ui : initial internal energy Uf : final internal energy Q : energy transferred to the system by heat W : work done on the system First Law of Thermodynamics First law of thermodynamics (cont’d) • The internal energy of any isolated system must remain constant. Molar specific heat at constant volume • Molar specific heat at constant volume Cv is the amount of heat Q needed to change the temperature by DT per mole at constant volume and is defined as: Q nCV n : number of mole DT V Internal energy of a monatomic ideal gas • We learned that the internal energy of a monatomic ideal gas is: U 3 nRT 2 • Therefore a change in U is proportional to a change in T: DU 3 nRDT 2 First Law of Thermodynamics Molar specific heat • From the first law of thermodynamics : DU U f U i Q W • At constant volume, W=0. Therefore the first law becomes: DU Q nCV DT • However, for an ideal gas: 3 DU nRDT 2 Therefore, whether its volume is constant or not, 3 CV R 2 First Law of Thermodynamics Molar specific heat (cont’d) • A gas with a larger specific heat requires more energy to realize a given temperature change. The size of molar specific heat depends on the structure of the gas molecule and how many ways it can store energy. A monatomic gas (He etc.) can store energy as motion in three different directions (3-dimension). Degrees of freedom = 3 A diatomic gas (H2 etc.) can store energy as motion in three different directions (3-dimension), and can tumble and rotate in two different directions. Degrees of freedom = 5 Other molecules can store energy as vibrations, which add more degrees of freedom. Each degree of freedom contribute to the molar specific heat by (1/2)R. So, for example, the molar specific heat of a diatomic gas such as oxygen O2 is : 5 x (1/2)R = (5/2)R. First Law of Thermodynamics Molar specific heat (cont’d) • See Table of molar specific heats Isobaric process • An isobaric process is a process where the pressure remains constant. W PDV • Isobaric process and first law: DU Q W DU Q PDV Q DU PDV First Law of Thermodynamics Isobaric process (cont’d) • Ideal gas in isobaric process Q DU PDV For a monatomic ideal gas : DU 3 nRDT 2 For an isobaric process : PV nRT PDV nRDT 3 5 nRDT nRDT nRDT 2 2 Q Define molar specific heat at constant pressure as: DT nC p p For an ideal gas in an isobaric process: Q Q nC p DT 5 5 nRDT C p R 2 2 C p CV R This is valid for all ideal gasses First Law of Thermodynamics Adiabatic process • An adiabatic process is a process where no energy enters or leaves the system by heat – the system is thermally isolated. For an adiabatic process, Q0 First law of thermodynamics DU W • It can also be shown that for an ideal gas in an adiabatic process : PV constant w here Cp CV First Law of Thermodynamics Isovolumetric process • An isovolumetric process, also called an isochoric process, is a process where the volume is constant. • Since the volume does not change, there is no work done. DU Q For an ideal gas : DU Q nCV DT First Law of Thermodynamics Isothermal process • An isothermal process is a process where the temperature is constant. • For an ideal gas, since the internal energy depends only on the temperature: DU 0 First law of thermodynamics W Q • It can be shown that the work done on an ideal gas is given by: Vf W nRT ln Vi First Law of Thermodynamics Examples • Example 12.4 : Expanding gas Suppose a system of monatomic ideal gas at 2.00x105 Pa and an initial temperature of 293 K slowly expands at constant pressure from a volume of 1.00 L to 2.50 L. (a) Find the work done on the environment. Wenv PDV (2.00 105 Pa)(2.50 10-3 m3 1.00 10 3 m 3 ) 3.00 10 2 J (b) Find the change in the internal energy of the gas. 2.50 10-3 m3 T f Ti (239 K) 733 K 3 3 PiVi Ti Vi 1.00 10 m Pf V f Tf Vf PiVi (2.00 105 Pa)(1.00 10-3 m3 ) n 8.21102 mol RTi (8.31 J/(K mol)(293 K) 3 DU nCV DT nRDT 4.50 10 2 J 2 First Law of Thermodynamics Examples • Example 12.4 : Expanding gas (cont’d) (c) Use the first law to obtain the energy transferred by heat. DU Q W Q DU W W Wenv 3.00 102 J Q 4.50 102 J (3.00 102 J) 7.50 102 J (d) Use the molar heat capacity at constant pressure to obtain Q. 5 Q nC p DT nRDT 7.50 10 2 J 2 (e) How would the answers change for a diatomic gas? 3 DU nCV DT 1nRDT 7.50 10 2 J 2 5 Q nC p DT 1nRDT 1.05 103 J 2 First Law of Thermodynamics Examples • Example 12.5 : Work and engine cylinder In a car engine operating at 1.80x103 rev/min, the expansion of hot, high-pressure gas against a piston occurs in about 10 ms. because energy transfer by heat typically takes a time on the order of minutes or hours, it is safe to assume that little energy leaves the hot gas during expansion. Estimate the work done by the gas on the piston during this adiabatic expansion by assuming the engine cylinder contains 0.100 moles of an ideal monatomic gas which goes from 1.20x103 K to 4.00x102 K typical engine temperatures, during the expansion. W DU Q DU 0 DU DU U f U i 3 nR(T f Ti ) 9.97 10 2 J 2 Wpiston W DU 9.97 102 J First Law of Thermodynamics Examples • Example 12.6 : An adiabatic expansion A monatomic ideal gas at a pressure 1.01x105 Pa expands adiabatically from an initial volume of 1.50 m3, doubling its volume. (a) Find the new pressure. Cp CV (5 / 2) R 5 (3 / 2) R 3 C P1V1 1.99 105 Pa m5 C P2V2 P2 3.19 104 Pa (b) Estimate the work done on the gas. First Law of Thermodynamics Examples • Example 12.7 : An isovolumetric process How much thermal energy must be added to 5.00 moles of monatomic ideal gas at 3.00x102 K and with a constant volume of 1.50 L in order to raise the temperature of the gas by 3.80x102 K? 3 Q DU nCV DT nRDT 4.99 103 J 2 • Example 12.8 : An isothermal expansion A balloon contains 5.00 moles of monatomic ideal gas. As energy is added to the system by heat, the volume increases by 25% at a constant temperature of 27.0oC. Find the work Wenv done by the gas in expanding the balloon, the thermal energy Q transferred to the gas, and the work W done on the gas. Wenv Vf nRT ln Vi 2.78 103 J Q W (note :V f 1.25Vi ) Second Law of Thermodynamics Heat engines • A heat engine takes in energy by heat and partially converts it to other forms, such as electrical and mechanical energy. • A heat engine, in general, carries some working substance through a cyclic process during which : (1) energy is transferred by heat from a source at a high temperature (2) work is done by the engine (3) energy is expelled by the engine by heat to a source at lower temperature. • A steam engine, the working substance is water. The water in the engine is carried through a cycle in which it first evaporates into steam in a boiler and then expands against a piston. After the steam is condensed with cooling water, it returns to the boiler, and the process is repeated. Second Law of Thermodynamics Heat engines (cont’d) • The engine absorbs energy Qh from the hot reservoir, does work Weng, then gives up energy Qc to the cold reservoir. Because the working substance goes through a cycle, always returning to its initial thermodynamic state – the initial and final internal energy is the same, so DU=0. DU 0 Q W Qnet W Weng Qh Qc Weng Qh Qc The work Wenv done by a heat engine equals the net energy absorbed by the engine. Second Law of Thermodynamics Heat engines (cont’d) • If the working substance is a gas, the work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram. • The thermal efficiency of a heat engine is defined by : e Weng Qh Qh Qc Qh 1 Qc Qh Second Law of Thermodynamics Examples • Example 12.10 : Efficiency of an engine During one cycle, an engine extracts 2.00x103 J of energy from a hot reservoir and transfers 1.50x103 J to a cold reservoir. (a) Find the thermal efficiency of the engine. e 1 Qc Qh 0.250 or 25.0% (b) How much work does this engine do in one cycle? Weng Qh Qc 5.00 102 J (b) How much power does the engine generate if it goes through four cycles in 2.50 s? W 4.00 (5.00 102 J) P 8.00 102 W Dt 2.50 s Second Law of Thermodynamics Examples • Example 12.11 : Analyzing an engine cycle A heat engine contains an ideal gas confined to a cylinder by a movable piston. The gas starts at A where T=3.00x102 K and B->C is an isothermal process. (a) Find the number n of moles of the gas and the temperature at B. n PAVA 0.203 mol RT A TB PBVB 9.00 10 2 K nR Second Law of Thermodynamics Examples • Example 12.11 : Analyzing an engine cycle (cont’d) (b) Find DU, Q, and W for isovolumetric process A->B. 3 DU AB nCV DT n R DT 1.52 103 J 2 DV 0 WAB 0 QAB DU AB 1.52 103 J Second Law of Thermodynamics Examples • Example 12.11 : Analyzing an engine cycle (cont’d) (c) Find DU, Q, and W for isothermal process B->C. DU AB nCV DT 0 VC WBC nRT ln 1.67 103 Pa VB DU BC 0 QBC WBC QBC WBC 1.67 103 J Second Law of Thermodynamics Examples • Example 12.11 : Analyzing an engine cycle (cont’d) (d) Find DU, Q, and W for isobaric process C->A. WCA PDV 1.01103 J (1 atm 1.01105 Pa) DU CA 3 nRT 1.52 103 J 2 QCA DU CA WCA 2.53 103 J Second Law of Thermodynamics Examples • Example 12.11 : Analyzing an engine cycle (cont’d) (e) Find the net change in internal energy DUnet DU net DU AB DU BC DU CA 0 (f) Find the energy input, Qh; the energy rejected, Qc; the thermal efficiency; and the net work performed by the engine. Qh QAB QBC 3.19 103 J Qc 2.53 103 J Qc e 1 0.207 Qh Weng (WAB WBC WCA ) 6.60 102 J Second Law of Thermodynamics Refrigeration and heat pump • A reversed heat engine is called refrigeration! Energy is injected into the engine called heat pump and that results in extraction of energy from the cold reservoir to the hot reservoir. Examples are refrigerator and air conditioner. • Coefficient of performance For a heat pump in the cooling mode COP (cooling mode) Qc W For a heat pump in the heating mode COP (heating mode) Qh W Second Law of Thermodynamics Example 12.12 : Cooling the leftovers • 2.00 L of leftover soup at T=323 K is placed in a refrigerator. Assume the specific heat of soup is the same as that of water and the density 1.25x103 kg/m3. The refrigerator cools the soup to 283 K. (a) If the COP of the refrigerator is 5.00, find the energy needed, in the form of work, to cool the soup. m V 2.50 kg Qc Q mcDT 4.19 105 J COP Qc W 5.00 W 8.38 104 J (b) If the compressor has a power rating 0.250 hp find the time needed to cool the food. P (0.250 hp )(746 W/1 hp) 187 W W Dt 448 s P Second Law of Thermodynamics Second law of thermodynamics • There are limits to the efficiency of heat engines. • An ideal engine which would convert all the input energy into work does not exist. • The Kelvin-Planck formulation of the second law of thermodynamics: No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of an equal amount of work. • This means that the efficiency e=Weng/|Qh| of engines must always be less than one. Some energy Qc must always be lost to the environment. • It is theoretically impossible to construct a heat engine with an efficiency 100%. We cannot get a greater amount of energy out of a cyclic process that we put in. Second Law of Thermodynamics Reversible and irreversible processes • No engine can operate with 100% efficiency, but different designs yield different efficiencies. • One design called Carnot cycle (engine) delivers the maximum possible efficiency. • A reversible process is a process in which every state along the same path is an equilibrium state. In a reversible process, the system can return to its initial condition (state) by going along the same path in reverse direction. • An irreversible process is a process which does not satisfy the condition for a reversible process. • Most natural processes are irreversible, but some of them are almost reversible. If a real process occurs so slowly that the system is virtually always in equilibrium, the process can be considered reversible. Second Law of Thermodynamics Carnot engine • Consider a heat engine operating in an ideal, reversible cycle called a Carnot cycle. Second Law of Thermodynamics Carnot engine (cont’d) • Consider a heat engine operating in an ideal, reversible cycle called a Carnot cycle. A->B : Isothermal expansion at Th. Qh from hot reservoir. WAB done by gas. B->C : Adiabatic expansion Th->Tc. No heat goes out or comes in. WBC done by gas. C->D : Isothermal compression at Tc. Qc to cold reservoir. WCD done on gas. D->A : Adiabatic compression Tc->Th. No heat goes out or comes in. WDA done on gas. Second Law of Thermodynamics Carnot engine (cont’d) • Ratio of heat input to output vs. ratio of temperatures Qc Tc Qh Th • Thermal efficiency of a Carnot enegine Qc Tc eC 1 1 Qh Th Th , Tc in K All Carnot engines operating reversibly between the same two temperatures have the same efficiency. • All real engines operate irreversibly, due to friction and brevity of their cycles, and are therefore less efficient that the Carnot engine. Second Law of Thermodynamics Example 12.13 : Steam engine • A steam engine has a boiler that operates at 5.00x102 K. The energy from the boiler changes water to steam which drives the piston. The temperature of the exhaust is that of the outside air, 300 K. (a) What is the engine’s efficiency if it is an ideal engine? Tc eC 1 0.400 (40%) Th (b) If the 3.50x103 J of energy is supplied from the boiler, find the work done by the engine on its environment. Qc Tc Tc Qc Qh 2.10 103 J Qh Th Th Weng Qh Qc 1.40 103 J Entropy Definition of entropy • Let Qr be the energy absorbed or expelled during a reversible, constant temperature process between two equilibrium states. Then the change in entropy during any constant temperature process connecting the two equilibrium states id defined by: Qr DS T SI unit: joules/kelvin (J/K) • A similar formula holds even when the temperature is not constant. • Although calculation of DS during a transition between two equilibrium states requires finding a reversible path that connects the states, the entropy change calculated on that reversible path is taken to be DS for the actual path. This is valid logic as the change in entropy DS depends only on the initial and final states and not on the path taken. • The entropy of the Universe increases in all natural processes. Entropy Meaning of entropy • In nature a disorderly arrangement is much more probable than an orderly one if the laws of nature are allowed act without interference. • Using statistical mechanics it can be concluded that isolated systems tend toward great disorder, and entropy is a measure of that disorder. S kB ln W kB : Boltzman constant W : a number proportional to the probability that system has a particular configuration. • The second law of thermodynamics is really a statement of what is most probable rather than of what must be. • The entropy of the Universe always increases Entropy Examples • Example 12.14 : Melting a piece of lead (a) Find the change in entropy of 300 g of lead when it melts at 327oC. Lead has a latent heat of fusion of 2.45x104 J/kg. Q mLf 7.35 10 J 3 T TC 273 6.00 10 2 K Q DS 12.3 J/K T (b) Suppose the same amount of energy is used to melt part of a piece of silver, which is already at its melting point of 961oC. Find the change in the entropy of the sliver. T TC 273 1.23 102 K Q DS 5.96 J/K T Entropy Examples • Example 12.15 : Ice, steam, and the entropy of the Universe A block of ice at 273 K is put in thermal contact with a container at 373 K, converting 25.0 g of ice to water at 273 K while condensing some of the steam to water at 373 K. (a) Find the change in entropy of the ice. Q mLf 8.33103 J DSice Qice 30.5 J/K Tice (b) Find the change in entropy of the steam. Thermal energy lost by the Qsteam Qice DS steam 22.3 J/K steam is equal to the thermal Tsteam Tsteam energy gained by the ice. (c) Find the change in entropy of the Universe. DSUni DSice DS steam 8.2 J/K Entropy Examples • Example 12.16 : A falling boulder A chunk of rock of mass 1.00x103 kg at 293 K falls from a cliff of height 125 m into a large lake, also at 293 K. Find the change in entropy of the lake, assuming that all of the rock’s kinetic energy upon entering the lake converts to thermal energy absorbed by the lake. PE mgh 1.23 106 J The rock’s kinetic energy at the time of entrance to the lake. Q PE DS 4.20 103 J/K T T Human Metabolism Application of thermodynamics to living organisms • Animals do work and give off energy by heat, and this lead us to believe the first law of thermodynamics can be applied to living organisms. • Let’s apply the first law in terms of the time rates of change of DU, Q, and W. On average, energy Q flows out of the body, and DU Q W work is done by the body on its surroundings: DU/Dt is negative Dt Dt Dt Q/Dt and W/Dt are negative. • Without supply of energy, the internal energy and the body temperature would decrease. But in reality, all animals acquire internal energy (chemical potential energy) by eating and breathing. • Overall the energy from oxidation of food ultimately supplies the work done by the body and energy lost from the body by heat. From this point of view, DU/Dt is the rate at which internal energy is added to our bodies by food, which balances the rate of energy loss by heat and work. Human Metabolism Measuring the metabolic rate • The metabolic rate DU/Dt is the rate at which chemical potential energy in food and oxygen are transformed into internal energy to balance the body losses of internal energy by work and heat. • The metabolic rate DU/Dt is directly proportional to the rate of oxygen consumption by volume. For an average diet, the consumption of one DVO2 DU liter of oxygen releases 4.8 kcal or 20 kJ of 4.8 Dt Dt energy. L/s Human Metabolism Metabolic rate, activity, and weight gain • Table below summarizes the measured rate of oxygen consumption in mL/(min kg) and the calculated metabolic rate for 65-kg male engaged in various activities. Human Metabolism Physical fitness and efficiency of the human body as a machine • One measure of a person’s physical fitness is his or her maximum capacity to use or consume oxygen. Table below gives some idea how well a person fit. • The body’s efficiency e is defined as the ratio of the mechanical power supplied by a human to the metabolic rate: W Dt e DU Dt Human Metabolism Example 12.17 : Fighting fat • In the course of 24 hours, a65-kg person spends 8 h at a desk puttering around the house, 1h jogging 5 miles, 5 h in moderate activity, and 8 h sleeping. What is the change in her internal energy during this period? DU Pi Dti (200 kcal/h)(10 h) (5 mi/h)(120 kcal/mi)(1 h) (400 kcal/h)(5 h) (70 kcal/h)(8 h) 5000 kcal