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Chapter 15: Thermodynamics

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Chapter 15: Thermodynamics Chapter 14 Thermodynamics Reference:chap.15 Alan Giambattista, Betty McCarthy Richardson and Robert C. Richardson, 2009, “College Physics”, 3rd ed., McGraw-Hill. Only for education •The First Law of Thermodynamics •Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) •Reversible and Irreversible Processes •Heat Engines •Refrigerators and Heat Pumps •The Carnot Cycle •Entropy (The Second Law of Thermodynamics) •The Third Law of Thermodynamics 普物AB—海大輪機系 1 張宏宜 The First Law of Thermodynamics The first law of thermodynamics says the change in internal energy of a system is equal to the heat flow into the system plus the work done on the system. UQWΔ = + 普物AB—海大輪機系 2 張宏宜 Thermodynamic Processes A state variable describes the state of a system at time t, but it does not reveal how the system was put into that state. Examples of state variables: pressure, temperature, volume, number of moles, and internal energy. A PV diagram can be used to represent the state changes of a system, provided the system is always near equilibrium. The area under a PV curve gives the magnitude of the work done on a system. W>0 for compression and W<0 for expansion. 普物AB—海大輪機系 3 張宏宜 Thermodynamic Processes To go from the state (Vi, Pi) by the path (a) to the state (Vf, Pf) requires a different amount of work then by path (b). To return to the initial point (1) requires the work to be nonzero. The work done on a system depends on the path taken in the PV diagram. The work done on a system during a closed cycle can be nonzero. 普物AB—海大輪機系 4 張宏宜 Thermodynamic Processes An isothermal process implies that both P and V of the gas change (PV∝T). 普物AB—海大輪機系 5 張宏宜 Thermodynamics for an Ideal Gas No work is done on a system when its volume remains constant (isochoric process). For an ideal gas (provided the number of moles remains constant), the change in internal energy is Q= UΔ = nCV Δ . T 普物AB—海大輪機系 6 張宏宜 Thermodynamics for an Ideal Gas For a constant pressure (isobaric) process, the change in internal energy is UQWΔ = + whereW P= − Δ V= − Δ nRandQ T = nCPΔ . T CP is the molar specific heat at constant pressure. For an ideal gas CP = CV+R. For a constant temperature (isothermal) process, ΔU = 0 and the work done on an ideal gas is ⎛ Vi ⎞ W= nRTln⎜ ⎟ . ⎝Vf ⎠ 普物AB—海大輪機系 7 張宏宜 Summary of Thermodynamic Process 普物AB—海大輪機系 8 張宏宜 Example (text problem 15.7) An ideal monatomic gas is taken through a cycle in the PV diagram. (a) If there are 0.0200 mol of this gas, what are the temperature and pressure at point C? From the graph: Pc = 98.0 kPa Using the ideal gas law PV T=c1180 c = K. c nR 普物AB—海大輪機系 9 張宏宜 Example (text problem 15.7) (b) What is the change in internal energy of the gas as it is taken from point A to B? This is an isochoric process so W = 0 and ΔU = Q. ⎛ 3 ⎞⎛PVBBAAPV⎞ UΔ Q = = nCV Δ T⎜ = ⎟ n⎜ R− ⎟ ⎝ 2 ⎠⎝ nR nR ⎠ 3 =PVPV() − 2BBAA 3 =VPP() − =200 − J 2 BA 普物AB—海大輪機系 10 張宏宜 Example (text problem 15.7) (c) How much work is done by this gas per cycle? The work done per cycle is the area between the curves on the PV diagram. Here W=½ΔVΔP = 66 J. (d) What is the total change in internal energy of this gas in one cycle? ⎛ 3 ⎞⎛PVf fPV i⎞ i UΔ = nCV Δ T⎜ = n⎟⎜ R− ⎟ ⎝ 2 ⎠⎝ nR nR ⎠ 3 =PVPV() −0 = 2 f f i i The cycle ends where 普物AB—海大輪機系 it began (ΔT = 0). 11 張宏宜 Example (text problem 15.8) An ideal gas is in contact with a heat reservoir so that it remains at constant temperature of 300.0 K. The gas is compressed from a volume of 24.0 L to a volume of 14.0 L. During the process, the mechanical device pushing the piston to compress the gas is found to expend 5.00 kJ of energy. How much heat flows between the heat reservoir and the gas, and in what direction does the heat flow occur? This is an isothermal process, so ΔU = Q + W = 0 (for an ideal gas) and W = -Q = -5.00 kJ. Heat flows from the gas to the reservoir. 普物AB—海大輪機系 12 張宏宜 Reversible and Irreversible Processes A process is reversible if it does not violate any law of physics when it is run backwards in time. For example an ice cube placed on a countertop (廚具檯面) in a warm room will melt. The reverse process cannot occur: an ice cube will not form out of the puddle of water (一灘水) on the countertop in a warm room. A collision between two billiard balls is reversible. Momentum is conserved if time is run forward; momentum is still conserved if time runs backwards. 普物AB—海大輪機系 13 張宏宜 Reversible and Irreversible Processes Any process that involves dissipation of energy is not reversible. Any process that involves heat transfer from a hotter object to a colder object is not reversible. The second law of thermodynamics (Clausius Statement): Heat never flows spontaneously from a colder body to a hotter body. 普物AB—海大輪機系 14 張宏宜 Heat Engines A heat engine is a device designed to convert disordered energy into ordered energy. The net work done by an engine during one cycle is equal to the net heat flow into the engine during the cycle (ΔU= 0). WQnet = net The efficiency of an engine is defined as donenet work by the W engine e = = net . heat input Qin Note: Qnet = Qin -Qout 普物AB—海大輪機系 15 張宏宜 Internal Combustion Engine 1. Intake stroke: The piston is pulled out, drawing the fuel-air mixture into the cylinder at atmospheric pressure. 2. Compression stroke: The piston is pushed back in, compressing the fuel-air mixture and work is done on the gas. 3. Ignition: A spark ignites the gases, quickly and dramatically raising the temperature and pressure. 4. Power stroke: The high pressure that results from ignition pushes the piston out. The gases do work on the piston and some heat flows out of the cylinder. 5. Exhaust stroke: a valve opened and the exhaust gases are pushed out of the cylinder. 普物AB—海大輪機系 16 張宏宜 Refrigerators and Heat Pumps In a heat engine, heat flows from hot to cold, with work as the output. In a heat pump, heat flows from cold to hot, with work as the input. The efficiency of a heat engine can be rewritten as net work output W e = = net heat input QH QQ− Q = HC=1 − C . QH QH 普物AB—海大輪機系 17 張宏宜 Example (text problem 15.16) (a) How much heat does an engine with efficiency of 33.3 % absorb in order to deliver 1.00 kJ of work? W1 . 00 kJ Q=net =3 . =00 kJ H e 0.333 (b) How much heat is exhausted by the engine? Q e=1 − C QH Q=C ()1 − e2H Q . = 00 kJ 普物AB—海大輪機系 18 張宏宜 Reversible Engines and Heat Pumps A reversible engine can be used as an engine (heat input from a hot reservoir and exhausted to a cold reservoir) or as a heat pump (heat is taken from cold reservoir and exhausted to a hot reservoir). From the second law of thermodynamics, no engine can have an efficiency greater than that of an ideal reversible engine that uses the same two reservoirs. The efficiency of this ideal reversible engine is TC er=1 − . TH 普物AB—海大輪機系 19 張宏宜 Carnot Cycle The ideal engine of the previous section is known as a Carnot engine. The Carnot cycle has four steps: 1. Isothermal expansion: takes in heat from hot reservoir; keeping the gas temperature at TH. 2. Adiabatic expansion: the gas does work without heat flow into the gas; gas temperature decreases to TC. 3. Isothermal compression: Heat QC is exhausted; gas temperature remains at TC. 4. Adiabatic compression: raises the temperature back to TH. 普物AB—海大輪機系 20 張宏宜 Carnot Cycle 普物AB—海大輪機系 21 張宏宜 Example (text problem 15.40) An engine operates between temperatures 650 K and 350 K at 65.0% of its maximum efficiency. (a) What is the efficiency of this engine? The maximum possible efficiency is TC 350 K =er −1 = 1 −0 . 462 = . TH 650 K The engine operates at e = 0.65er = 0.30 or 30% efficiency. 普物AB—海大輪機系 22 張宏宜 Example (text problem 15.40) (b) If 6.3×103 J is exhausted to the low temperature reservoir, how much work does the engine do? The heats exchanged at the reservoirs are related to each other through QC=1( − e) H Q . WQQnet=HC − QC ⎛ e ⎞ = −QC =⎜ ⎟ Q2C = . 7 kJ ()1− e ⎝1− e ⎠ 普物AB—海大輪機系 23 張宏宜 Entropy Heat flows from objects of high temperature to objects at low temperature because this process increases the disorder of the system. Entropy is a measure of a system’s disorder. Entropy is a state variable. If an amount of heat Q flows into a system at constant temperature, then the change in entropy is Q ΔS = . T Every irreversible process increases the total entropy of the universe. Reversible processes do not increase the total entropy of the universe. The second law of thermodynamics (Entropy Statement): The entropy of the universe never decreases. 普物AB—海大輪機系 24 張宏宜 Example (text problem 15.48) An ice cube at 0.0 °C is slowly melting. What is the change in the ice cube’s entropy for each 1.00 g of ice that melts? To melt ice requires Q = mLf joules of heat.
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