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UH PHYS 1302 - Ch18

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Chapter 18 - The Laws of Thermodynamics1 The Zeroth Law of ThermodynamicsIf object A is in thermal equilibrium with object C, and object B is separately inthermal equilibrium with object C, then objects A and B will be in thermal equilib-rium.Objects are in thermal equilibrium when they have the same temperature.2 The First Law of ThermodynamicsThe change in a system’s internal energy (∆U = Uf− Ui) is related to the heat Qand the work W as follows:∆U = Q − W (1)The first law is just conservation of energy, specifically including heat.Sign conventions:• Q > 0 for heat gained by the system• Q < 0 for heat lost by the system• W > 0 for work done by the system• W < 0 for work done on the system(Problem 18.3) A swimmer does 6.7 × 105J of work and gives off 4.1 × 105J of heatduring a workout. Determined ∆U, W , and Q for the swimmer.L. Whitehead 1 Phys 1302The work is done by the swimmer, so W = 6.7 × 105J. The heat is lost by theswimmer, so Q = −4.1 × 105J.∆U = Q − W = −4.1 × 105− 6.7 × 105= −10.8 × 105JThe state of a system is determined by the temperature, pressure, and volume.3 Thermal ProcessesAll of the processes that we’re going to talk about are assumed to be quasi-staticreversible, which means they occur so slowly that at any given time the system andits surroundings are in equilibrium. We also assume there are no dissipative forcesin the system. Another way of saying it is that these processes are all reversible.For a process to be reversible, it must be possible to return both the system andits surroundings to exactly the same states they were in before the process began.(Friction is an example of an irreversible process.)Constant Pressure ProcessSuppose we have gas held in a cylinder with cross-sectional area A. The pressureis held constant at P0, but the volume of the gas expands moving the piston fromposition xito position xf(the gas is doing work on the piston). The initial volumeof the gas is Vi= Axiand the final volume of the gas is Vf= Axf. The force theL. Whitehead 2 Phys 1302gas exerts on the piston is F = P0A, so the work done by the gas is:W = F (xf− xi) = P0A(xf− xi) = P0(Vf− Vi)In general, for a constant pressure process that changes the volume ∆V , the workdone by the gas is:W = P ∆V (2)In the pressure vs volume plot above, the area under the curve is P0(Vf− Vi), which isequal to the work done by the gas. This applies to any process: the work done by anexpanding gas is equal to the area under the pressure vs volume curve representingthe process.Constant Volume ProcessSuppose we have a gas held in a container of fixed volume. Suppose we add heat,causing the pressure to increase. Since the container has a fixed volume, the gas doesno work.W = 0 (3)∆U = Q − W = Q (4)Notice that the area under the pressure vs volume curve is zero, as expected for zerowork done.Isothermal Process (Constant Temperature)L. Whitehead 3 Phys 1302For an ideal gas at constant temperature, P V = NkT = constant, so P is inverselyproportional to V : P = constant/V .The work done by an expanding gas is equal to the area under the pressure vs volumecurve. For an isotherm curve, the area under the curve is:W = NkT lnVfVi= nRT lnVfVi(5)Calculus derivation (don’t show in class):W =ZF · dx=ZF dx=ZP Adx=ZP dV=ZNkTVdV= NkTZVfVidVV= NkT (ln Vf− ln Vi)= NkT lnVfViL. Whitehead 4 Phys 1302Adiabatic ProcessAn adiabatic process is one in which no heat flows in or out of the system.Q = 0 (6)∆U = Q − W = −W (7)Suppose we have a well-insulated cylinder that does not allow heat flow in or out.When the piston is pushed down, the volume of the gas decreases, but the tempera-ture and pressure both increase. When the volume is allowed to expand, the pressureand temperature both decrease.An adiabatic process can occur if the system is thermally insulated. An adiabaticprocess can also occur when the process happens so rapidly that there is no time forheat to flow.(Problem 18.24) During an adiabatic process, the temperature of 3.92 moles of amonatomic idea gas drops from 485◦C to 205◦C. For this gas, find (a) the workit does, (b) the heat it exchanges with its surroundings, and (c) the change in itsinternal energy.Because its an adiabatic process, the heat exchange is Q = 0.The internal energy of an ideal gas is given by U = (3/2)nRT . The change inL. Whitehead 5 Phys 1302temperature is ∆T = Tf− Ti= 205 − 485 = −280K∆U = −WW = −∆UW = −(Uf− Ui)W = −32nR(Tf− Ti)W = −32(3.92moles)(8.31J/(mole K))(−280K)W = 1.37 × 104J∆U = −W = −1.37 × 104J4 Specific Heats for an Ideal Gas: Constant Pres-sure, Constant Volume[skipping]5 The Second Law of ThermodynamicsWhen objects of different temperatures are brought into thermal contact, the spon-tanous flow of heat that results is always from the high-temperature object to thelow-temperature object. Spontaneous heat flow never proceeds in the reverse direc-tion.The second law is sometimes referred to as the “arrow of time”. If you saw a videoof a water droplet on someone’s hand spontaneously turning into a snowflake anddrifting upwards, you would know the video is running backward. That process couldnot happen spontaneously.L. Whitehead 6 Phys 13026 Heat Engines and the Carnot CycleA heat engine is a device that converts heat into work.In a steam engine, heat is supplied to vaporize water in the boiler. The steam en-ters the engine and expands, moving a piston. The movement of the piston suppliesmechanical work to the external world.The steam then goes into a condenser whereit gives off heat and condenses to liquid form.In a general heat engine, heat is supplied by a hot reservoir (the boiler in the steamengine). Some of that heat is converted to work, and the rest is given off as wasteheat to the cold reservoir (the condenser in the steam engine). If Qhis the magnitudeof the heat supplied by the hot reservoir and Qcis the magnitude of the heat givento the cold reservoir, conservation of energy indicates that the work W done is:W = Qh− Qc(8)The efficiency of the engine is defined as the fraction of the supplied heat that isconverted to work:e =WQh=Qh− QcQh= 1 −QcQh(9)Carnot’s Theorem:Suppose we have an engine operating between two constant-temperature reservoirs.For the engine to have the maximum possible efficiency, all processes must be re-versible.L. Whitehead 7 Phys 1302This maximum efficiency (which is only obtained by reversible engines) depends onlyon the


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UH PHYS 1302 - Ch18

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