Lecture 10: Carnot theorem

Transcription

1 ecture 0: Carnot teorem Feb 7, 005 Equivalence of Kelvin and Clausius formulations ast time we learned tat te Second aw can be formulated in two ways. e Kelvin formulation: No process is possible wose sole result is te complete conversion of eat into work. It can be also epressed in a sligtly different form: It is impossible to transform eat systematically into mecanical work in a cycle involving only one bat e Clausius formulation: No process is possible wose sole effect is transfer of eat from a colder to a otter body. It was mentioned tat tese two formulations are equivalent, wic can be formally proved as follows. Suppose te Second aw in te Kelvin s formulation does not old, i.e. one can produce some work using only one bat of some temperature. is work can be transformed into eat, for eample by friction. So using te work wic te engine produces one can increase te temperature of any oter system of arbitrary temperature. Since may be iger ten it turns tat one transfers eat from colder system to te ot one, wic is violation of te Second aw in te Clausius formulation. For instance, one can use te work to run anoter engine wic operates te inverse Carnot cycle (see te figure below). is combined macine produces no work and its net effect is transferring eat from cold bat to te ot one. e combined macine absorbs positive eat c from te cold bat, and delivers te eat d = to te ot bat. Clearly, d > 0 because = = c. (Note: in tis lecture we use symbols,, etc. for absolute values of corresponding quantities.)

2 = d c c c c et us made te opposite assumption, i.e. tat, in violation of te Clausius formulation, one manages to transfer a certain amount of eat from cold bat to te ot one. Suppose it is eactly te eat anoter eat engine absorbs from te ot bat during isotermal epansion. en te ot bat receives and gives up te same amount of eat, and remains virtually uncanged. As a result, te net effect is producing work at te epanse of eat etracted from a single (cold) bat, wic is a violation of te Second aw in te Kelvin formulation. e above argument prove tat te Clausius and Kelvin statements are equivalent. First Carnot teorem: General proof en we considered te Carnot cycle for an ideal gas we ave found tat efficiency of te cycle η = ( c )/ depends only on temp. of two eat bats and does not depend of gas caracteristics, say gas eat capacity. is suggests tat te result is actually general and valid for arbitrary gas. e first Carnot teorem: all reversible engines operating between te same two bats as te same efficiency η = ( c )/. Consider two Carnot cycle engines using different gases but working between te same two temperatures. Suppose te engines ave different efficiencies η and η. e Carnot teorem states tat it is impossible. et us assume for a moment tat te teorem is wrong and tat efficiencies are different, say η < η. One can demonstrate tat tis assumption leads to a contradiction. et us operate te first engine counterclockwise.

3 () () is would requires some eternal work. e can use te work produced by te second engine to feed te first one. e reasoning is especially simple for te special case wen =. Since η = / te assumption η < η gives >. Also, recall tat η = / = c en inequalities η < η and > gives = c. c > c. e total eat received from te cold bat bat is c = c c > 0 and te eat obtained by te ot bat is > 0. e total work produced by te combined engine is zero, and te net result is tat some amount of eat is pumped out of cold bat and is given up to te ot bat wic is violation of te Second law (in te Clausius formulation). In general. But one can easily bypass tis difficulty in a following way. Suppose / is a rational number and terefore can be written as / = n /n were n i some integers. en n = n. It is important to understand tat tere is no need to run bot macine simultaneously. In te combined macine te first and second engine may work in a sequence. For eample, first te second engine performs n cycles and does work n, for eample lifting te piston. en te 3

4 potential energy of te piston, wic is eactly n, is spent on running te first engine. If n and n satisfy te condition n = n, ten te first engine will performed eactly n cycles. et us consider all quantities related to te first engine, suc as, c, during not one but n cycles: = n, c = n c = n, and similar for te second engine = n, c = n c = n. Since =, one can repeat te above proof for te case = just canging te notations, and using te quantities related not to one cycle but to n cycle for te first engine and n cycles for te second one. 3 Second Carnot teorem e Carnot cycle is te only reversible cycle wic involve only two temperatures. But tere are many irreversible cycles operating between two temperatures. If one canges D parameters of te system fast, say by a sudden move of te piston, evolution of te gas becomes essentially irreversible involving many complicated irreversible processes suc as sound propagation. Consider te Carnot cycle and some oter irreversible cycle, bot operating between te same two temperatures. Carnot c X e Second Carnot teorem states tat η c η. et us assume for a moment tat te teorem wrong and tat η c < η. One can apply te same trick as before, reversing te Carnot cycle. Considering te combination of two macines 4

5 c Carnot c X c Carnot X and using te same argument as before, one can see tat our assumption η c < η is inconsistent wit te Second aw, and terefore is wrong. Note: One can ask wat prevent us from applying te argument in opposite direction, tat is inverting not te Carnot engine but te oter macine? But te X macine can not be reversed by definition because its operation involves fast irreversible processes leading to dissipation of energy. e first and second Carnot s teorems can be formulated togeter as follows If tere are several eat engines operating between te same two temperatures, ten all reversible engines ave te same efficiency, wile irreversible ones ave efficiencies wic can never eceed te efficiency of te reversible engines is teorem is of paramount importance and as implications wic go far beyond te area of eat engines. 5

6 4 ermodynamic temperature Recall tat we accepted operational or better to say empirical definition of temperature. e empirical temperature we used so far is based on properties of rarefied gases. e used gases since teir D properties beave identically in te limit of low density. et us call suc temperature ideal-gas temperature. Now we ave found a quantity wic does not depend on properties of any substance at all. Namely, te Carnot teorem suggests tat te efficiency of te Carnot cycle depends only on empirical temperatures of two bats, or η = = φ(, ), / = φ(, ) f(, ). Here te eplicit form of te function f depends on definition of temperature we use. For te ideal gas temperature we used so far f(, ) = /, but for oter empirical temperature scales it may be different. Actually, one can prove (see below) tat te function f can not be arbitrary and always can be factorized as f(, ) = g( ) g( ). en one can define temperature as θ = g( ), and terefore / = θ /θ. e temperature θ defined in tis way is called termodynamic temperature and as an advantage tat it does not depend of properties of any particular system. Clearly, we still ave a freedom to coose te D temperature wit an arbitrary factor. But tis ambiguity can be eliminated wen one assign a certain value to a temperature at a certain reference point. For eample one can take θ as a temperature of melting ice and assign to it te value en all oter temperatures will be unambiguously defined θ = θ. Suppose we need to measure a D temperature of a substance X. o do it, we must use a eat engine and operate te Carnot cycle taking X as one bat and melting ice as te oter. en we must measure amounts of eat obtained from X, and released to te Ice Bat i. en θ = 73.5 i 6

7 emma: f(, ) = = g( ) g( ) Consider a combined macine in te figure below. One can always arrange tat te eat released by te second engine eactly te same as eat received by te second one. C C 3 3 / = f(, ), 3 / = f(, 3 ) But tis equivalent to a single macine operating between and 3, 3 / = f(, 3 ). Since 3 / = 3 / / we ave f(, 3 ) = f(, )f(, 3 ). e left side does not depend on wic terefore must cancel out in te rigt side. is is possible only if f(, y) factorize as g(y)/g(): g( ) g( 3 ) g( ) g( ) = g( 3) g( ) One can prove tat termodynamic and empirical ideal-gas temperatures equal to eac oter. e ave defined te D using te ratio = θ θ 7

8 were and are eat received from (or injected to) te bat one and two, respectively, and θ i are teir D. On te oter and, using te for ideal gas temperature we obtained or η = = =, =. Comparing tis wit our definition of D temperature = θ θ one observes tat te empirical ideal-gas may differ from te D temperature only by a constant factor = aθ. Since D temperature itself is defined up to te arbitrary factor, one can cose a =, so tat bot temperature scales coincide. 8