Introduction to Thermodynamics and Statistical Physics phần 5 pdf

2.5. Processes in Ideal Gas V p isobaric i s o t h e r m a l i s e n t r o p i c i s oc hor i c V p isobaric i s o t h e r m a l i s e n t r o p i c i s oc hor i c Fig. 2.2. Four processes for which dN =0. thermal equilibrium throughout the entire process. This can be achieved b y varying the external parameters at a rate that is sufficiently slow to allow the system to remain very close to thermal equilibrium at any moment during the process. The four exam ple to be analyzed below are (see fig. 2.2): • Isothermal process - temperature is constan t • Isobaric pr ocess - pressure is constant • Isochoric process - volume is constant • Isentropic process - entropy is con stant Note that in general, using the definition of the heat capacity at constant volume given by Eq. (2.67) together with Eq. (1.87) one finds that c V = µ ∂U ∂τ ¶ N,V . (2.95) Furthermore, as can be seen from Eq. (2.85), the energy U of an ideal gas in the classical limit is independent on the volume V (it can be expressed as afunctionofτ and N only). Th us w e conclude that for processes for which dN = 0 the change in energy dU can be expressed as dU = c V dτ. (2.96) Eyal Buks Thermodynamics and Statistical Physics 61 Chapter 2. Ideal Gas 2.5.1 Isothermal Process Since τ is constant one finds using Eq. (2.96) that ∆U = 0. Integrating the relation dW = pdV and using Eq. (2.55) one finds Q = W = V 2 Z V 1 pdV = Nτ V 2 Z V 1 dV V = Nτ log V 2 V 1 . (2.97) 2.5.2 Isobaric Process Integrating the relation dW = pdV for this case where the pressure is con- stant yields W = V 2 Z V 1 pdV = p (V 2 − V 1 ) . (2.98) The change in energy ∆U can be found by in tegrating Eq. (2.96) ∆U = τ 2 Z τ 1 c V dτ. (2.99) The heat added to the system Q can be found using Eq. (2.94) Q = W + ∆U = p (V 2 − V 1 )+ τ 2 Z τ 1 c V dτ. (2.100) Note that if the temperature de pendence of c V can be ignored to a good approximation one has ∆U = c V (τ 2 − τ 1 ) . (2.101) Eyal Buks Thermodynamics and Statistical Physics 62 2.5. Processes in Ideal Gas 2.5.3 Isochoric Process In this case the v olume is constant, thus W =0.ByintegratingEq.(2.96) one finds that Q = ∆U = τ 2 Z τ 1 c V dτ. (2.102) Also in this case, if the temperature dependence of c V can be ignored to a good approximation one has Q = ∆U = c V (τ 2 − τ 1 ) . 2.5.4 Isentropic Process In this case the entropy is constant, thus dQ = τdσ =0,andtherefore dU = −dW ,thususingtherelationdW = pdV and Eq. (2.96) one has c V dτ = −pdV, (2.103) or using Eq. (2 .55) c V dτ τ = −N dV V . (2.104) This relation can be rewritten using Eq. (2.89) as dτ τ =(1−γ) dV V . (2.105) where γ = c p c V . (2.106) The l ast result can be e asily inte grated if the t emperature dependence of the factor γ can be ignored to a good approximation. For that case one has log τ 2 τ 1 =log µ V 2 V 1 ¶ 1−γ . (2.107) Thus τ 1 V γ−1 1 = τ 2 V γ−1 2 , (2.108) or using Eq. (2 .55) p 1 V γ 1 = p 2 V γ 2 . (2.109) Eyal Buks Thermodynamics and Statistical Physics 63 Chapter 2. Ideal Gas τ h τ l Q h Q l W τ h τ l Q h Q l W Fig. 2.3. Carnot heat engine. In other words both quantities τV γ−1 and pV γ remain unchanged during this process. Using the last result allow s integrating the relation dW = pdV −∆U = W = V 2 Z V 1 pdV = p 1 V γ 1 V 2 Z V 1 V −γ dV = p 1 V γ 1 ³ V −γ+1 1 − V −γ+1 2 ´ γ − 1 = p 2 V 2 − p 1 V 1 1 − γ = N (τ 2 − τ 1 ) 1 − γ = −c V (τ 2 − τ 1 ) . (2.110) 2.6CarnotHeatEngine In this section we discuss an example of a heat engine proposed by Carnot that is based on an ideal classical gas. Each cycle is made of four steps (see Figs. 2.3 and 2.4) 1. Isothermal expansion at temperature τ h (a → b) 2. Isentropic expansion from temperature τ h to τ l (b → c) 3. Isothermal compression at temperature τ l (c → d) 4. Isentropic compression from temperature τ l to τ h (d → a) All four steps are assumed to be sufficiently slow to maintain the gas in thermal equilibrium throughout the en t ire cycle. The engine exchan ges heat Eyal Buks Thermodynamics and Statistical Physics 64 2.6. Carnot Heat Engine V p a b c d V a V b V d V c p a p b p d p c τ h τ l V p a b c d V a V b V d V c p a p b p d p c τ h τ l Fig. 2.4. A cycle of Carnot heat engine. with the en v ironment during both isothermal pr ocesses. Using Eq. (2.97) one finds that the heat extracted from the hot reservoir Q h at temperature τ h during step 1 (a → b)isgivenby Q h = Nτ h log V b V a , (2.111) and t he heat extracted fro m the cold thermal reservoir Q l at temperature τ l during step 3 (c → d)isgivenby Q l = Nτ l log V d V c , (2.112) where V n isthevolumeatpointn ∈ {a, b, c, d}.NotethatQ h > 0since the system undergoes expansion in step 1 whereas Q l < 0 since the system undergoes compression during step 3. Both thermal reservoirs are assumed to be very large systems that c an exc hange heat with the e ngine without changing their temperature. No heat is exchanged during the isentropic steps 2 and 4 (since dQ = τdσ). Thetotalworkdonebythesystempercycleisgivenby W = W ab + W cd + W bc + W da = Nτ h log V b V a + Nτ l log V d V c + N (τ l − τ h ) 1 − γ + N (τ h − τ l ) 1 − γ = N µ τ h log V b V a + τ l log V d V c ¶ , (2.113) Eyal Buks Thermodynamics and Statistical Physics 65 Chapter 2. Ideal Gas where the work in both isothermal processes W ab and W cd is calculated using Eq. (2.97), whereas the work in both isentropic processes W bc and W da is calculate using Eq. (2.110). Note that the following holds W = Q h + Q l . (2.114) This is expected in view of Eq. (2.94) since the gas returns after a full cycle to its initial state and therefore ∆U =0. The efficiency of the h eat engine is d efined as the ratio between the work done by the system and the heat extracted from the hot reservoir per cycle η = W Q h =1+ Q l Q h . (2.115) Using Eqs. (2.111) and (2.113) one finds η =1+ τ l log V d V c τ h log V b V a . (2.116) Employing Eq. (2.108) for both isen tropic processes yields τ h V γ−1 b = τ l V γ−1 c , (2.117) τ h V γ−1 a = τ l V γ−1 d , (2.118) th us by dividing these equations one finds V γ−1 b V γ−1 a = V γ−1 c V γ−1 d , (2.119) or V b V a = V c V d . (2.120) Using this result one finds that the efficiency of Carnot heat engine η C is given by η C =1− τ l τ h . (2.121) 2.7 Limits Imposed Upon the Efficiency Is it possible to construct a heat engine that operates between the same heat reservoirs at temperatures τ h and τ l that will hav e efficiency larger than the value given by Eq. (2.121)? As we will see below the answ er is no. This conclusion is obtained by no ticing that the total entropy remains unchanged in each of the four steps that constructs the Carnot’s cycle. Consequently, Eyal Buks Thermodynamics and Statistical Physics 66 2.7. LimitsImposedUpontheEfficiency τ Q W M τ Q W M Fig. 2.5. An idle heat engine (Perpetuum Mobile of the second kind). the entire process is reversible, namely, by varyin g the external param eters in the opposite direction, the process can be reversed. We consider below a general model of a heat engine. In a continuos oper- ation the heat engine repeats a basic cycle one after another. We make the following assumptions: • At the end of each cycle the heat engine returns to the same macroscopic state that it was in initially (otherwise, continuous operation is impossible). • The w ork W done per cycle by the heat engine does not change the entropy of the environment (this is the case when, for example, the work is used to lift a weight - a process that only changes the center of mass of the weight, and therefore causes no entropy change). Figure (2.5) shows an ideal heat engine that fully transforms the heat Q extracted from a thermal reservoir into work W ,namelyQ = W .Suchan idle engine has a unity efficiency η = 1. Is it possible to realized such an idle engine? Such a process does not violate the la w of energy conservation (first law of thermodynamics). However, as we will see below it violates the second law of thermodynamics. Note also that the opposite process, namely a process that transforms work into heat w ithout losses is possible, as c an be seen from the example seen in Fig. (2.6). In this system the weigh normally goes dow n and consequen tly the blender rotates and heats the liquid in the container. In principle, the opposite process at which the wei gh goes up and the liquid cools down doesn’t violate the law of energy conservatio n , h owever, it violates the second law (Perpetuum Mobile ofthesecondkind),aswewill see below. To show that the idle heat engine shown in Fig. (2.5) can not be realized we employ the second law and require that the total change in entrop y ∆σ per cycle is non-negative ∆σ ≥ 0 . (2.122) The only c hange in entropy per cycle is due to the heat that is subtracted from the heat bath Eyal Buks Thermodynamics and Statistical Physics 67 Chapter 2. Ideal Gas Fig. 2.6. Transforming work into heat. ∆σ = − Q τ , (2.123) thus since Q = W (energy conservation) we find that W τ ≤ 0 . (2.124) Namely, the work done by the heat engine is non-positive W ≤ 0. This result is kno w n as Kelvin’s principle. Kelvin’s principle: In a cycle process, it is impossible to extract heat from a heat reservoir and convert it all into work. As we will see below, Kelvin’s pr inciple is equivalent to Clausius’s principle that states: Clausius’s principle: It is impossible that at the end of a cycle process, heat has been transferred from a colder to a hotter thermal reservoirs without applying any work in the process. A refrigerator and an air conditioner (in cooling mode) are examples of systems that transfer heat from a colder to a hotter thermal reservoirs. Ac- cording to Clausius’s principle such systems require that work is consumed for their operation. Theorem 2.7.1. Kelvin’s principle is equivalent to Clausius’s principle. Proof. Assume that Clausius’s principle does not hold. Th us the system shown in Fig. 2.7(a) that transfers heat Q 0 > 0 from a cold thermal reser- voir at temperature τ l to a hotter one at temperature τ h >τ l is possible. In Fig. 2.7(b) a heat engine is added that extracts heat Q>Q 0 from the hot thermal reservoir, delivers heat Q 0 to the cold one, and performs work W = Q − Q 0 . The comb ina t ion of both systems extracts heat Q − Q 0 from the hot thermal r eservoir and converts it all into work, in contradiction with Kelvin’s principle. Assume tha t Kelvin’s principle does not hold. Th us the system shown in Fig. 2.8(a) that extracts heat Q 0 from a thermal reservoir at t emperature τ h and conv erts it all into work is possible. In Fig. 2.8(b) a refrigerat or is added Eyal Buks Thermodynamics and Statistical Physics 68 2.7. LimitsImposedUpontheEfficiency τ h τ l -Q 0 Q 0 M (a) τ h τ l Q-Q 0 W=Q-Q 0 M -Q 0 Q 0 M (b) τ h τ l -Q 0 Q 0 M (a) τ h τ l Q-Q 0 W=Q-Q 0 M -Q 0 Q 0 M (b) Fig. 2.7. The assumption that Clausius’s principle does not hold. that emplo ys the work W = Q 0 to remove heat Q from a colder thermal reservoir at temperature τ l <τ h and to deliver heat Q 0 + Q to the hot thermal reservoir. The combination of both systems transfers heat Q from a colder to a hotter thermal reservoirs without consuming any work in the process, in contradiction with Clausius’s principle. As we have seen unity efficiency is impossible. What is the largest possible efficiency of an heat en gine? Theorem 2.7.2. The efficiency η of a heat engine operating between a hotter and colder heat r eservoirs at temperatur e τ h and τ l respectively can not exceed the value η C =1− τ l τ h . (2.125) Proof. A heat engine (labeled as ’I’) is seen in Fig. 2.9. A Carnot heat engine operated in the reverse direction (labeled as ’C’) is added. Here we exploit the fact the Carnot’s cycle is rev ersible. The efficiency η of the heat engine ’I’ is giv en by η I = W Q 0 h , (2.126) whereas the efficiency of the reversed Carnot heat engine ’C’ is given by E q. (2.121) Eyal Buks Thermodynamics and Statistical Physics 69 Chapter 2. Ideal Gas τ h τ l Q 0 M (a) W= Q 0 τ h τ l Q 0 M (b) W= Q 0 Q M -Q 0 - Q τ h τ l Q 0 M (a) W= Q 0 τ h τ l Q 0 M (b) W= Q 0 Q M -Q 0 - Q Fig. 2.8. The assumption that Kelvin’s principle does not hold. τ h τ l Q h C W Q’ l I Q’ h Q l τ h τ l Q h C W Q’ l I Q’ h Q l Fig. 2.9. Limit imposed upon engine efficiency. η C = W Q h =1− τ l τ h . (2.127) For the combined system, the Clausius’s principle requires that Q 0 h − Q h > 0 . (2.128) thus η I ≤ η C =1− τ l τ h . The same argument that was employed in the proof above can be used to deduce the following corollary: Eyal Buks Thermodynamics and Statistical Physics 70 [...]... , (2. 151 ) where p is the pressure, V is the volume, is the temperature, N is the number of particles and b is a constant Eyal Buks Thermodynamics and Statistical Physics 74 2.8 Problems Set 2 a) Calculate the dierence cp cV between the heat capacities at constant pressure and at constant volume b) Consider an isentropic expansion of the gas from volume V1 and temperature 1 to volume V2 and temperature... Buks Thermodynamics and Statistical Physics 75 Chapter 2 Ideal Gas p p1 b c p2 a V2 V1 V Fig 2.10 an isentropic process, and in the third one the pressure is constant p2 Assume that the heat capacities CV and Cp are temperature independent Show that the eciency of this engine is given by =1 p2 (V1 V2 ) , V2 (p1 p2 ) (2. 157 ) where = Cp /Cv 20 Consider a refrigerator consuming work W per cycle to. .. at pressure p and temperature The walls of the box have N0 absorbing sites, each of which can absorb 0, 1, or 2 atoms of the gas The energy of an unoccupied site and the energy of a site occupying one atom is zero The energy of a site occupying two atoms is Show that the mean number of absorbed atoms is given by hNa i = N0 + 22 e , 1 + + 2 e (2.147) where = 1/ and = à M 2}2 ả3/2 5/ 2 p (2.148)... probability pn to nd n molecules in a small volume v (namely, v . good approximation one has ∆U = c V (τ 2 − τ 1 ) . (2.101) Eyal Buks Thermodynamics and Statistical Physics 62 2 .5. Processes in Ideal Gas 2 .5. 3 Isochoric Process In this case the v olume is constant,. N µ τ h log V b V a + τ l log V d V c ¶ , (2.113) Eyal Buks Thermodynamics and Statistical Physics 65 Chapter 2. Ideal Gas where the work in both isothermal processes W ab and W cd is calculated using Eq. (2.97),. entropy per cycle is due to the heat that is subtracted from the heat bath Eyal Buks Thermodynamics and Statistical Physics 67 Chapter 2. Ideal Gas Fig. 2.6. Transforming work into heat. ∆σ = − Q τ ,