Lecture 45 Entropy·Clausius theorem· Entropy as a state function· Second law statement using entropy·Calculating entropy increase
Lecture 45 Entropy • Clausius theorem • Entropy as a state function • Second law statement using entropy • Calculating entropy increase
Clausius theorem.SupposeasystemabsorbsheatSQattemperatureT. Since the value of does not depend on the details of how the heat is transferred,wecanassumeitisfromaCarnotengine,whichinturnabsorbsheatO.fromaheat reservoir with constanttemperature To.Arbitrary.ForCarnotcycleWSystem0= 800→ 8Q0= ToSQSQCarnotTTo-WoEngine·Thereforeinonecycle,thetotalheatabsorbedfromthereservoirisoQ0HeatC SQReservolrQo = ToToT·Since aftera cycle,the systemand the Carnotengine as a whole return to itsinitialstatus,thedifferenceoftheinternalenergyiszero.Qo = AEint + W + Wo= W + Wo = Wtotal2
Clausiustheorem • Suppose a system absorbs heat 𝛿𝑄 at temperature 𝑇. • Since the value of 𝛿𝑄 𝑇 does not depend on the details of how the heat is transferred, we can assume it is from a Carnot engine, which in turn absorbs heat 𝛿𝑄0 from a heat reservoir with constant temperature 𝑇0. • For Carnot cycle 𝛿𝑄 𝑇 = 𝛿𝑄0 𝑇0 → 𝛿𝑄0 = 𝑇0 𝛿𝑄 𝑇 • Therefore in one cycle, the total heat absorbed from the reservoir is 𝑄0 = 𝑇0 ර 𝛿𝑄 𝑇 • Since after a cycle, the system and the Carnot engine as a whole return to its initial status, the difference of the internal energy is zero. 𝑄0 = Δ𝐸𝑖𝑛𝑡 + 𝑊 + 𝑊0 = 𝑊 + 𝑊0 = 𝑊𝑡𝑜𝑡𝑎𝑙 2
AccordingtotheKelvin-Planckstatementof SecondLawofthermodynamics, we cannot drain heat from one reservoirand convert them entirely into work without making anyother changes, soWtotal ≤ 0Therefore,T SQWhichis called ClausiusInequalityIf the systemis reversible, then reverse its path and do theexperimentagain wecangetCSQ≤0Thusforreversiblecaser Q0T3
According to the Kelvin-Planck statement of Second Law of thermodynamics, we cannot drain heat from one reservoir and convert them entirely into work without making any other changes, so 𝑊𝑡𝑜𝑡𝑎𝑙 ≤ ０ Therefore， ර 𝛿𝑄 𝑇 ≤ 0 Which is called Clausius Inequality. If the system is reversible, then reverse its path and do the experiment again we can get − ර 𝛿𝑄 𝑇 ≤ 0 Thus for reversible case ර 𝛿𝑄 𝑇 = 0 3
The Carnot Cyclelisothermal2Ti=T2-ThIVadiabaticP-WoutV4V2ladiabaticQn = nRThln%,Qc = nRT:ln13IIlisothermailVT-T,-TdouPiVY= p4V→ piVi = nRT, P4V4 = nRT4VV-1Ti-piViT2-VY-1,T3 T4VYP4V4V4/V1 = V3 /V2 → V2 /V1 = V3/V4QnQcconstantThTcThis implies that there is a function of statewhich changes over the Carnotcycle when heat is added and subtracted. Clausius called this state functionentropy.4
𝑄ℎ = 𝑛𝑅𝑇ℎln 𝑉2 𝑉1 , 𝑄𝑐 = 𝑛𝑅𝑇𝑐 ln 𝑉4 𝑉3 𝑝1𝑉1 𝛾 = 𝑝4𝑉4 𝛾 → 𝑝1𝑉1 = 𝑛𝑅𝑇1 , 𝑝4𝑉4 = 𝑛𝑅𝑇4 𝑇1 𝑇4 = 𝑝1𝑉1 𝑝4𝑉4 = 𝑉4 𝛾−1 𝑉1 𝛾−1 , 𝑇2 𝑇3 = 𝑉3 𝛾−1 𝑉4 𝛾−1 , 𝑉 Τ4 𝑉1 = 𝑉 Τ3 𝑉2 → 𝑉 Τ2 𝑉1 = 𝑉 Τ3 𝑉4 𝑄ℎ 𝑇ℎ = 𝑄𝑐 𝑇𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 This implies that there is a function of state which changes over the Carnot cycle when heat is added and subtracted. Clausius called this state function entropy. 4 The Carnot Cycle
EntropyForreversibleprocesses8Q = dEint +SW= CydT + pdV = nCydT + nRT VIn order to calculus the integral, we should know how T and V changeduring the processRewritetheequationasdTdv8Q+nRd(ncyInT + nRlnV)ncyTTVDefine8QdsTWhich is a total differentiation. We call the function S as entropy.5
Entropy For reversible processes 𝛿𝑄 = 𝑑𝐸𝑖𝑛𝑡 + 𝛿𝑊 = 𝐶𝑉𝑑𝑇 + 𝑝𝑑𝑉 = 𝑛𝑐𝑉𝑑𝑇 + 𝑛𝑅𝑇 𝑑𝑉 𝑉 In order to calculus the integral, we should know how 𝑇 and 𝑉 change during the process Rewrite the equation as 𝛿𝑄 𝑇 = 𝑛𝑐𝑉 𝑑𝑇 𝑇 + 𝑛𝑅 𝑑𝑉 𝑉 = 𝑑 𝑛𝑐𝑉ln𝑇 + 𝑛𝑅ln𝑉 Define 𝑑𝑆 = 𝛿𝑄 𝑇 Which is a total differentiation. We call the function 𝑆 as entropy. 5