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Chapter 3 Second Law of Thermodynamics

# Chapter 3 Second Law of Thermodynamics - Chapter 3 The...

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Chapter 3 - The Second Law of Thermodynamics (Feb. 2008) We’ll use a different sequence than the book. First, we will get familiar with a new state function, the entropy S. After a brief interlude about the efficiency of heat engines, we will discuss how S > 0 is a signature of spontaneous processes in isolated systems. 3.1) Entropy S, a State Function Related to q rev (2 nd half of 20-2) dU = q rev + w rev with w rev = - P dV, so for an ideal gas q rev = f(T)/2 nR dT + P dV = f(T)/2 nR dT + n RT/V dV As shown in Homework 4.2), for an ideal gas, dS = q rev /T = 1/2 nR f(T)/T dT + n R 1/V dV is an exact differential. {Generalize this result: Consider an isolated system (dU = 0, w = 0, q = 0) consisting of two rigid subsystems w A = w B = 0 A is an ideal gas, B is any system in thermal contact with A. Heat produced in A is absorbed in B and vice versa: q rev,A + q rev,B = q = 0 Since T A = T B , q rev,B /T = - q rev,A /T , the same exact differential for both systems.} The Second Law generalizes this result, postulating that S is a state function for any system. dS = q rev /T S is a state function like U, so we can calculate it for any state of the system. 3.2) S(T, V) for an Ideal Gas dS is an exact differential S is a state function S(T, V) can be found Let’s start with some simple integrals: For isothermal volume change: U = 0, so q rev = - w rev S = 1 2 dS = 1 2 q rev /T = - 1 2 w rev /T = = nR V1 V2 1/V dV = nR (lnV 2 – lnV 1 ) = nR ln(V 2 /V 1 )

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For constant-volume temperature change: w = 0, so q rev = dU = C V (T) dT S = 1 2 dS = 1 2 q rev /T = 1 2 dU/T = = T1 T2 f/2 n R/T dT = f/2 nR (lnT 2 – lnT 1 ) = f/2 nR ln(T 2 /T 1 ) (for T-independent f; otherwise, this is more complicated). Logarithmic dependence is typical for S = k B ln = R/N A ln Combine these results to S(T,V) = nR ln(V/V 0 ) + f/2 nR ln(T/T 0 ) + S(T 0 ,V 0 ) for an ideal gas with constant f (this works so simply only because the T and V dependencies are unrelated) Calculate dS = ( S/ T) V dT + ( S/ V) T dV Use the chain rule of derivatives: ( S/ T) V = f/2 nR 1/(T/T 0 ) 1/T 0 = f/2 nR 1/T ( S/ V) T = nR 1/(V/V 0 ) 1/V 0 = nR 1/V Put these together: dS = f/2 nR 1/T dT + nR 1/V dV Just as derived from the First Law. Problem for T = 0 (dS becomes infinite). That this may not happen will be discussed in conjunction with the Third Law of thermodynamics. 3.3) Entropy and Disorder In statistical mechanics, entropy is recognized as a measure of disorder/randomness/number of configurations of the system with the same energy. Formally,
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Chapter 3 Second Law of Thermodynamics - Chapter 3 The...

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