l16 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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Lecture 16 5.60/ 20 .110/2.772 1 Absolute Entropy Third Law of thermodynamics Absolute Entropies Absolute entropy of an ideal gas Start with fundamental equation pdV TdS dU = dS = dU + pdV T for ideal gas: dT C dU V = and V nRT p = dV V nR T dT C dS V + = At constant T, dT=0 dS T = pdV T For an ideal gas, pV = nRT dS T = nRdV V At constant T () ( ) Vdp pdV nRT d pV d = = = 0 plugging into dS T : dS T =− nRdp p This allows us to know how S(p) if T held constant. Integrate! 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
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Lecture 16 5.60/ 20 .110/2.772 2 For an arbitrary pressure p, S ( p , T ) = S ( p o , T ) nRdp p p o p ³ = S ( p o , T ) nR ln p p o § © ¨ · ¹ ¸ where p o is some reference pressure which we set at 1 bar. ± S(p,T) = S o (T) – nR lnp (p in bar) S as P But to finish, we still need S o ( T ) ! Suppose we had ) K 0 ( S o (standard molar entropy at 0 Kelvin) dT C dH Vdp TdS dH p = + = for ideal gas dp T V dT T C dS Vdp TdS dT C p p = + = p R T S T p S ln ) ( ) , ( 0 = p S ° 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
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Lecture 16 5.60/ 20 .110/2.772 3 Then using S T § © · ¹ p = C p T we should be able to get S o ( T ) . Integrating over dS eqn, assuming C
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l16 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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