Thermodynamics filled in class notes_Part_54

Thermodynamics filled in class notes_Part_54 - 115 4.3....

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Unformatted text preview: 115 4.3. LEGENDRE TRANSFORMATIONS Now since ∂h ∂s ∂h ∂P = T, (4.121) = v, (4.122) P s one has P Po T = To v = RTo Po R/cP s cP , exp exp s cP R/cP −1 P Po (4.123) . (4.124) Dividing Eq. (4.123) by Eq. (4.124) gives T v Pv = = P , R RT , (4.125) (4.126) which is the thermal equation of state. Substituting from Eq. (4.123) into the canonical equation for h, Eq. (4.118), one also finds for the caloric equation of state h h cP T + ( h o − cP T o ) , = = cP (T − To ) + ho , (4.127) (4.128) which is useful in itself. Substituting in for T and To , h = cP P v Po vo − R R + ho . (4.129) Using, Eq. (4.78), h ≡ u + P v , we get u + P v = cP Pv Po vo − R R + uo + Po vo . (4.130) So u= u= u= u= u= u= cP cP − 1 Pv − − 1 Po vo + uo , R R cP − 1 (P v − Po vo ) + uo , R cP − 1 (RT − RTo ) + uo , R (cP − R) (T − To ) + uo , (cP − (cP − cv )) (T − To ) + uo , cv ( T − T o ) + u o . (4.131) (4.132) (4.133) (4.134) (4.135) (4.136) So one canonical equation gives us all the information one needs. Often, it is difficult to do a single experiment to get the canonical form. CC BY-NC-ND. 18 November 2011, J. M. Powers. 116 4.4 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Heat capacity Recall that cv = cP = ∂u ∂T ∂h ∂T , (4.137) . (4.138) v P Then perform operations on the Gibbs equation du = T ds − P dv, ∂u ∂s =T , ∂T v ∂T v ∂s cv = T . ∂T v (4.139) (4.140) (4.141) Likewise, dh = T ds + vdP, ∂s ∂h =T , ∂T P ∂T P ∂s cP = T . ∂T P (4.142) (4.143) (4.144) One finds further useful relations by operating on the Gibbs equation: du = T ds − P dv, ∂u ∂s =T − P, ∂v T ∂v T ∂P − P. =T ∂T v (4.145) (4.146) (4.147) So one can then say u = u(T, v ), ∂u ∂u dT + du = ∂T v ∂v ∂P = cv dT + T ∂T (4.148) dv, (4.149) T v −P dv. (4.150) For an ideal gas, one has ∂u ∂v =T T ∂P ∂T v = 0. CC BY-NC-ND. 18 November 2011, J. M. Powers. −P =T R v − RT , v (4.151) (4.152) ...
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