Thermodynamics filled in class notes_Part_51

Thermodynamics filled in class notes_Part_51 - 109 4.2. TWO...

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Unformatted text preview: 109 4.2. TWO INDEPENDENT VARIABLES Since x and z are independent, so are dx and dz , and the coefficients on each in Eq. (4.53) must be zero. Therefore from the coefficient on dx in Eq. (4.53) ∂x ∂y ∂y − 1 = 0, z ∂x z ∂x ∂y = 1. ∂y z ∂x z (4.54) (4.55) So ∂x ∂y 1 z , (4.56) = 0, = (4.57) ∂y ∂x z and also from the coefficient on dz in Eq. (4.53) ∂x ∂y z ∂y ∂z + x ∂x ∂z ∂x ∂z y =− y ∂x ∂y z ∂y ∂z ,. (4.58) x So ∂x ∂z ∂y ∂x y z ∂z ∂y = −1. x (4.59) If one now divides Eq. (4.49) by a fourth differential, dw , one gets dx = dw ∂x ∂y z ∂x dy + dw ∂z y dz . dw (4.60) Demanding that z be held constant in Eq. (4.60) gives ∂x ∂w = z ∂x ∂w z ∂y ∂w z ∂x ∂w z ∂w ∂y = = z ∂x ∂y z ∂x ∂y z ∂x ∂y z ∂y ∂w , (4.61) z , (4.62) . (4.63) If x = x(y, w ), one then gets dx = ∂x ∂y dy + w ∂x ∂w dw. (4.64) y CC BY-NC-ND. 18 November 2011, J. M. Powers. 110 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Divide now by dy while holding z constant so ∂x ∂y = z ∂x ∂y ∂x ∂w + w ∂w ∂y y . (4.65) z These general operations can be applied to a wide variety of thermodynamic operations. Example 4.4 Apply Eq. (4.65) to a standard P − v − T system and let ∂x ∂y ∂T ∂v = z . (4.66) s So T = x, v = y , and s = z . Let now u = w. So Eq. (4.65) becomes ∂T ∂v ∂T ∂v = s ∂T ∂u + u v ∂u ∂v . (4.67) s Now by definition ∂u ∂T cv = so Now by Eq. (4.26), one has ∂T ∂u ∂u ∂v s = v , (4.68) 1 . cv (4.69) v = −P , so one gets ∂T ∂v ∂T ∂v = s u − P . cv For an ideal gas, u = u(T ). Inverting, one gets T = T (u), and so ∂T ∂v s =− P . cv (4.70) ∂T ∂v u = 0, thus (4.71) For an isentropic process in an ideal gas, one gets dT dv dT T = = = ln T To T To = = CC BY-NC-ND. 18 November 2011, J. M. Powers. RT P =− , cv cv v R dv − , cv v dv −(k − 1) , v vo (k − 1) ln , v vo k−1 . v − (4.72) (4.73) (4.74) (4.75) (4.76) ...
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