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ME 209 2/15 Concept of Exact Differential-I Basic Thermodynamics Property Relations • Consider a property surface. Let it be a p,v,T surface

A • Let point 1 and 2 be close to each 1 Kannan Iyer other and the point 1-2 can be 2 [email protected] reached by infinite number of B paths. If we look at dv = V 2-V1, it should be independent of the path • Let 1-A and B-2 be isotherms and 1-B and A-2 be isobars Department of Mechanical Engineering ∂v ∂v ∂v ∂v dv = dp + dT = dT + dp Indian Institute of Technology, Bombay 1−2 ∂ ∂ ∂ ∂ p T1 T p2 T 1p p T2 Path 1-A-2 Path 1-B-2

3/15 4/15 Concept of Exact Differential-II Concept of Exact Differential-III

∂v ∂v ∂v ∂v dv = dp + dT = dT + dp • As the points are close by, we can write the previous 1−2 ∂ ∂ ∂ ∂ p T1 T p2 T p1 p T2 equation as ∂  ∂  ∂  ∂   v   v   ∂ ∂   ∂ ∂  ⇒ = v v v v ∂T  ∂p  ∂p  ∂T   dp + dT  −  dT + dp  = 0  T   p   ∂ ∂   ∂ ∂  p T  p T1 T p2   T 1p p T2  ∂2 ∂ 2 ⇒ v = v  ∂v ∂v   ∂v ∂v  ∂T∂p ∂p∂T ⇒  − dp =  − dT  ∂p ∂p   ∂T ∂T   T2 T1   p2 1p  • The value of a mixed differential is independent of the order of differentiation  ∂ ∂   ∂ ∂   v − v   v − v   ∂p ∂p   ∂T ∂T  • The result is a consequence of assuming dv is ⇒  T2 T1  =  p2 p1  independent of direction. Those differentials that dT dp satisfy this property are called exact differentials

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5/15 6/15 Concept of Exact Differential-IV Rules of Partial Derivatives-I • Every property change is an exact differential and Every exact differential represents change of a • Now let us look at the relation between partial property derivatives ∂x ∂x dx = dy + dz ∂ ∂ • The whole thing can be generalised as, given three y z z y variables x, y and z and they have a relation of the • Similarly, we can write form = + ∂y ∂y dz M dx)y,x( N dy)y,x( dy = dx + dz ∂x ∂z then the differential dz is exact, if z x • Substituting the expression for dy in second equation ∂M ∂N ⇒ = into the first equation, we get ∂y ∂x ∂x  ∂y ∂y  ∂x dx =  dx + dz  + dz • In thermodynamics, knowing that properties are exact, ∂y  ∂x ∂z  ∂z we shall equate the cross derivatives z z x y

7/15 8/15 Rules of Partial Derivatives-II Rules of Partial Derivatives-III

• Collecting the coefficients of dx and dz, we can write • Now if we go to two neighboring states such that dx = 0  ∂ ∂   ∂ ∂ ∂  and dz ≠ 0, then, we can write,  − x y  =  x y + x  dx 1  dz ∂y ∂x  ∂y ∂z ∂z  ∂x ∂y ∂x  z z   z x y  = − ∂y ∂z ∂z • Now if we go to two neighboring states such that dz = 0 z x y and dx ≠ 0, then it is necessary to have • Now applying the reciprocal rule, we can write

∂x ∂y ∂x ∂y ∂x 1 ∂x ∂y −1 1− = 0 Or =1 Or = = ∂x ∂y ∂z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ Or = −1 y z x z y x z y y y z z z z z x ∂y ∂z ∂x ∂ ∂ z x y x z x y

• We can call the above as Reciprocal Rule • The above can be called as Cyclic Rule

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9/15 10/15 Thermodynamic Functions-I Thermodynamic Functions-II • Having laid the foundation for deriving property • Let us begin with the two Tds relations that we have relations, we shall now go on to define two new derived earlier thermodynamic functions called Helmholtz and Gibbs du = Tds − pdv 1 dh = Tds + vdp 2 functions • Differentiating the two functions defined in previous • The Helmholtz function A and Gibbs function G are slide, we can write defined as da = du − Tds − sdT 3 dg = dh − Tds −sdT 4 A = U − TS G = H − TS • Their intensive counterparts are a and g • Substituting for du-Tds from Eq. (1) in Eq. (3) and a = u − Ts g = h − Ts similarly substituting for dh-Tds from Eq. (2) in Eq. (4), we get • Now we shall begin manipulating these functions da = −pdv − sdT 5 dg = vdp −sdT 6

11/15 12/15 Thermodynamic Functions-III Thermodynamic Functions-IV • Now let us manipulate Eq. (1), where u = u(s,v) • From Eq. (5) da = −pdv − sdT du = Tds − pdv ∂a ∂a • Chain rule implies • We can write, = − ,p = −s 9 ∂v ∂T ∂u ∂u T v du = ds + dv ∂ ∂ = − s v v s • From Eq. (6) dg vdp sdT • Comparing the above two equations, we can write, ∂g ∂g ∂ ∂ • We can write, = ,v = −s 10 u u 7 ∂ ∂ = ,T = −p p T p ∂ ∂ T s v v s • Thus we have obtained the basic thermodynamic • From Eq. (2) dh = Tds + vdp properties, p,v,T and s have been defined as the ∂h ∂h derivative of u, h, a and g. Due to this aspect, u, h, a • We can write, = ,T = v 8 ∂ ∂ and g are also called thermodynamic potentials s p p s

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13/15 14/15 Maxwell Relations Auxiliary Relations-I • Now we shall use the exact differential rule and relate • Now we shall equate T is Eqs. (7) and (8) to give the derivatives ∂ ∂ T = − p ∂u ∂h du = Tds − pdv ⇒ 11 ⇒ = 15 ∂v ∂s ∂ ∂ s v s v s p ∂T ∂v dh = Tds + vdp ⇒ = 12 • Similarly equating -p is Eqs. (7) and (9) to give ∂p ∂s s p ∂ ∂ ⇒ u = − a ∂p ∂s 16 = − − ⇒ = 13 ∂v ∂v da pdv sdT ∂ ∂ s T T v v T • Equating v is Eqs. (8) and (10) to give ∂v ∂s dg = vdp − sdT ⇒ = − 14 ∂ ∂ T p p ∂h ∂g T ⇒ = 17 ∂ ∂ • The relations in Eqs. (11) – (14) are called Maxwell p s p T Relations

15/15 Auxiliary Relations-II • Equating -s is Eqs. (8) and (10) to give ∂ ∂ ⇒ a = g ∂ ∂ 18 T v T p

• The fundamental question that arises is what is the use of all these relations? • They provide means to construct property tables from the measured p, v and T data and some additional measurements • The aim is to measure minimum quantities and construct property data

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