Math Background for Thermodynamics ∑
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MATH BACKGROUND FOR THERMODYNAMICS A. Partial Derivatives and Total Differentials Partial Derivatives Given a function f(x1,x2,...,xm) of m independent variables, the partial derivative ∂ f of f with respect to x , holding the other m-1 independent variables constant, , is defined by i ∂ xi xj≠i ∂ f fx( , x ,..., x+ ∆ x ,..., x )− fx ( , x ,..., x ,..., x ) = 12ii m 12 i m ∂ lim ∆ xi x →∆ 0 xi xj≠i i nRT Example: If p(n,V,T) = , V ∂ p RT ∂ p nRT ∂ p nR = = − = ∂ n V ∂V 2 ∂T V VT,, nTV nV , Total Differentials Given a function f(x1,x2,...,xm) of m independent variables, the total differential of f, df, is defined by m ∂ f df = ∑ dx ∂ i i=1 xi xji≠ ∂ f ∂ f ∂ f = dx + dx + ... + dx , ∂ 1 ∂ 2 ∂ m x1 x2 xm xx2131,...,mm xxx , ,..., xx ,..., m-1 where dxi is an infinitesimally small but arbitrary change in the variable xi. nRT Example: For p(n,V,T) = , V ∂ p ∂ p ∂ p dp = dn + dV + dT ∂ n ∂ V ∂ T VT,,, nT nV RT nRT nR = dn − dV + dT V V 2 V B. Some Useful Properties of Partial Derivatives 1. The order of differentiation in mixed second derivatives is immaterial; e.g., for a function f(x,y), ∂ ∂ f ∂ ∂ f ∂ 22f ∂ f = or = ∂ y ∂ xx ∂ ∂ y ∂∂yx ∂∂xy y x x y 2 in the commonly used short-hand notation. (This relation can be shown to follow from the definition of partial derivatives.) 2. Given a function f(x,y): ∂ y 1 a. = etc. ∂ f ∂ f x ∂ y x ∂ f ∂ y ∂ x b. = −1 (the cyclic rule) ∂ x ∂ f ∂ y yxf Proof: The total differential of f is ∂ f ∂ f df = dx + dy ∂ x ∂ y yx We could, in principle, solve the equation f = f(x,y) for y in terms of x and f; i.e., we could obtain the function y(x,f). Its total differential would be given by ∂ y ∂ y dy = dx + df ∂ x ∂ f fx Substituting the above expression for df into the right-hand side of the expression for dy, we obtain ∂ y ∂ y ∂ f ∂ f dy = dx + dx + dy ∂ x ∂ f ∂ x ∂ y fxyx or ∂ y ∂ f ∂ y ∂ y ∂ f 0 = + dx + - 1dy (I) ∂ f ∂ x ∂ x ∂ f ∂ y xy f xx Since dx and dy are arbitrary infinitesimal changes, the only way that the right-handside of (I) can be zero for all choices of dx and dy is if the coefficients of dx and dy are each separately equal to zero. Setting the coefficient of dy equal to zero gives ∂ y ∂ f ∂ y 1 = 1 or = q.e.d. ∂ f ∂ y ∂ f ∂ f xx x ∂ y x 3 Moreover, since any of the three variables f, x, and y could have been chosen as the dependent variable, it must also be true that ∂ x 1 ∂ x 1 = and = ∂ f ∂ f ∂ y ∂ y y f ∂ x ∂ x y f Setting the coefficient of dx equal to zero in eqn. (I) gives ∂ y ∂ f ∂ y = - ∂ f ∂ x ∂ x xy f Thus, ∂ y ∂ f ∂ f ∂ x xy 1 ∂ x = -1 or since = , ∂ y ∂ y ∂ y f ∂ x ∂ x ff ∂ f ∂ y ∂ x − = 1 q.e.d. ∂ x ∂ f ∂ y yxf 1 ∂ V 1 ∂ V Example: Using the cyclic rule and the definitions α = and β = - , V ∂ T V ∂ p p T ∂ p α show that = . ∂T β V Solution: From the cyclic rule, ∂ p ∂ V ∂T = −1. ∂T ∂ p ∂ V VTp Thus, ∂ V - ∂ T ∂ p -1 p -Vα α = = = = q.e.d. ∂T ∂ V ∂T ∂ V -Vβ β V ∂ p ∂ V ∂ p Tp T 3. Given two functions of x and y: f(x,y) and g(x,y), ∂ f ∂ f ∂ y a. = (chain rule) ∂ g ∂ y ∂ g xxx 4 ∂ f ∂ f ∂ f ∂ y b. = + ∂ x ∂ x ∂ y ∂ x gxgy Proof: The total differential of f considered to be a function of x and y is ∂ f ∂ f df = dx + dy , ∂ x ∂ y yx while the total differential of y considered to be a function of x and g is ∂ y ∂ y dy = dx + dg . ∂ x ∂ g gx Substituting this expression for dy into the expression for df above gives ∂ f ∂ f ∂ y ∂ y df = dx + dx + dg ∂ x ∂ y ∂ x ∂ g yxgx ∂ f ∂ f ∂ y ∂ f ∂ y = + dx + dg. ∂ x ∂ y ∂ x ∂ y ∂ g yxg xx But if f is considered to be a function of x and g, rather than of x and y, its total differential is given by ∂ f ∂ f df = dx + dg. ∂ x ∂ g gx These two expressions for df must be identical for any choice of the arbitrary infinitesimal changes dx and dg. Therefore, it must be true that ∂ f ∂ f ∂ f ∂ y ∂ f ∂ f ∂ y = + and = q.e.d. ∂ x ∂ x ∂ y ∂ x ∂ g ∂ y ∂ g gxgy xxx Examples: 1. For a van der Waals gas (i.e., a model system which obeys the van der Waals equation of state), the pressure p and the internal energy U, as functions of n, V, and T, are given by 2 2 nRT na 3 na p = − and U = nRT − Vnb− V 2 2 V respectively, where a and b are constants. Use these equations and the chain rule to derive an ∂ U equation for in terms of n, V, and T. ∂ p nT, 5 ∂ U ∂ V ∂ U ∂ U ∂ V nT, Solution: = = ∂ p ∂ V ∂ p ∂ p nT,,, nT nT ∂ V nT, naV22/ na = = 22na2 nRT na RTV 2 − − V 32()Vnb− V ()Vnb− 2 ∂ U 2. Given that the constant-volume heat capacity C ≡ , show that V ∂ T V ∂U ∂U = CV + α . ∂T V ∂V p T ∂ U ∂ U ∂ U ∂ V Solution: = + ∂T ∂T ∂ V ∂ T pVTp ∂ U ∂ V But = C and = Vα. Therefore, ∂ T V ∂T V p ∂ U ∂ U = CV + α q.e.d. ∂ T V ∂ V p T ------------------------------------------------------------------------------------------------------- Note: In this section, we have considered functions of two independent variables. However, all the relations derived hold for functions of more than two variables provided all the additional independent variables are held constant; e.g., for a function f(w,x,y,z), ∂ ∂ f ∂ ∂ f = , ∂x ∂ yy ∂ ∂ x wxz,, wyz,, wyz,, wxz,, ∂ f ∂ y ∂ x = − 1, etc. ∂ x ∂ f ∂ y wyz,, wxz ,, f ,, wz 6 C. Exact and Inexact Differentials; Line Integrals Linear Differentials An infinitesimal quantity m dz = ∑ Mi (, x12 x ,...,) xmi dx i=1 = M11 dx + M 2 dx 2 + ... + Mmm dx is called a linear differential; the right-hand side of the equation is called a linear differential form in m variables. We shall be concerned primarily with linear differentials in two variables; in this case, let us write dz = Mdx + Ndy where M and N are, in general, functions of x and y. Exact Differentials: dz = Mdx + Ndy is an exact differential if and only if there exists a function of x and y, F(x,y), such that dF = dz for all values of x and y. A more practically useful test for exactness is the following: dz = Mdx + Ndy is an exact differential ∂ M ∂ N if and only if = . ∂ y ∂ x xy It is easy to see that this relation must be true for an exact differential dz because in that case ∂ F ∂ F M = and N = and ∂ x ∂ y y x ∂ M ∂ ∂ F ∂ ∂ F ∂ N = = = . ∂ yy ∂ ∂ xx ∂ ∂ y ∂ x xy x y x y ∂ M ∂ N It can also be shown that the converse is true; i.e., if = , then dz is exact. ∂ y ∂ x xy Inexact Differentials dw = M′dx + N′dy is an inexact differential if and only if there exists no function F(x,y), such that dF = dw for all values of x and y. Equivalently, dw is inexact if and only if ∂ M '' ∂ N ≠ . ∂ y ∂ x xy Examples: The linear differential dz = xy2dx + x2ydy is an exact differential because it is the total differential of the function F = ½ x2y2 + C, where C is an arbitrary constant. Here, ∂ M ∂ ∂ ∂ N = ()xy22 = 2. xy = ()xy = ∂ yy ∂ ∂ x ∂ x xx yy 7 2 2 On the other hand, the linear differential dw = x y dx + xy dy is inexact, as can most easily be demonstrated by noting that ∂ M ∂ ∂ N ∂ = ()xy222 = x2 while = ()xy = y ∂ yy ∂ ∂ xx ∂ xx y y Line Integrals Consider a curve C in the X-Y plane connecting two points (x′,y′) and (x′′,y′′) and imagine that we chop the curve up into n segments as shown at the left. Then the line integral I of the linear differential dz = Mdx + Ndy on the curve C is defined by n []∆∆+ I = lim∑ MxyxNxyy(,ii ) i (, ii ) i n→∞ i=1 where M(x,y) and N(x,y) are given functions and M(xi,yi) and N(xi, yi) are the values of these functions at the midpoint of the ith segment. The notation we shall use for this integral is ('','')xy I =[∫ Mdx+ Ndy] (',')xy C In order to carry out this integration, one must eliminate one of the variables x or y, which are related via the equation for the curve C, and then evaluate the resulting ordinary definite integral. Examples: 1. Calculate the line integral of the inexact differential dw = x2ydx + xydy along the curve y = x from (0,0) to (1,1). (,)11 Solution: I =[∫ x2 dx+ xydy] But y = x and dy = dx. Therefore, (,)00 yx= 1 1 2 2 3 2 I = 2∫ x dx = 3 x = 3 0 0 2. Calculate the line integral of dw = x2ydx + xydy along the curve y = x2 from (0,0) to (1,1). (,)11 Solution: I =[∫ x2 dx+ xydy] But y = x2 and dy = 2xdx.