Chapter 5 Simple Applications of Macroscopic Thermodynamics

Preliminary Discussion Classical, Macroscopic, Thermodynamics: Drop the statistical mechanics notation for average quantities. We know that all variables are averages only. We’ll discuss relationships between macroscopic variables using the Laws of Thermodynamics. Internal Energy = E Entropy = S Temperature = T Gases: External parameter = Volume Volume = V Pressure = p General system: External parameter = x Parameter = x Generalized Force = X

TdS = dE + pdV Combined 1st & 2nd Laws Assume external parameter = V in order to have a specific case to discuss. 1st & 2nd Laws of Thermodynamics 1st Law: đQ = dE + pdV 2nd Law: đQ = TdS Combined 1st & 2nd Laws TdS = dE + pdV 5 Variables: T, S, E, p, V It can be shown that: 3 of these can always be expressed as functions of 2 others. That is, there are 2 independent variables & 3 dependent variables. Which 2 are chosen independent is arbitrary.

Brief, Pure Math Discussion Consider 3 variables: x, y, z. Suppose that we know that x & y are independent variables. Then, there must be a function z = z(x,y). The total differential of z has the form: dz  (∂z/∂x)ydx + (∂z/∂y)xdy (a) Suppose instead that we want to take y & z as independent variables. Then, there must be a function x = x(y,z). The total differential of x has the form: dx  (∂x/∂y)zdy + (∂x/∂z)ydz (b) Using (a) & (b) together, the partial derivatives in (a) & those in (b) can be related to each other. Assume all functions are analytic. So, the 2nd cross derivatives are equal: Such as (∂2z/∂x∂y) = (∂2z/∂y∂x) etc.

It’s exact differential is df  y1dx1 + y2dx2 By definition: Mathematics Summary Consider a function of 2 independent variables: f  f(x1,x2). It’s exact differential is df  y1dx1 + y2dx2 By definition: Because f(x1,x2) is an analytic function, it is always true that The applications all use this with the 1st & 2nd Laws of Thermodynamics: TdS = dE + pdV

dE = TdS – pdV (1) Properties of the Internal Energy E Choose S & V as independent variables: E = E(S,V) ∂E ∂E dE (2) Comparison of (1) & (2) clearly shows that ∂E ∂E Applying the general result with 2nd cross derivatives gives: Maxwell Relation I! 6

H  H(S,p)  E + pV  Enthalpy If we choose S & p as independent variables, it is convenient to define the following energy: H  H(S,p)  E + pV  Enthalpy Use the combined 1st & 2nd Laws. Rewrite them in terms of dH: dE = TdS – pdV = TdS – [d(pV) – Vdp] or dH = TdS + Vdp But also: (1) (2) Comparison of (1) & (2) clearly shows that Applying the general result with 2nd cross derivatives gives: Maxwell Relation II! 7

Summary 2. dH = TdS + Vdp 3. dF = - SdT – pdV 4. dG = - SdT + Vdp Energy Functions 1. Internal Energy: E  E(S,V) 2. Enthalpy: H = H(S,p)  E + pV 3. Helmholtz Free Energy: F = F (T,V)  E – TS 4. Gibb’s Free Energy: G = G(T,p)  E – TS + pV Combined 1st & 2nd Laws 1. dE = TdS – pdV 2. dH = TdS + Vdp 3. dF = - SdT – pdV 4. dG = - SdT + Vdp lll lll lll lll 8 8

The 4 Most Common Maxwell Relations Summary The 4 Most Common Maxwell Relations 9 9

Maxwell’s Relations: Table (E → U) 10 10