1 The Second Law of Thermodynamics (II)

2 The Fundamental Equation We have shown that: dU = dq + dw plus dw rev = -pdV and dq rev = TdS We may write: dU = TdS – pdV (for constant composition)

3 In chapter 3 we discussed total integrals. Properties of the internal energy We can express U as a function of S and V, i.e. U = f ( S,V ) If z = f (x,y) then: dU = TdS – pdVc.f. We have discovered that

4 Properties of the internal energy Recall the test for exactness: If the differential is exact then: All state functions have exact differentials

5 Properties of the internal energy Therefore: Where: Because this is exact we may write: We have obtained our first Maxwell relation!

6 Relationships between state functions: Be prepared! U and S are defined by the first and second laws of thermodynamics, but H, A and G are defined using U and S. The four relationships are: We can write the fundamental thermodynamic equation in several forms with these equations dU = TdS – PdV dH = TdS + VdP dA = -SdT - PdV dG = -SdT + VdP Gibbs Equations

7 Properties of the internal energy Also consider dH = TdS + Vdp, and writing H = f ( S,p ) Where: Because this is exact we may write: We have obtained our second Maxwell relation!

8 The Maxwell Relations (a)  U = q + w (b)  S = q rev /T (c) H = U + pV (d) A = U – TS (e) G = H - TS dU = TdS – pdV 2.dH = TdS + Vdp 3.dA = -SdT - pdV 4.dG = -SdT + Vdp

9 The Maxwell Relations: The Magic Square V AT G PHS U “Vat Ug Ship” Each side has an energy ( U, H, A, G ) Partial Derivatives from the sides Thermodynamic Identities from the corners Maxwell Relations from walking around the square

10 Example: Calculate the change in enthalpy if the pressure on one mole of liquid water at 298 K is increased from 1 atm to 11 atm, assuming that V and α are independent of pressure. At room temperature α for water is approximately 3.0 × K -1. (The expansion coefficient) The volume of 1 mole of water is about L.

11 Properties of the Gibbs energy G = H - TS dG = dH –TdS - SdT dG = dU + pdV + Vdp –TdS - SdT dU = TdS –pdV dG = TdS – pdV + pdV + Vdp –TdS - SdT dG = Vdp - SdT G = f ( p, T ) dH = dU +pdV + Vdp H = U + pV

12 Properties of the Gibbs energy dG = Vdp - SdT V is positive so G is increasing with increasing p G T (constant p) Slope = -S G P (constant T) Slope = V S is positive (-S is negative) so G is decreasing with increasing T

13 Dependence of G on T Using the same procedure as for the dependence of G on p we get: To go any further we need S as a function of T ? Instead we start with: G = H - TS -S = (G – H)/T

14 Dependence of G on T Let G/T = x This is the Gibbs-Helmholtz Equation

15 Dependence of G on T Two expressions: Gibbs-Helmholtz Equation Changes in entropy or, more commonly, changes in enthalpy can be used to show how changes in the Gibbs energy vary with temperature. For a spontaneous (  G < 0) exothermic reaction (  H < 0) the change in Gibbs energy increases with increasing temperature.  G/T T (constant p) Slope = -  H/T 2 = positive for exothermic reaction Very negative Less negative

16 Dependence of G on p It would be useful to determine the Gibbs energy at one pressure knowing its value at a different pressure. dG = Vdp - SdT We set dT = 0 and integrate:

17 Dependence of G on p Liquids and Solids. Only slight changes of volume with pressure mean that we can effectively treat V as a constant. Often V  p is very small and may be neglected i.e. G for solids and liquids under normal conditions is independent of p.

18 Dependence of G on p Ideal Gases. For gases V cannot be considered a constant with respect to pressure. For a perfect gas we may use:

19 Dependence of G on p Ideal Gases. We can set p i to equal the standard pressure, p  ( = 1 bar). Then the Gibbs energy at a pressure p is related to its standard Gibbs energy, G , by:

20 Dependence of G on p Exercise 5.8(b) When 3 mol of a perfect gas at 230 K and 150 kPa is subjected to isothermal compression, its entropy decreases by 15.0 J K -1. Calculate (a) the final pressure of the gas and (b)  G for the compression.

21 Dependence of G on p Real Gases. For real gases we modify the expression for a perfect gas and replace the true pressure by a new parameter, f, which we call the fugacity. The fugacity is a parameter we have simply invented to enable us to apply the perfect gas expression to real gases.

22 Dependence of G on p Real Gases. We may show that the ratio of fugacity to pressure is called the fugacity coefficient: Where  is the fugacity coefficient Because we are expressing the behaviour of real gases in terms of perfect gases it is of little surprise that  is related to the compression factor Z: We may then write

23 Summary 1.The four Gibbs equations. 2.The four Maxwell relations. (The Magic Square!) 3.Properties of the Gibbs energy Variation of G with T The Gibbs-Helmholtz equation. Variation of G with p Fugacity

24 Exercise: For the state function A, derive an expression similar to the Gibbs-Helmholtz equation.

25 Exercise 5.15 (a) (first bit) Evaluate (  S/  V) T for a van der Waals gas.

26 Preparation for Chapter 6: So far we have only considered G = f ( p, T ). To be completely general we should consider Gas a function of p, T and the amount of each component, n i. G = f ( p,T, n i ) Then: where  is the chemical potential.