1. Chapter 3 : Slide 1 Chapter 3 The First Law of Thermodynamics The Machinery

2. Chapter 3 : Slide 2 OUTLINE Partial Derivatives and the First Law 1. Partial Derivatives and the Internal Energy, U: Internal Pressure,  T, and the Expansion Coefficient,  2.Partial Derivatives and the Enthalpy, H: Isothermal Compressibility,  T, and the Joule-Thomson Coefficient,  3. The Relation Between C p and C V.

3. Chapter 3 : Slide 3 Preamble (little bit of a refresher): A state function is also known as a state property! State property is an intuitive name for what we mean i.e. a state property is something like pressure, temperature, internal energy. The “state” in state property reminds us that the properties we are talking about are only state dependent that is to say that the difference between a state property in one state and another state is independent of how we got from the old state to the new state.

4. Chapter 3 : Slide 4 Preamble (little bit of a refresher): There is such a thing as a path property or path function. An example of a path property is work, w. The amount of work done in going from an initial state, i, to a final state, f, is totally dependent on the path. This means we CANT write: Instead we have to write: Wrong! We say dw is an inexact differential

5. Chapter 3 : Slide 5 Preamble (little bit of a refresher): A state function is also known as a state property! The term “state function” reminds us that the properties of a system are inter-related e.g. for a one component system: p = f ( n, V, T ) recall pV = nRT And also: U = f ( n, V, T ) or U = f ( n, V, p ) or U = f ( n, p, T ) The multivariable nature of these state functions means that the use of partial derivatives is very common in thermodynamics!

6. Chapter 3 : Slide 6 Preamble (little bit of a refresher): Homework from this lecture: Exercises 3.4 – 3.7 3.4(a) part (a) Show that x 2 y+3y 2 has an exact differential.

7. Chapter 3 : Slide 7 Partial Derivatives and the Internal Energy For a closed system of constant composition: U = f ( V, T ) Constant T, change V to V+dV, then U changes to: Constant V, change T to T+dT, then U changes to: Change V to V+dV and T to T+dT, then U changes to: Because dU = U'-U, then:

8. Chapter 3 : Slide 8 Partial Derivatives and the Internal Energy A small (infinitesimal) change in internal energy is proportional to small changes in volume and temperature. Recall : We introduce a new term,  T, the internal pressure.

9. Chapter 3 : Slide 9 Partial Derivatives and the Internal Energy For NH 3,  T,m = 840 Pa at 300K and 1 bar and C V,m = 27.32 J K -1 mol -1. Calculate the change in molar internal energy as NH 3 is heated from 298 K to 300 K while being compressed from 1 L to 0.9 L

10. Chapter 3 : Slide 10 Partial Derivatives and the Internal Energy Consider an isothermal, closed system of constant composition, i.e. T is constant i.e dT = 0  T gives an indication of how the internal energy changes with respect to changes in volume.  T gives a measure of the strength of the cohesive force in the sample.

11. Chapter 3 : Slide 11 Partial Derivatives and the Internal Energy Let us now consider the change in internal energy that accompanies a change in T when p is kept constant i.e. now U = f ( p, T ). We had written: Lets divide by dT And impose constant p

12. Chapter 3 : Slide 12 Partial Derivatives and the Internal Energy We introduce a new term, , the expansion coefficient. A large  indicates that the samples volume responds strongly to changes in temperature. Exactly the same as on the previous slide

13. Chapter 3 : Slide 13 Partial Derivatives and the Enthalpy Just like U, the enthalpy, H, is a state function, we can write H= f (p, T). Recall : We may manipulate this to give:  T is the isothermal compressibility  is the Joule-Thomson coefficient

14. Chapter 3 : Slide 14 Partial Derivatives and the Enthalpy  T is the isothermal compressibility  is the Joule-Thomson coefficient Exactly the same as on the previous slide

15. Chapter 3 : Slide 15 3.16(b) The isothermal compressibility of lead at 293 K is 2.21  10 -6 atm -1. Calculate the pressure that must be applied in order to increase its density by 0.08 %.

16. Chapter 3 : Slide 16 The Perfect gas and  T, ,  T,  PropertyPDValue for Perfect GasInfo: Internal Pressure  T =0 Strength/nature of interactions between molecules Expansion Coefficient  =1 / T The higher T, the less responsive is its volume to a change in temperature Isothermal Compressibility  T =1 / p The higher the p, the lower its compressibility Joule-Thomson Coefficient  =0 Another indication of molecular interactions.

17. Chapter 3 : Slide 17 The Perfect gas and  T, ,  T,  PropertyPDValue for Perfect GasInfo: Internal Pressure  T =0 Strength/nature of interactions between molecules

18. Chapter 3 : Slide 18 The JOULE EXPERIMENT To measure  T i.e. test U as a function of V for a real gas. Joule detected no temperature change i.e. q = 0. Any work done? NO, so w = 0. Interpretation  U = q + w = 0 (1st law). U does not change with a change in V, i.e.  T = 0. This is not a very sensitive experiment because water has a much higher C p than air.

19. Chapter 3 : Slide 19 James Joule and the Gothic Era.

20. Chapter 3 : Slide 20 The Perfect gas and  T, ,  T,  PropertyPDValue for Perfect GasInfo: Internal Pressure  T =0 Strength/nature of interactions between molecules Expansion Coefficient  =1 / T The higher T, the less responsive is its volume to a change in temperature Isothermal Compressibility  T =1 / p The higher the p, the lower its compressibility Joule-Thomson Coefficient  =0 Another indication of molecular interactions.

21. Chapter 3 : Slide 21 THE JOULE-THOMPSON EXPERIMENT A further test of intermolecular forces in real gases. Imagine a sample of gas pushed through a porous plug, in an isolated tube (adiabatic system). The temperature is measured on each side of the plug. Analysis w = p i V i - p f V f Since  U = U f - U i = w ( because q = 0 ), U f + p f V f = U i + p i V i i.e.  H = 0 This is a constant enthalpy (isenthalpic) process.

22. Chapter 3 : Slide 22 If  > 0 then this indicates that the gas cools when expanded (-ve  p). If  < 0 then this indicates that the gas cools when compressed (+ve  p). The "inversion temperature" indicates where on a T,p plot µ flips from +ve to –ve. For a real gas µ is non-zero (except at the inversion temperature) and thus H shows some variation with p. The Linde process THE JOULE-THOMPSON EXPERIMENT

23. Chapter 3 : Slide 23 3.13(b) A vapor at 22 atm and 5 o C was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the Joule-Thomson Coefficient, , at 5 o C, assuming it remains constant over this temperature range.

24. Chapter 3 : Slide 24 The relation between C V and C p Recall : Therefore : For a perfect gas C p differs from C V by the work needed to change the volume of the system. This work is (a) the work required to push back the atmosphere and (b) the work of stretching the bonds of the material.

25. Chapter 3 : Slide 25 The relation between C V and C p For any substance: