CHAPTER 4. THERMODYNAMICS: THE FIRST LAW

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4-2 how the change of state was carried out. Therefore, the changes in the values of the state functions are determined only by the initial and final states of the system. For example, consider the four states of a system composed of an ideal gas, as depicted in Fig. 4-1. path 1 State 2 P 2 = 2 atm V 2 = 22 L T 2 = 546 K State 1 P 1 = 1 atm V 1 = 22 L T 1 = 273 K path 3 State 4 P 4 = 2 atm V 4 = 11 L T 4 = 273 K path 2 State 3 P 3 = 1 atm V 3 = 11 L T 3 = 136 K Figure 4-1. Transformation of a system from state 1 to state 4 via three different paths. Three different paths lead from state 1 to state <strong>4.</strong> However, the changes in all three of the state functions listed (P, V, and T) are the same for each path. Going from state 1 to state 4 via path 1, path 2, or path 3 results in P = 1 atm, V = 11 L, and T = 0 K. In other words, the final value of a state function does not depend on the path taken between the initial and final state. On the other hand, suppose that instead of different states of a system, the above (state) labels represented different cities on a map and the three paths represented different routes between them. It is obvious that distance is not a state function since the distance between city 1 and city 4 depends on the route taken. In the city analogy, is altitude a state function? Of course distance is not a thermodynamic variable, but two very important thermodynamic variables that are not state functions are work and heat. C. Work and heat
is = 4-3 Mechanical work is defined as a force applied in the direction of motion times the distance moved (displacement). Specifically, dw ' PF@ Pds ' Fdscosθ , where θ the angle between the force vector and the displacement vector. An example of the application of this formula is the calculation of the work done in raising a mass m to a height h against the force of gravity: F = ma = mg dw = Fds cos θ mgds w ' h 0 mgds ' mgh. A particularly important form of work in chemical thermodynamics is the work associated with a pressure-volume change. The diagram in Fig. 4-2 shows a cross sectional view of a piston undergoing a compression. Figure 4-2. A piston system undergoing a compression. The applied force can be expressed as an external pressure times the cross sectional area of the piston; F = P A. Therefore, ex but dw = Fds = P Ads, ex
Page 1: 4-1 CHAPTER 4. THERMODYNAMICS: THE
Page 5 and 6: expansion processes which start at
Page 7 and 8: 4-7 when heat is added to a system,
Page 9 and 10: 4-9 !3 = 30.8 kJ/mol - (8.314x10 kJ
Page 11 and 12: 4-11 In this case, n = 3 and so D =
Page 13 and 14: F. Thermochemistry 4-13 A very impo
Page 15 and 16: 4-15 Applying this formula directly
Page 17 and 18: orbitals orbitals 4-17 o H (C ) = 7
Page 19 and 20: ). w = !nRT ln (V /V ) 2 1 4-19 whe
Page 21 and 22: 4-21 An adiabatic expansion is show
Page 23 and 24: 4-23 Review Questions 1. Define ope
Page 25 and 26: Answers: 4-25 4. (a) w = 0, (b) w =
Page 27: = 4-27 where where c = C /R. V P 1

CHAPTER 4. THERMODYNAMICS: THE FIRST LAW

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