By P W Atkins; Julio De Paula
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Extra resources for Physical Chemistry 7th ed [SOLUTION MANUAL]
Example text
However, if both pistons are diathermic, the result is different. The solution for both pistons being diathermic follows. See Fig. 1. 1 Initial equilibrium state. 3 L = constant Final equilibrium state. The diathermic walls allow the passage of heat. Consequently, at equlibrium all chambers will have the same temperature T1 = T2 = T3 = 348 K. 17 bar = p2 = p3 = The chambers will have equal volume. 11 dw = ∂w dx + ∂x y,z ∂w dy + ∂y x,z ∂w dz ∂z x,y dw = (y + z) dx + (x + z) dy + (x + y) dz This is the total differential of the function w, and a total differential is necessarily exact, but here we will demonstrate its exactness showing that its integral is independent of path.
0 × 103 J The heat is different for all three paths; heat is not a state property. 27 U is independent of path Since U (A → B) = q(ACB) + w(ACB) = 80 J − 30 J = 50 J U = 50 J = q(ADB) + w(ADB) (a) q(ADB) = 50 J − (−10 J) = +60 J (b) q(B → A) = U (B → A) − w(B → A) = −50 J − (+20 J) = −70 J The system liberates heat. 6] na × 3b 1 1 − Vr,2 Vr,1 − The van der Waals constants a and b can be eliminated by defining wr = 8 Vr,2 − 1/3 wr = − nTr ln 9 Vr,1 − 1/3 −n 1 1 − Vr,2 Vr,1 Along the critical isotherm, Tr = 1 and Vr,1 = 1, Vr,2 = x.
13) − (−1543)] kJ mol−1 = −5798 kJ mol−1 Comment. 19(b). 24(b) Because entropy is a state function, we can choose any convenient path between the initial and final states. 5 kJ THE SECOND LAW: THE CONCEPTS 65 Solutions to problems Assume that all gases are perfect and that data refer to 298 K unless otherwise stated. 4 J K−1 mol−1 Since Stotal > 0, the transition l → s is spontaneous at −5◦ C (b) A similar cycle and analysis can be set up for the transition liquid → vapour at 95◦ C.
