By Peter William Atkins
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Therefore, U = 0 for paths ACB and ADB as well and the fact that CV ,m = 23 R is not needed for the solution. 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.
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.
So, p n−1 Vi n−1 pf n (3) = n=0 Vf pi Substitution of eqn 3 into eqn 2 and using ‘1’ and ‘2’ to represent the initial and final states, respectively, yields RT1 × (4) w = n−1 p2 p1 n−1 n −1 for n = 0 and n = 1 THE FIRST LAW: THE CONCEPTS 37 In the case for which n = 0, eqn 1 gives w= C × 0−1 1 Vf−1 − 1 = −C(Vf − Vi ) Vi−1 = −(pV 0 )any state × (Vf − Vi ) = −(p)any state (Vf − Vi ) (5) w = −p V for n = 0, isobaric case In the case for which n = 1 w=− C dV = − Vn p dV = − Vi Vf Vi = (RT )any state ln Vf w = (pV n )any state ln (6) w = RT ln (7) V1 V2 = RT ln C Vf dV = −C ln V Vi Vi Vf = (pV )any state ln p2 p1 for n = 1, isothermal case To derive the equation for heat, note that, for a perfect gas, Tf p f Vf − 1 = CV Ti −1 q + w = CV Ti Ti p i Vi Vin−1 = CV Ti − 1 [because pV n = C] n−1 Vf = CV Ti q = CV Ti pf pi pf pi n−1 n RTi = CV Ti − n−1 = n−1 n −1 −1 − [using eqn 3 (n = 0)] RTi n−1 × pf pi CV 1 − RTi R n−1 pf pi n−1 n −1 n−1 n CV pf 1 − RTi Cp − C V n−1 pi CV 1 = − RTi Cp n − 1 C −1 = = (n − 1) − (γ − 1) RTi (n − 1) × (γ − 1) pf pi n−γ = RTi (n − 1) × (γ − 1) n−1 n n−1 n pf pi −1 pf pi CV pf pi −1 −1 n−1 n 1 1 − RTi γ −1 n−1 n−1 n pf pi = V U = q + w = CV (Tf − Ti ).