Question

physical chemistry

5. Consider a solid with equation of state: V = V.(1 - KT(P-B) + al) Where: KT = 2.70 x 10 atm 1 a = 1.04 x 103 deg 1 B = 0.011 atm One mole of solid with volume Vo of 0.0100 e is placed in a piston and cylinder and allowed to come to equilibrium at 30.0°K and 1.011 atm. Starting with the solid in this initial state, the following process is then carried out: The system is placed in a vacuum jacket, and the pressure is suddenly increased to 20,000 atm and held constant until equilibrium is reached at a temperature of 30.5K. The sample is then removed from the vacuum and expanded by moving along the curve: P= 3.961 x 105 - 3.846 x 107V until it is restored to the initial state. Determine Aq for the second step.

Answer 1

page:oi Answer: Given information, Consider soled with equation of state. . Va Volt-ky (P-B) + FT) where, kg a 2-tox106 atm' da houx 103 degal Po ao ou atm . P: 20000 atm :: one mole of solid toith volume of Vo is 0.01000 is placed in a piston and cylinden and allowed to come to Equellebrium at sok and 1.01 alm. starting with the sold in thie's initial state. The following proces is then carried out. for the second step: V = V0 (4- Kp (P-B) +47) oooill-(2.70 X106 aterii) (20000 -0.011) + 1. Axloblebat

page 102 2001lle-(.70xlők atm") % 19999.989atis + louxio do" *30.5k e Dio1 lll- 010539999703 +0.03172. 201014 10.9460000247 +0.03172) = Doll ( 0.9174200294) viz q. 71 x 1031 Va a VO (1-ky (P-B)+ dt) Coll (1-(2.70x106)(1.011 – 0.011)+ louxto?dey! X30.5k ooolll 1-19.40x166 X1 + 0.031512 0.018 (1-2.70xc06 + 0.03179) = 0.0d ( 0.99973 + 0.03172) 20.01€ 11.0344173) Va.? 0.01031717 l Since, Temperatine is constant, duco.

page:03 * Determine sq for second step: - tet, y qacho 2nRT In valve Hene, nel mol .R= 8.314 Il molik I. 30.5K then Aq: imot x 84314 J/motik (305%) in 0.01031317 0.0093197 8.314J X 30.5 x 0.05377. 2 13.818J l. Aq = 13.8.185