Solutions to Exam 2 (FALL 2005)

(1)   (a)  a(p_T+p)V

(b) nR

(c) (R+Bp²)/R

(2) (a) w = 0; q = -935 J; DU = -935; DH = -1559 J; DS = -3.59 J

      (b) w = 624 J; q = -1559 J; DU = -935 J; DH = -1559 J; DS = -5.98 J

      (c) w = 11487 J; DH = DU = 0; q = -11487 J; DS = -38.29 J

(3) DS = -5.142 J/K mol;  (and, in the process, Tf = 367.46 K)

(4) 0.96 J/K

(5) 245.428 J/K mol