The standard enthalpy of formation of CaSO 4 (s) at 25 ◦ C is − 1432 . 7 kJ mol − 1 .The equation that describes this formation reaction is: (A) CaSO 4 (s) −−→ Ca(s) S(s) 2 O 2 (g) (B) Ca(s) S(s) 2 O 2 (g) −−→ CaSO 4 (s) (C) Ca 2 (aq) SO 2 − 4 (aq) −−→ CaSO 4 (s) (D) CaSO 4 (s) −−→ Ca(s) S(s) 4 O(g) (E) Ca(s) S(s) 4 O(g) −−→ CaSO 4 (s)

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole