At 25°C, the standard enthalpy of formation of HF(aq) is −320.1 kJ/mol. Of OH−(aq), it is −229.6 kJ/mol. Of F−(aq), it is −329.9 kJ/mol, and of H2O(l), it is −285.8 kJ/mol. (a) Calculate the standard enthalpy of neutralization of HF(aq):
HF(aq) + OH−(aq) → F−(aq) + H2O(l)
(b) Using the value of -56.2 kJ as the standard enthalpy change for the reaction
H+(aq)+OH−(aq)→H2O(l)
calculate the standard enthalpy change for the reaction
HF(aq)→H+(aq)+F−(aq)

i don't know people like to fight everywhere rip.

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