Assuming the temperature and volume remain constant, changes to the pressure in the reaction vessel will directly correspond to changes in the number of moles based on the ideal gas law (pv=nrt). suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in part a, and the temperature and volume were constant at values of 295 k and 2.00 l, respectively. if the pressure was 5.81 atm prior to the reaction, what would be the expected pressure after the reaction was completed? express the pressure in atmospheres to three significant digits.

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