A group of scientists produces a sample of potassium by extracting it from a giant truckload of bananas. 97.00 kg of the potassium is K-39 (38.897 amu), 21.00 kg of the potassium is K-40 (39.955 amu), and 11.00 kg of the potassium is K-41 (40.965 amu). What is the average atomic mass of the potassium in this sample?

law of conservation of matter

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