A gas balloon has a volume of 80.0 mL at 300K , and a pressure of 50.0 kPa. if the pressure changes to 80.0 kPa and the temperature rises to 320 K, what is the new volume of balloon in milliliters

53.3

Explanation:

V2=(T2 x P1 x V1)/(T1 x P2)

(320x50x80)/(300x80)

53.3

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