At a constant pressure and a temperature of 280 k the gas in a cylinder has a volume of 10.0 l. if the volume of the gas is increased to 20.0 l what is the new temperature 28 k 560 k 140 k 2800 k

At constant pressure a known moles of gas has volume which is directly proportional to temperature (Charle's Law)

Thus V α T

Therefore

V1/V2 = T1 /T2

V1 = initial volume = 10 L

T1 = initial temperature= 280 K

V2 = final volume = 20L

T2 = final temperature = ?

T2 = T1V2 / V1 = 280 X 20 / 10 = 560 K

Thus the new temperature will also doubles up as volume is increased by two times

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