Question

1 mol of Ar is expanded from 2L to 8L and from 25C to 100C instantaneously. Find the change in entropy of this process. THE HINT GIVEN IN CLASS WAS: P = 1mol* 0.082 atmL/molK* 298K / 2L = 12.218 atm =12.05 bar)

Answer 1

pressure(P) = nRT/V

                    = 1*0.0821*298/2

                     = 12.233 atm

work done(w) = - p*DV

                       = -12.233*(8-2)

                        = -73.298 l.atm

         ( 1 l.atm = 101.3 joule)

                             = -78.298*101.3

                              = -7931.6 joule

          q = -w = 7931.6 j

change in temperature(DT) = 100-25 = 75

change in entropy(DS) = q/DT = 7931.6/75 = 105.75 j/c