Question

A sample containing 5.20 g 02 gas has a volume of 31.0 L . Pressure and temp remain constant .

1- what is the new volume if O.300 mole O2 gas is added. V =

2- what is the volume after 7.50 g he is added to the o2 gas already in the container?

Answer 1

no of moles of O2 n1 = W/G.M.Wt

                                = 5.2/32    = 0.1625 moles

                                 V1 = 31L

                                 n2 = 0.3

             PV1 = n1RT

            PV2 = n2RT

             V1/V2 = n1/n2

             31/V2   = 0.1625/0.3

                  V2   = 31*0.3/0.1625   = 57.23L

2.     

no of moles of O2 n1 = W/G.M.Wt

                                = 5.2/32    = 0.1625 moles

                                 V1 = 31L

no of moles of O2 n2     = 7.5/32 = 0.2344moles

                             V2     =

       V1/V2 = n1/n2

             31/V2   = 0.1625/0.2344

                  V2   = 31*0.2344/0.1625    = 44.72L