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When 0.653 L of Ar at 1.20 atm and 227°C is mixed with 0.278 L of...

When 0.653 L of Ar at 1.20 atm and 227°C is mixed with 0.278 L of O2 at 501 torr and 127°C in a 400. mL flask at 27°C, what is the pressure in the flask?

Solutions

Expert Solution

pv = nRT

no of moles of argon nAr = PV/RT

                                             = 1.2*0.653/0.0821*500

                                             = 0.019moles

no of moles of O2 nO2    = PV/RT

                                     p     = 501/760 = 0.659atm

                                     T       = 127 +273 = 400K

                                    nO2    = 0.659*0.278/0.0821*400 = 0.00557moles

total no of moles nAr + nO2     = 0.019 +0.00557 = 0.02457moles

      PV = nRT

       P   = nRT/V

              = 0.024457*0.0821*300/0.4   = 1.5atm >>>> anser

pressure in the flask = 1.5atm

queen_honey_blossom answered 1 year ago
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