Question

After 0.600 L of Ar at 1.50 atm and 248°C is mixed with 0.200 L of O₂ at 333 torr and 123°C in a 400.−mLflask at 23°C, what is the pressure in the flask?

Answer 1

Ar

P = 1.5atm

V   = 0.6L

T   = 248+273   = 521K

PV = nRT

n    = PV/RT

      = 1.5*0.6/0.0821*521    = 0.021moles

O2

V = 0.2L

P   = 333 torr = 333/760   = 0.438atm

T   = 123 + 273    = 396K

PV = nRT

n   = PV/RT

      = 0.438*0.2/0.0821*396    = 0.00296mole

Total no of moles of mixture   = 0.021+0.00296   = 0.02396moles

T = 23 + 273 = 296K

V   = 400ml = 0.4L

PV = nRT

P   = nRT/V

       = 0.02396*0.0821*296/0.4   = 1.455atm >>>>>answer

the pressure in the flask   = 1.455atm