Question

A flask containing 425 mL of Ar at 5.67 atm and a flask containing 325 mL of Xe at 6.50 atm at room temperature (25℃) are connected by a valve. When the valve is opened: What is the partial pressure of Ar? What is the partial pressure of Xe? What is the total pressure? What is the mole fraction of Ar? What is the mole fraction of Xe?

Answer 1

we find moles or Ar using PV = nRT

P = 5.67 atm , R = 0.08206 Latm/molK , V = 425 ml = 0.425 L , T = 25C = 25+273 = 298 Kwe find n

5.67 atm x 0.425 L = n x 0.08206 Latm/molK x 298 K

n = 0.09854 mol = mol of Ar

for Xe ,   6.5 atm x 0.325 L = n x 0.08206 L/atm/molK x 298 K

n ( Xe) = 0.08639 mol

total gas moles when both gases mixed = 0.09854+0.08639 = 0.18493 mol

mol fraction of Ar = ( moles of Ar) / ( total gas moles) = ( 0.09854 /0.18493) = 0.533

mol fraction of Xe= 1-0.533 = 0.467

total volume = 425+325 = 750 ml = 0.75L

total Pressure be P   , P x 0.75L = 0.18493 mol x 0.08206 Latm/molK x 298 K

P = 6.03 atm

Partial Presure of Xe P(Ar) = mol fraction of Ar x Total pressure

              = ( 0.533 ) ( 6.03 )

         = 3.21 atm

Partial Pressure of Xe P ( Xe) = 6.03-3.21 = 2.82 atm