Question

If a gaseous mixture is made by combining 3.02 g of Ar and 2.81 g of Kr in an evacuated 2.50 L container at 25.0 °C, what are the partial pressures of each gas and what is the total pressure exerted by the gaseous mixture?    (All three answers must be in units of atm)

P_Ar

P_Kr

P_Total

Answer 1

As we know that the gases are ideal and we have the relation PV=nRT

P=total pressure of the system

V= volume of the system=2.5

n= total moles in the system

R= universal gas constant=0.0821

T= temperature of system=25+273=298 K

Atomic mass of Kr=83.8 g

Atomic mass of Ar=40 g

Moles of Kr=2.81/83.8=0.0335

Moles of Ar=3.02/40=0.0755

Hence total moles=0.0335+0.0755=0.109

Hence using the ideal gas equation

P=(0.109*0.0821*298)/2.5

P=1.067 atm

PAr= PTotal*Mole fraction of Ar=1.067*(0.0755/0.109)=0.739 atm

PKr= PTotal*Mole fraction of Kr=1.067*(0.0335/0.109)=0.328 atm

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