Question

If a gaseous mixture is made by combining 3.38 g of Ar and 1.14 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?

PAr=?

PKr=?

Ptotal=/

Answer 1

Lets first calculate the moles of each gas

Moles of Ar =3.38 g /39,95 g = 0.0846 moles of Ar

Moles of Kr = 1.14 g/ 83.80g = 0.0136 moles of Kr

Now we have the formula

PV = nRT

R = universal gas constant = 0.0821 atm.L/mol.K

so

Partial Pressure

P_Ar = nRT/V

= {0.0846 moles 0.0821 atm.L/mol.K (273.15+25) K} / 2.50 L

= 2.0708/ 2.5

= 0.8283 atm

P_Ar = 0.8283 atm

Also for

P_Kr = nRT/V

= 0.0136 moles 0.0821 atm.L/mol.K (273.15+25) K} / 2.50 L

= 0.3329 / 2.5

= 0.1331 atm

P_Kr = 0.1331 atm

So

P_total = P_{Ar +} P_Kr

   =0.8283 atm + 0.1331 atm

P_total = 0.9614 atm