Question

If a gaseous mixture is made by combining 3.17 g Ar 3.17 g Ar and 1.30 g Kr 1.30 g Kr in an evacuated 2.50 L container at 25.0 ∘ C, 25.0 ∘C, what are the partial pressures of each gas, P Ar PAr and P Kr , PKr, and what is the total pressure, P total , Ptotal, exerted by the gaseous mixture?

Answer 1

The partial pressures of argon is 0.77 atm and krypton is 0.16 atm.

The total pressure exerted by the gaseous mixture is 0.93 atm

Step-by-step explanation:

Mass of argon gas =
`3.17 g`

Mass of krypton gas =
`1.30 g`

Mixture in terms of mole fractions:

Moles of argon gas =
`n_1=(3.17 g)/(40g/mol)=0.079mol`

Moles of krypton gas =
`n_2=(1.30 g)/(84 g/mol)=0.016 mol`

Mole fraction of argon gas =

=
`\chi_1=(n_1 )/(n_1+n_2)=(0.079 mol)/(0.079mol+ 0.016mol)=0.83`

Mole fraction of carbon dioxide gas =

=
`\chi_2=(n_2 )/(n_1+n_2)=(0.016 mol)/(0.079 mol+ 0.016mol)=0.17`

Pressure of the exerted by the gaseous mixture= P

Temperature of the mixture = T = 25.0 °C =25.0+273K=298.0 K

Volume of the container in which mixture is kept =
`V=2.50 L`

Moles of gases,n =
`n_1+n_2=0.079 mol + 0.016 mol= 0.095 mol`

PV=nRT

`P* 2.50L=0.095 mol* 0.0821 atm L/mol K* 298.0 K`

`P=(0.095 mol* 0.0821 atm L/mol K* 298.0 K)/(2.50 L)=0.93 atm`

Partial pressure of argon gas =
`p_1`

Partial pressure of krypton gas =
`p_2`

`p_1=P* \chi_1=0.93 atm* 0.83=0.77 atm`

`p_2=P* \chi_2=0.93 atm * 0.17 =0.16 atm`

The partial pressures of argon is 0.77 atm and krypton is 0.16 atm.

The total pressure exerted by the gaseous mixture is 0.93 atm