Question

A mixture containing 0.765 mol He(g), 0.243 mol Ne(g), and 0.118 mol Ar(g) is confined in a 10.00-L vessel at 25 C

A. Calculate the partial pressure of He in the mixture.

B. Calculate the partial pressure of Ne in the mixture.

C. Calculate the partial pressure of Ar in the mixture.

D. Calculate the total pressure of the mixture.

Answer 1

The total pressure of the mixture is given by the sum of partial pressures of each individual gas (Dalton’s Law)

P_total = P_He +P_Ne +P_Ar

according to ideal gas equation PV = NRT

P =NRT/V

Total moles of gas = 0.765 + 0.243 +0.118 = 1.126

Total pressure =1.126 mol * 0.08206 L atm mol^-1 K^-1 * 298 K / 10 L

=2.7535 atm

The partial pressure of individual gas within the overall mixture

Pi = P_total x_i

where x_i  is the mole fraction , x_i = no. of moles of i / total no. of moles gas

A)  

Partial pressure of He, P_He = Ptotal *X_He

=2.7535atm * 0.765 / 1.126 = 1.8707 atm

B)

  

Partial pressure of Ne, P_Ne = Ptotal *X_Ne

=2.7535atm * 0.243 / 1.126 = 0.5942 atm

c)

Partial pressure of Ar, P_Ar = Ptotal *XAr

=2.7535atm * 0.118   / 1.126 = 0.2885 atm

D)

Total pressure =1.126 mol * 0.08206 L atm mol^-1 K^-1 * 298 K / 10 L

=2.7535 atm