Question

A closed, rigid 2.5 L flask contains a mixture of Ne(g) and F2(g) with a total pressure of 3.32 atm at 0oC. The system is heated to 15oC. The change in entropy for this process was 0.345 J/K. What is the XNe in the flask

Answer 1

Explanation:

The heat capacities of Ne and F₂ are:

Cv(F₂) ≅ (5/2)R

Cv(Ne) ≅ (3/2)R.

Using ideal gas equation,

PV = nRT

n(total) = PV/RT

= (3.32 * 2.5)/(0.08206 × 273.15)

= 0.3703 mol

For one mole heated at constant volume, the entropy change is:

∆S = ∫dq/T

= ∫(Cv/T)dT from T1 to T2 = Cvln(T2/T1) = Cvln(288.15/273.15)

= 0.05346•Cv

So, for 0.3703 mol,

∆S = (0.3703 × 0.05346)Cv

(0.3703 × 0.05346)Cv = 0.345

Cv = 17.43 J/mol-K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R

= 2.5 × 8.314 J/mol•K

= 12.471 J/mol•K

For pure F₂, Cv = (5/2)R

= 2.5 • 8.314 J/mol•K

= 20.785 J/mol•K

Cv(Ne/F₂) = Cv(Ne) + Cv(Ne/F₂)

17.43 = X × 12.471 + (1 – X) × 20.785

(20.875 – 8.314) × X = 17.43

X = 0.415

1 – X = 0.585

moles Ne = (0.415 × 0.3703 mol)

= 0.154 mol