Question

A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When the mixture is heated to 15°C, the entropy of the mixture increases by 0.345 J/K. What amount (in moles) of each substance (Ne and F2) is present in the mixture? (heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R)

Answer 1

Answer: moles of Ne = 0.149 moles

moles of F₂ = 0.221 moles

Step-by-step explanation:

The process occurs at costant volume.

Neon is a monoatomic gas, Cv =3/2R, Flourine is a diatomic gas, Cv = 5/2R

n = PV/RT ; where n is total = number of moles, P is total pressure = 3.32atm, V is volume = 2.5L, R is molar gas constant = 0.08206 L-atm/mol/K = 8.314 J/mol/K, T is temperature = 0.0°C = 273.15K

n(total) = (3.32 atm)(2.5 L)/(0.08206 L-atm/mol/K)(273.15) = 0.3703 mol

For one mole heated at constant volume,

Change in entropy, ∆S = ∫dq/T = ∫(Cv/T)dT

From T1 to T2, Cvln(T2/T2) = Cvln(288.15/273.15) = 0.05346•Cv

So, for 0.3703 moles,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

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

For pure Ne, Cv = (3/2)R = 1.5 • 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

If Y is the mole fraction of Ne, Y can be determined by setting the observed entropy change (∆S) to the weighted average of the entropy changes expected for the two different gases in the mixture:

17.43 J/mol/K = Y * 12.471 J/mol/K + (1 – Y) * 20.785 J/mol/K

⇒ 20.785 J/mol/K – 8.314 J/mol/K * Y = 17.43J/mol/K

Y = 0.403 ; 1 – Y = 0.597

moles of Ne = (0.403)(0.3703 mol) = 0.149 moles

moles of F₂ = (0.597)(0.3703 mol) = 0.221 moles