Answer:
There are 0.14911 moles of neon and 0.22089 moles of fluorine
Explanation:
Denoting 1 as initial state , 2 as final state and assuming ideal gas behaviour for neon and fluorine, then for the initial state we can apply:
P*V= n*R*T
n = P*V/(R*T)
where
P₁=absolute pressure= 3.32 atm
V=volume=2.5 L (constant)
n=number of moles
R= ideal gas constant =0.082 atm*L/mol K
T₁= absolute temperature =0°C= 273K
replacing values
n = P₁*V/(R*T₁) = 3.32 atm*2.5 L/(0.082 atm*L/mol K*273 K) = 0.37 moles of gas mixture
then for a constant volume process, from the second law of thermodynamics and assuming reversible process:
dQ= n*cv*dT
ΔS = ∫dQ/T = ∫ n*cv*dT/T = n*cv*ln(T₂/T₁)
therefore since T₂=15°C = 288 K
ΔS = n*Cv*ln(T₂/T₁) → Cv= ΔS /[n*ln(T₂/T₁)] = 0.345 J/K/( 0.37 mol* ln( 288 K/ 273 K) = 17.432 J/mol K
then denoting n as neon and f as fluorine
Cv = ∑ xi Cvi = xn*Cvn + xf Cvf = 3/2R * xn + 5/2R*(1- xn) = 5/2R - R*xn
therefore since also R=8.314 J/mol K
xn = (5/2R - Cv)/R = 5/2-Cv/R = 5/2 - 17.432 J/mol K/(8.314 J/mol K) = 0.403
therefore
n neon = n*xn = 0.37 mol* 0.403 = 0.14911 moles of neon
n fluorine = n - n neon = 0.37 mol - 0.14911 mol = 0.22089 moles of fluorine
