Aclosed 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

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

at the bottom

explanation:

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.