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:

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