Respuesta :
Answer:
moles Ne = 0.154 mol
moles F₂ = 0.217 mol
Explanation:
Step 1: Data given
Volume of the vessel system = 2.5 L
Total pressure = 3.32 atm at 0.0 °C
The mixture is heated to 15.0 °C
The entropy of the mixture increases by 0.345 J/K
The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R
Step 2: Define the gas
Neon is a monoatomic gas, composed of Ne atoms
⇒ Cv(Ne) ≅ (3/2)R
Fluorine is a diatomic gas, composed of F₂ molecules.
⇒ Cv(F₂) ≅ (5/2)R
Step 3: Calculate moles of gas
p*V = n*R*T
⇒ with p = the total pressure = 3.32 atm
⇒ with V = the total volume = 2.5 L
⇒ with n = the number of moles of gas
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 273.15 Kelvin
n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol
Step 4: Calculate moles of Ne and F2
For one mole heated at constant volume,
∆S = Cv*ln(288.15/273.15) = 0.05346*Cv
⇒ ∆S for 0.3703 mol,
∆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 X is the mole fraction of Ne, we can find X by:
17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K
⇒ 20.875 – 8.314 * X = 17.43
X = 0.415 , 1 – X = 0.585
moles Ne = (0.415)(0.3703 mol) = 0.154 mol
moles F₂ = (0.585)(0.3703 mol) = 0.217 mol
There are 0.221 moles of fluorine and 0.150 moles of neon in the closed vessel.
How to determine the amount of moles in a closed vessel
In this question we must apply the concepts of entropy ([tex]S[/tex]), in joules per Kelvin and internal energy ([tex]U[/tex]), in joules. Let suppose that each gas behaves ideally. First, we need to determine the amount of moles ([tex]n[/tex]), in moles, inside the vessel:
[tex]n = \frac{P\cdot V}{R_{u}\cdot T}[/tex] (1)
Where:
- [tex]V[/tex] - Volume of the vessel, in liters.
- [tex]T[/tex] - Temperature, in Kelvin.
- [tex]P[/tex] - Pressure, in atmospheres.
- [tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.
If we know that [tex]R = 0.082\,\frac{atm\cdot L}{mol\cdot K}[/tex], [tex]P = 3.32\,atm[/tex], [tex]V = 2.5\,L[/tex] and [tex]T = 273.15\,K[/tex], then the amount of moles is:
[tex]n = \frac{(3.32\,atm)\cdot (2.5\,L)}{\left(0.082\,\frac{atm\cdot L}{\mol\cdot K} \right)\cdot (273.15\,K)}[/tex]
[tex]n = 0.371\,mol[/tex]
Then, the total entropy of the system ([tex]S[/tex]) is represented by the following equation: (Please notice that fluorine is a diatomic gas and neon is a monoatomic gas):
[tex]S = \left[\frac{3\cdot (n-x)\cdot R_{u}}{2}+\frac{5\cdot x\cdot R_{u}}{2} \right]\cdot \ln \frac{T_{2}}{T_{1}}[/tex] (2)
Where:
- [tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, in Kelvin.
- [tex]R_{u}[/tex] - Ideal gas constant, in joules per mole-Kelvin.
- [tex]n[/tex] - Amount of moles of the system, in moles.
- [tex]x[/tex] - Amount of moles of fluorine, in moles.
If we know that [tex]S = 0.345\,\frac{J}{K}[/tex], [tex]n = 0.371\,mol[/tex], [tex]R_{u} = 8.371\,\frac{J}{mol\cdot K}[/tex], [tex]T_{1} = 273.15\,K[/tex] and [tex]T_{2} = 288.15\,K[/tex], then the amount of moles of fluorine is:
[tex]0.345 = \left[\frac{3\cdot (0.371-x)\cdot (8.371)}{2} + \frac{5\cdot x\cdot (8.371)}{2} \right]\cdot \ln \frac{288.15}{273.15}[/tex]
[tex](4.658-12.557\cdot x + 20.928\cdot x)\cdot 0.053 = 0.345[/tex]
[tex](4.658+8.371\cdot x)\cdot 0.053 = 0.345[/tex]
[tex]4.658 + 8.371\cdot x = 6.509[/tex]
[tex]8.371\cdot x = 1.851[/tex]
[tex]x = 0.221[/tex]
There are 0.221 moles of fluorine and 0.150 moles of neon in the closed vessel. [tex]\blacksquare[/tex]
To learn more on entropy, we kindly invite to check this verified question: https://brainly.com/question/15025401
