Aclosed vessel of volume 2.6 l contains a mixture of neon and fluorine gas. the total pressure of the mixture in the vessel is 3.0 atm at 0.0 °c. when the mixture is heated to 14.0 °c, the entropy of the mixture increases by 0.30 j ⋅ k⁻¹. calculate the number of moles of neon and fluorine in the mixture.

The number of moles neon in the mixture is 0.148 moles

The number of fluorine in the mixture is 0.200 moles

Explanation:

Step 1: Data given

Volume = 2.6 L

Total pressure = 3.0 atm at 0.0 °C

When the mixture is heated to 14.0 °C, the entropy of the mixture increases by 0.30 J/K

Step 2: Calculate total moles

Neon is a monoatomic gas, composed of Ne atoms.

Fluorine is a diatomic gas, composed of F₂ molecules.

The heat capacities are different because F₂ has rotational degrees of freedom and for that reason, Cv(F₂) ≅ (5/2)R while Cv(Ne) ≅ (3/2)R.

n(total) = PV/RT = (3.00 atm)(2.6 L)/(0.08206 L-atm/mol*K)(273.15) = 0.348 mol

For one mole heated at constant volume,  

Step 3: Define ∆S

∆S =  Cv*ln(287.15/273.15) = 0.04998*Cv

So, for 0.348 mol,  

∆S = (0.348 mol)(0.04998)Cv = 0.30 J/K

⇒ Cv = 17.25 J/mol*K for the Ne/F₂ mixture.

Step 4: Calculate Cv

For pure Neon (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

Step 5: Calculate moles of Ne and F2

if X is the mole fraction of Ne, we can find X by setting the observed ∆S to the weighted average of the ∆S's expected for the two different gases in the mixture:

17.25 J/mol*K = X * 12.471 J/mol*K + (1 – X) * 20.785 J/mol•K

17.25 = x12.471 + 20.785 - 20.785x

-3.535 = -8.314x

x = 0.425

1 - x = 0.575

moles Ne = (0.425)(0.348 mol) = 0.148 mol

moles F₂ = (0.575)(0.348 mol) = 0.200 mol

The number of moles neon in the mixture is 0.148 moles

The number of fluorine in the mixture is 0.200 moles

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol