Entropy of Vaporization Calculation for 5 Moles of Argon at 87.5 K and 150 K

[dancingdodo27]

Homework Statement

5 moles of liquid argon undergoes vaporization at its normal boiling point (87.5 K) and the resulting argon gas is subsequently heated to 150 K under constant volume conditions. Calculate the change of entropy for this process. The standard enthalpy change of vaporization, ∆ ⊖ is +6.5 kJ mol-1 for argon. Treat gaseous argon as an ideal gas in your calculations.

Homework Equations

ΔS= Δ_vapH/ T
ΔS= q/T
ΔS= nC_vln(Tf/Ti)

The Attempt at a Solution

For entropy of vaporisation at 87.5 K
Δ_vapH= 6.5 kJ mol^-1 for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol^-1
Using Trouton's Rule ΔS₁= 32500/87.5 = 371.43 J K_-1

Entropy from 87.5 K to 150 K
ΔS₂= 5⋅ 3/2R⋅ ln(150/87.5)
ΔS₂= 33.61 J K^-1

Total entropy change= ΔS_sys= ΔS₁ + ΔS₂
ΔS_sys= 371.43 + 33.61 = 405.04 J K^-1

 

[Chestermiller]

Homework Statement

5 moles of liquid argon undergoes vaporization at its normal boiling point (87.5 K) and the resulting argon gas is subsequently heated to 150 K under constant volume conditions. Calculate the change of entropy for this process. The standard enthalpy change of vaporization, ∆ ⊖ is +6.5 kJ mol-1 for argon. Treat gaseous argon as an ideal gas in your calculations.

Homework Equations

ΔS= Δ_vapH/ T
ΔS= q/T
ΔS= nC_vln(Tf/Ti)

The Attempt at a Solution

For entropy of vaporisation at 87.5 K
Δ_vapH= 6.5 kJ mol^-1 for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol^-1
Using Trouton's Rule ΔS₁= 32500/87.5 = 371.43 J K_-1

Entropy from 87.5 K to 150 K
ΔS₂= 5⋅ 3/2R⋅ ln(150/87.5)
ΔS₂= 33.61 J K^-1

Total entropy change= ΔS_sys= ΔS₁ + ΔS₂
ΔS_sys= 371.43 + 33.61 = 405.04 J K^-1

Your methodology looks correct. I haven't checked your arithmetic.