- #1

dancingdodo27

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## 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

_{vap}H/ T

ΔS= q/T

ΔS= nC

_{v}ln(Tf/Ti)

## The Attempt at a Solution

For entropy of vaporisation at 87.5 K

Δ

_{vap}H= 6.5 kJ mol

^{-1}for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol

^{-1}

Using Trouton's Rule ΔS

_{1}= 32500/87.5 = 371.43 J K

_{-1}

Entropy from 87.5 K to 150 K

ΔS

_{2}= 5⋅ 3/2R⋅ ln(150/87.5)

ΔS

_{2}= 33.61 J K

^{-1}

Total entropy change= ΔS

_{sys}= ΔS

_{1}+ ΔS

_{2}

ΔS

_{sys}= 371.43 + 33.61 = 405.04 J K

^{-1}