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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= ΔvapH/ T
ΔS= q/T
ΔS= nCvln(Tf/Ti)
The Attempt at a Solution
For entropy of vaporisation at 87.5 K
ΔvapH= 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= ΔSsys= ΔS1 + ΔS2
ΔSsys= 371.43 + 33.61 = 405.04 J K-1