The enthalpy of vapourisation of water is 186.5 KJ/mol,then what is it's entropy of vapourisation

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Solution

When water change from liquid to vapour state we will calculate enthalpy of vapourisation.

Water boiling point T = 100 degree celcius = 373K

At boiling point or during vaporisation

the reversible process liquid ⇋vapour is in equilibrium at one atmospheric pressure.

∴ΔG=0

ΔG = ΔH−TΔS

ΔS = ΔH/T

ΔH = 186.5 KJ/Mol

T =373K

So ΔS = 186.5/373 KJ/(MolK)

ΔS = 0.5 KJ/(MolK)

Answer : Entropy of vapourisation = 0.5KJ/(Mol*K)

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Enthalpy of vaporization for water is 186.5 kJ mol^-1.

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