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The molar heats of fusion and vaporization of argon are 1.3 kJ/mol and 6.3 kJ/mol respectively, and argon's melting point and boiling point are -190°C and -186 °C, respectively. Calculate the entropy changes for the fusion and vaporization of argon.

User Jspurim
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Answer:

The entropy changes for the fusion and vaporization of argon is 15.634 J/mol K and 72.289 J/mol K respectively.

Step-by-step explanation:

The molar heats of fusion =
\Delta H_(fus)= 1.3 kJ/mol

Melting point of argon = -190°C = 83.15 K

Entropy changes for the fusion =
\Delta S_(fus)


\Delta S_(fus)=(\Delta H_(fus))/(83.15 K)=(1.3 kJ/mol)/(83.15 K)=0.015634 kJ/mol K=15.634 J/mol K

The molar heats of vaporization of argon =
\Delta H_(vap)=6.3 kJ/mol

Boiling point of argon = -186°C = 87.15 K

Entropy changes for the vaporization=
\Delta S_(vap)


\Delta S_(vap)=(\Delta H_(vap))/(87.15K)=(6.3 kJ/mol)/(87.15 K)=0.072289kJ/mol K=72.289 J/mol K

User Bachposer
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