Answer:
The entropy changes for the fusion and vaporization of argon is 15.634 J/mol K and 72.289 J/mol K respectively.
Explanation:
The molar heats of fusion =[tex]\Delta H_{fus}= 1.3 kJ/mol[/tex]
Melting point of argon = -190°C = 83.15 K
Entropy changes for the fusion = [tex]\Delta S_{fus}[/tex]
[tex]\Delta S_{fus}=\frac{\Delta H_{fus}}{83.15 K}=\frac{1.3 kJ/mol}{83.15 K}=0.015634 kJ/mol K=15.634 J/mol K[/tex]
The molar heats of vaporization of argon = [tex]\Delta H_{vap}=6.3 kJ/mol[/tex]
Boiling point of argon = -186°C = 87.15 K
Entropy changes for the vaporization= [tex]\Delta S_{vap}[/tex]
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{87.15K}=\frac{6.3 kJ/mol}{87.15 K}=0.072289kJ/mol K=72.289 J/mol K[/tex]
