Answer: The entropy change of the tetrahydrofuran is 70.8 J/K
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of tetrahydrofuran = 8.2 g
Molar mass of tetrahydrofuran = 72 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of tetrahydrofuran}=\frac{8.2g}{72g/mol}=0.114mol[/tex]
To calculate the entropy change for different phase at same temperature, we use the equation:
[tex]\Delta S=n\times \frac{\Delta H_{vap}}{T}[/tex]
where,
[tex]\Delta S[/tex] = Entropy change = ?
n = moles of tetrahydrofuran = 0.114 moles
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporization = 32.0 kJ/mol = 32000 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = [tex]66.0^oC=[66+273]K=339K[/tex]
Putting values in above equation, we get:
[tex]\Delta S=\frac{0.114mol\times 32000J/mol}{339K}\\\\\Delta S=10.8J/K[/tex]
Hence, the entropy change of the tetrahydrofuran is 70.8 J/K
