Answer : The entropy change of the system is, 19.5 J/K
Solution :
First we have to calculate the moles of benzene.
[tex]\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=\frac{17.5g}{78.11g/mole}=0.224moles[/tex]
Now we have to calculate the entropy change of the system.
Formula used :
[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S[/tex] = entropy change of the system = ?
[tex]\Delta H[/tex] = enthalpy of vaporization = 30.7 kJ/mole
n = number of moles of benzene = 0.224 mole
[tex]T_b[/tex] = normal boiling point of benzene = [tex]80.1^oC=273+80.1=353.1K[/tex]
Now put all the given values in the above formula, we get the entropy change of the system.
[tex]\Delta S=\frac{0.224mole\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K[/tex]
Therefore, the entropy change of the system is, 19.5 J/K
