Answer:
375 K
Explanation:
Using the experssion shown below as:
[tex]\Delta G^0=\Delta H^0_{vap}-T\Delta S^0_{vap}[/tex]
At vaporization point, the liquid and the gaseous phase is in the equilibrium.
Thus,
[tex]\Delta G^0=0[/tex]
So,
[tex]Delta H^0_{vap}=T\Delta S^0_{vap}[/tex]
Given that:
[tex]Delta H^0_{vap}=55.5\ kJ/mol[/tex]
Also, 1 kJ = 10³ J
So,
[tex]Delta H^0_{vap}=55500\ J/mol[/tex]
[tex]\Delta S^0_{vap}=148\ J/K.mol[/tex]
So, temperature is :
[tex]T=\frac{Delta H^0_{vap}}{\Delta S^0_{vap}}[/tex]
[tex]T=\frac{55500}{148}[/tex]
T= 375 K
