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The entropy of vaporization of benzene is 85/(K mol) at 100 K.What is the enthalpy of vaparisation in kJ Explain
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The entropy of vaporization of benzene is 85/(K mol) at 100 K.What is ...
Enthalpy of Vaporization of Benzene

The enthalpy of vaporization, also known as the heat of vaporization, is the amount of heat required to convert a substance from a liquid phase to a gaseous phase at a constant temperature and pressure. In this case, we need to determine the enthalpy of vaporization of benzene.

Given Information:
Entropy of vaporization (ΔSvap) = 85 J/(K mol) at 100 K

To calculate the enthalpy of vaporization (ΔHvap), we can use the following equation:

ΔHvap = T * ΔSvap

where ΔHvap is the enthalpy of vaporization, T is the temperature in Kelvin, and ΔSvap is the entropy of vaporization.

Conversion of Units:
The given entropy of vaporization is in J/(K mol), but we need to convert it to kJ/(K mol) for consistency with the enthalpy unit.

85 J/(K mol) = 0.085 kJ/(K mol)

Calculation:
Now, we can calculate the enthalpy of vaporization using the given information:

ΔHvap = T * ΔSvap
= 100 K * 0.085 kJ/(K mol)
= 8.5 kJ/mol

Therefore, the enthalpy of vaporization of benzene is 8.5 kJ/mol.

Explanation:
The enthalpy of vaporization represents the energy required to break intermolecular forces and convert a substance from the liquid phase to the gaseous phase. It is related to the entropy of vaporization, which describes the degree of disorder or randomness in the system during the phase change.

In this case, benzene is being vaporized at a temperature of 100 K. The given entropy of vaporization, 85 J/(K mol), indicates that there is a high degree of disorder or randomness in the system during the vaporization process. This means that the benzene molecules are becoming more dispersed and moving freely in the gaseous phase.

The enthalpy of vaporization is calculated by multiplying the temperature by the entropy of vaporization. The resulting value, 8.5 kJ/mol, represents the heat energy required to vaporize one mole of benzene at 100 K.

Conclusion:
The enthalpy of vaporization of benzene at 100 K is 8.5 kJ/mol. This value indicates the amount of heat energy required to convert liquid benzene into gaseous benzene at a constant temperature and pressure.
Community Answer
The entropy of vaporization of benzene is 85/(K mol) at 100 K.What is ...
8.5 kJ
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