The enthalpy of vaporisation of a liquid is `30 kJ mol^(-1)` and entropy of vaporisation is `75 J mol^(-1) K^(-1)`. The boiling point of the liquid at `1atm` is :

Options

A. 250 K
B. 400 K
C. 450 K
D. 600 K

Explanation

To find the boiling point of the liquid at 1 atm, we can use the relationship between enthalpy of vaporization (ΔH) and entropy of vaporization (ΔS) at equilibrium, which is given by the Gibbs free energy equation:

1. **Identify the Given Values:**
- Enthalpy of vaporization, ΔH = 30 kJ/mol
- Entropy of vaporization, ΔS = 75 J/mol·K

2. **Convert Units:**
Since ΔH is given in kilojoules and ΔS in joules, we need to convert ΔH into joules for consistency:
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