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) given by the equation:

\[ \Delta H = T \Delta S \]

Where:
- ΔH = Enthalpy of vaporization (in joules)
- T = Temperature (in Kelvin)
- ΔS = Entropy of vaporization (in joules per Kelvin)

### Step-by-Step Solution:

1. **Convert Enthalpy of Vaporization to Joules**:
Given that the enthalpy of vaporization is \(30 \, \text{kJ mol}^{-1}\), we need to convert this to joules:
\[ \Delta H = 30 \, \text{kJ mol}^{-1} = 30 \times 1000 \, \text{J mol}^{-1} = 30000 \, \text{J mol}^{-1} \]

2. **Use the Entropy of Vaporization**:
The entropy of vaporization is given as \(75 \, \text{J mol}^{-1} \text{K}^{-1}\):
\[ \Delta S = 75 \, \text{J mol}^{-1} \text{K}^{-1} \]

3. **Rearrange the Equation to Solve for Temperature (T)**:
We can rearrange the equation \( \Delta H = T \Delta S \) to solve for T:
\[ T = \frac{\Delta H}{\Delta S} \]

4. **Substitute the Values into the Equation**:
Substitute the values of ΔH and ΔS into the equation:
\[ T = \frac{30000 \, \text{J mol}^{-1}}{75 \, \text{J mol}^{-1} \text{K}^{-1}} \]

5. **Calculate the Temperature**:
Now perform the division:
\[ T = \frac{30000}{75} = 400 \, \text{K} \]

### Final Answer:
The boiling point of the liquid at 1 atm is **400 K**.

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