The enthalpy of vaporization of a substance is 840 J per mol and its boiling point is `-173^@C`. Calculate its entropy of vaporization.

Options

A. `8.4 "Jmol"^(-1)K^(-1)`
B. `49 "Jmol"^(-1)K^(-1)`
C. `21 "Jmol"^(-1)K^(-1)`
D. `12 "Jmol"^(-1)K^(-1)`

Answer

To calculate the entropy of vaporization (ΔS) of a substance given its enthalpy of vaporization (ΔH) and boiling point, we can use the relationship between these quantities. The formula we will use is:

\[ \Delta S = \frac{\Delta H}{T} \]

where:
- ΔS = entropy of vaporization
- ΔH = enthalpy of vaporization
- T = temperature in Kelvin

### Step 1: Convert the boiling point from Celsius to Kelvin
The boiling point is given as -173°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating this gives:

\[ T = -173 + 273.15 = 100.15 \text{ K} \approx 100 \text{ K} \]

### Step 2: Use the enthalpy of vaporization
The enthalpy of vaporization (ΔH) is given as 840 J/mol.

### Step 3: Substitute the values into the entropy formula
Now we can substitute ΔH and T into the formula for ΔS:

\[ \Delta S = \frac{\Delta H}{T} = \frac{840 \text{ J/mol}}{100 \text{ K}} \]

### Step 4: Calculate ΔS
Perform the calculation:

\[ \Delta S = \frac{840}{100} = 8.4 \text{ J/K/mol} \]

### Conclusion
The entropy of vaporization of the substance is 8.4 J/K/mol.

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