Calculate the temperature in centrigrade when vaporisation of water in equilibrium at one atmospheric pressure ( Enthalpy of vaporisation `= 40 . 63 xx 10^(3) J mol^(-1) , Delta_(vap ) S= 108.8 J K^(-1) mol^(-1))`

Answer

To calculate the temperature in degrees Celsius at which the vaporization of water is in equilibrium at one atmospheric pressure, we can use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). At equilibrium, ΔG = 0.

The equation we will use is:

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

At equilibrium, ΔG = 0, so we can rearrange the equation to find the temperature (T):

\[ 0 = \Delta H - T \Delta S \]

This simplifies to:

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

### Step 1: Convert the enthalpy of vaporization from kJ/mol to J/mol
Given:
- ΔH (enthalpy of vaporization) = 40.63 x 10³ J/mol

This value is already in J/mol, so we can use it directly.

### Step 2: Use the given entropy of vaporization
Given:
- ΔS (entropy of vaporization) = 108.8 J K⁻¹ mol⁻¹

### Step 3: Substitute the values into the temperature equation
Now we can substitute the values of ΔH and ΔS into the equation:

\[ T = \frac{40.63 \times 10^3 \, \text{J/mol}}{108.8 \, \text{J K}^{-1} \text{mol}^{-1}} \]

### Step 4: Calculate the temperature in Kelvin
Calculating the above expression:

\[ T = \frac{40630}{108.8} \approx 373.0 \, \text{K} \]

### Step 5: Convert the temperature from Kelvin to Celsius
To convert from Kelvin to Celsius, use the formula:

\[ T(°C) = T(K) - 273.15 \]

Substituting the value we found:

\[ T(°C) = 373.0 - 273.15 \approx 99.85 \, °C \]

### Final Answer:
The temperature in degrees Celsius when vaporization of water is in equilibrium at one atmospheric pressure is approximately **99.85 °C**.

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