The normal boiling point of watr is 373 K (at 760 mm), vapour pressure of water at 298 K is 23 mm. If entyalpy of vaporization is `40.656` kj//mol.the boiling point of water at 23 mm pressure will be :

Options

A. 250 K
B. `294.4 K`
C. `51.6 K`
D. `12.5 K`

Explanation

To find the boiling point of water at a pressure of 23 mm Hg using the Clausius-Clapeyron equation, we can follow these steps:

### Step 1: Write down the Clausius-Clapeyron equation
The Clausius-Clapeyron equation relates the change in vapor pressure with temperature to the enthalpy of vaporization. It is given by:

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_v}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Where:
- \( P_1 \) = vapor pressure at temperature \( T_1 \)
- \( P_2 \) = vapor pressure at temperature \( T_2 \)
- \( \Delta H_v \) = enthalpy of vaporization
- \( R \) = universal gas constant
- \( T_1 \) = temperature corresponding to \( P_1 \)
- \( T_2 \) = temperature corresponding to \( P_2 \)

### Step 2: Identify the known values
From the problem statement:
- Normal boiling point of water \( T_2 = 373 \, K \) (at \( P_2 = 760 \, mm \, Hg \))
- Vapor pressure of water at \( T_1 = 298 \, K \) is \( P_1 = 23 \, mm \, Hg \)
- Enthalpy of vaporization \( \Delta H_v = 40.656 \, kJ/mol = 40656 \, J/mol \)
- Universal gas constant \( R = 8.314 \, J/(K \cdot mol) \)

### Step 3: Rearrange the equation to solve for \( T_1 \)
We need to rearrange the Clausius-Clapeyron equation to solve for \( T_1 \):

\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_v}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Substituting the known values:

\[ \ln\left(\frac{760}{23}\right) = -\frac{40656}{8.314} \left(\frac{1}{373} - \frac{1}{T_1}\right) \]

### Step 4: Calculate \( \ln\left(\frac{P_2}{P_1}\right) \)
Calculating the left side:

\[ \frac{760}{23} \approx 33.04 \]
\[ \ln(33.04) \approx 3.496 \]

### Step 5: Substitute and solve for \( T_1 \)
Now substituting back into the equation:

\[ 3.496 = -\frac{40656}{8.314} \left(\frac{1}{373} - \frac{1}{T_1}\right) \]

Calculating the right side:

\[ -\frac{40656}{8.314} \approx -4895.4 \]

So we have:

\[ 3.496 = 4895.4 \left(\frac{1}{373} - \frac{1}{T_1}\right) \]

### Step 6: Isolate \( \frac{1}{T_1} \)
Now, isolate \( \frac{1}{T_1} \):

\[ \frac{1}{373} - \frac{1}{T_1} = \frac{3.496}{4895.4} \]

Calculating the right side:

\[ \frac{3.496}{4895.4} \approx 0.000714 \]

So,

\[ \frac{1}{T_1} = \frac{1}{373} - 0.000714 \]

Calculating \( \frac{1}{373} \):

\[ \frac{1}{373} \approx 0.002684 \]

Now substituting:

\[ \frac{1}{T_1} = 0.002684 - 0.000714 \approx 0.00197 \]

### Step 7: Calculate \( T_1 \)
Finally, take the reciprocal to find \( T_1 \):

\[ T_1 \approx \frac{1}{0.00197} \approx 507.6 \, K \]

### Conclusion
The boiling point of water at 23 mm pressure is approximately \( 507.6 \, K \).