The normal boiling point of water is 373 k. Vapour pressure of water at temperature T is 19 mm hg. If enthalpy of vapourization is 40.67 kJ/mol, then temperature T would be
(use : log 2 = 0.3, R : 8.3` Jk^(-1)mol^(-1)`):

Options

A. 250 K
B. 291.4 K
C. 230 K
D. 290 K

Answer

To solve the problem, we will use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization. Here are the steps:

### Step 1: Write down the Clausius-Clapeyron equation
The Clausius-Clapeyron equation is given by:

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

Where:
- \( P_1 \) = vapor pressure at temperature \( T_1 \) (normal boiling point)
- \( P_2 \) = vapor pressure at temperature \( T_2 \)
- \( \Delta H_{vap} \) = enthalpy of vaporization
- \( R \) = universal gas constant
- \( T_1 \) = normal boiling point in Kelvin
- \( T_2 \) = temperature we need to find in Kelvin

### Step 2: Identify the values from the problem
From the problem, we have:
- \( T_1 = 373 \, \text{K} \) (normal boiling point of water)
- \( P_1 = 760 \, \text{mmHg} \) (atmospheric pressure)
- \( P_2 = 19 \, \text{mmHg} \) (vapor pressure at temperature \( T \))
- \( \Delta H_{vap} = 40.67 \, \text{kJ/mol} = 40670 \, \text{J/mol} \) (conversion to Joules)
- \( R = 8.3 \, \text{J/(K mol)} \)

### Step 3: Substitute the values into the equation
Substituting the known values into the Clausius-Clapeyron equation:

\[ \ln \left( \frac{19}{760} \right) = -\frac{40670}{8.3} \left( \frac{1}{T_2} - \frac{1}{373} \right) \]

### Step 4: Calculate \( \ln \left( \frac{19}{760} \right) \)
Calculating the left side:

\[ \frac{19}{760} \approx 0.025 \]
\[ \ln(0.025) \approx -3.688 \]

### Step 5: Calculate the right side
Now, we calculate the right side:

\[ -\frac{40670}{8.3} \approx -4896.39 \]

### Step 6: Set up the equation
Now we can set up the equation:

\[ -3.688 = -4896.39 \left( \frac{1}{T_2} - \frac{1}{373} \right) \]

### Step 7: Solve for \( \frac{1}{T_2} \)
Rearranging gives:

\[ \frac{1}{T_2} - \frac{1}{373} = \frac{-3.688}{-4896.39} \]

Calculating the right side:

\[ \frac{3.688}{4896.39} \approx 0.000752 \]

So we have:

\[ \frac{1}{T_2} = 0.000752 + \frac{1}{373} \]

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

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

Adding these gives:

\[ \frac{1}{T_2} \approx 0.002684 + 0.000752 \approx 0.003436 \]

### Step 8: Find \( T_2 \)
Taking the reciprocal:

\[ T_2 \approx \frac{1}{0.003436} \approx 291.4 \, \text{K} \]

### Conclusion
Thus, the temperature \( T \) is approximately \( 291.4 \, \text{K} \).