Calculate the boiling point of water at `700mm` pressure of `Hg`. The heat of vaporisation of water is `540 "cal' g`.

Explanation

To calculate the boiling point of water at a pressure of 700 mm of Hg, we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the heat of vaporization. Here are the steps to solve the problem:

### Step 1: Understand the given data
- Heat of vaporization of water (ΔH) = 540 cal/g
- Molar mass of water = 18 g/mol
- Pressure at boiling point (P1) = 700 mm Hg
- Standard boiling point of water (P2) = 760 mm Hg (at 1 atm)

### Step 2: Convert ΔH to per mole
Since the heat of vaporization is given per gram, we need to convert it to per mole:
\[ \Delta H = 540 \, \text{cal/g} \times 18 \, \text{g/mol} = 9720 \, \text{cal/mol} \]

### Step 3: Use the Clausius-Clapeyron equation
The Clausius-Clapeyron equation is given by:
\[ \ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Where:
- \(P_1\) = 700 mm Hg (pressure at boiling point)
- \(P_2\) = 760 mm Hg (standard pressure)
- \(T_2\) = 373 K (boiling point at standard pressure)
- \(R\) = 2 cal/(K·mol) (gas constant in appropriate units)

### Step 4: Substitute the known values into the equation
\[ \ln\left(\frac{760}{700}\right) = \frac{9720}{2} \left(\frac{1}{T_1} - \frac{1}{373}\right) \]

### Step 5: Calculate the left side
Calculate the natural logarithm:
\[ \ln\left(\frac{760}{700}\right) \approx \ln(1.0857) \approx 0.0823 \]

### Step 6: Rearrange the equation to solve for \(T_1\)
Substituting the values into the equation:
\[ 0.0823 = 4860 \left(\frac{1}{T_1} - \frac{1}{373}\right) \]

### Step 7: Solve for \(\frac{1}{T_1}\)
Rearranging gives:
\[ \frac{1}{T_1} - \frac{1}{373} = \frac{0.0823}{4860} \]

Calculating the right side:
\[ \frac{1}{T_1} - \frac{1}{373} \approx 0.00001693 \]

Adding \(\frac{1}{373}\) to both sides:
\[ \frac{1}{T_1} \approx 0.00001693 + \frac{1}{373} \approx 0.002684 \]

### Step 8: Calculate \(T_1\)
Taking the reciprocal gives:
\[ T_1 \approx \frac{1}{0.002684} \approx 372.5 \, \text{K} \]

### Step 9: Convert to Celsius if necessary
To convert Kelvin to Celsius:
\[ T_1 \approx 372.5 - 273.15 \approx 99.35 \, \text{°C} \]

### Final Answer
The boiling point of water at 700 mm Hg is approximately **372.5 K** or **99.35 °C**.