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 to the change in temperature. The equation is given by:

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

Where:
- \( P_1 \) = standard pressure (760 mm of Hg)
- \( P_2 \) = pressure at which we want to find the boiling point (700 mm of Hg)
- \( \Delta H_{vap} \) = heat of vaporization (540 cal/g)
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T_1 \) = boiling point of water at standard pressure (373 K)
- \( T_2 \) = boiling point of water at 700 mm of Hg (unknown)

### Step 1: Convert the heat of vaporization to appropriate units
First, we need to convert the heat of vaporization from calories to Joules because the gas constant \( R \) is in J/(mol·K).

\[ \Delta H_{vap} = 540 \text{ cal/g} \times 4.184 \text{ J/cal} = 2268.96 \text{ J/g} \]

### Step 2: Set up the equation
Now, we can set up the equation using the values we have:

\[ \log \frac{700}{760} = \frac{2268.96}{8.314} \left( \frac{1}{373} - \frac{1}{T_2} \right) \]

### Step 3: Calculate \( \log \frac{700}{760} \)
Calculate the logarithm:

\[ \log \frac{700}{760} = \log(0.9211) \approx -0.035 \]

### Step 4: Substitute and rearrange the equation
Now we substitute this value into the equation:

\[ -0.035 = \frac{2268.96}{8.314} \left( \frac{1}{373} - \frac{1}{T_2} \right) \]

Calculating the left side:

\[ \frac{2268.96}{8.314} \approx 273.6 \]

So the equation becomes:

\[ -0.035 = 273.6 \left( \frac{1}{373} - \frac{1}{T_2} \right) \]

### Step 5: Solve for \( \frac{1}{T_2} \)
Now we can isolate \( \frac{1}{T_2} \):

\[ \frac{1}{373} - \frac{1}{T_2} = -\frac{0.035}{273.6} \]

Calculating the right side:

\[ -\frac{0.035}{273.6} \approx -0.000128 \]

So we have:

\[ \frac{1}{T_2} = \frac{1}{373} + 0.000128 \]

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

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

Adding the two values:

\[ \frac{1}{T_2} \approx 0.002684 + 0.000128 = 0.002812 \]

### Step 6: Calculate \( T_2 \)
Now we take the reciprocal to find \( T_2 \):

\[ T_2 \approx \frac{1}{0.002812} \approx 355.5 \text{ K} \]

### Step 7: Convert to Celsius
To convert Kelvin to Celsius, we subtract 273.15:

\[ T_2 \approx 355.5 - 273.15 \approx 82.35 \text{ °C} \]

### Final Answer
The boiling point of water at 700 mm of Hg is approximately **82.35 °C**.