A certain quantity of a gas occupied `100ml` when collected over water at `15^(@)C` and `750mm` pressure . It occupies `91.9ml` in dry state at `NTP`. Find the `V.P.` of water at `15^(@)C`

Options

A. `20mm`
B. `13.2mm`
C. `18mm`
D. 63mm

Explanation

To solve the problem step by step, we will use the ideal gas law and the concept of vapor pressure. Here’s how we can approach it:

### Step 1: Understand the Given Information
- The gas occupies 100 ml when collected over water at 15°C and 750 mm pressure.
- The same gas occupies 91.9 ml in dry state at Normal Temperature and Pressure (NTP).
- We need to find the vapor pressure (V.P.) of water at 15°C.

### Step 2: Convert Temperature to Kelvin
The temperature needs to be converted from Celsius to Kelvin:
\[ T_1 = 15°C + 273 = 288 \, K \]
At NTP, the temperature is:
\[ T_2 = 20°C + 273 = 293 \, K \]

### Step 3: Define the Pressures
Let \( P \) be the vapor pressure of water at 15°C. The pressure of the dry gas can be calculated as:
\[ P_1 = 750 \, \text{mm} - P \]
At NTP, the pressure \( P_2 \) is:
\[ P_2 = 760 \, \text{mm} \]

### Step 4: Use the Ideal Gas Law
According to the ideal gas law, we can equate the moles of gas at two different states:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substituting the known values:
\[ \frac{(750 - P) \times 100}{288} = \frac{760 \times 91.9}{293} \]

### Step 5: Solve for \( P \)
Cross-multiply to eliminate the fractions:
\[ (750 - P) \times 100 \times 293 = 760 \times 91.9 \times 288 \]
Calculating the right side:
\[ 760 \times 91.9 \times 288 = 19891200 \]
So we have:
\[ (750 - P) \times 29300 = 19891200 \]
Now, divide both sides by 29300:
\[ 750 - P = \frac{19891200}{29300} \]
Calculating the left side:
\[ \frac{19891200}{29300} \approx 678.5 \]
Thus:
\[ 750 - P \approx 678.5 \]
Now, solve for \( P \):
\[ P \approx 750 - 678.5 = 71.5 \, \text{mm} \]

### Step 6: Conclusion
The vapor pressure of water at 15°C is approximately:
\[ P \approx 71.5 \, \text{mm} \]