A given mass of gas occupies `919.0` mL of dry state at N.T.P. The mass when collect over water at `15^(@)C` and `0.987 "bar"` pressure occupies one liter volume. Calculate the vapor pressure of water at `15^(@)C`.

Explanation

To solve the problem, we need to calculate the vapor pressure of water at 15°C using the given data about the gas collected over water. Here’s a step-by-step solution:

### Step 1: Understand the Conditions
- At NTP (Normal Temperature and Pressure), the volume of the gas is given as 919 mL.
- At NTP, the temperature is 273 K (0°C) and the pressure is 1.013 bar.

### Step 2: Convert Given Volumes
- The volume of gas collected over water at 15°C and 0.987 bar is given as 1 liter, which is equal to 1000 mL.

### Step 3: Convert Temperature to Kelvin
- The temperature at which the gas is collected is given as 15°C.
- Convert this to Kelvin:
\[ T_2 = 15 + 273 = 288 \text{ K} \]

### Step 4: Use the Ideal Gas Law
Using the ideal gas equation in the form of Boyle's Law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \(P_1 = 1.013 \text{ bar}\) (pressure at NTP)
- \(V_1 = 919 \text{ mL}\) (volume at NTP)
- \(T_1 = 273 \text{ K}\) (temperature at NTP)
- \(P_2\) is the pressure of the dry gas we need to find.
- \(V_2 = 1000 \text{ mL}\) (volume at experimental conditions)
- \(T_2 = 288 \text{ K}\) (temperature at experimental conditions)

### Step 5: Rearranging the Equation
We can rearrange the equation to solve for \(P_2\):
\[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \]

### Step 6: Substitute the Values
Substituting the values into the equation:
\[ P_2 = \frac{1.013 \text{ bar} \times 919 \text{ mL} \times 288 \text{ K}}{273 \text{ K} \times 1000 \text{ mL}} \]

### Step 7: Calculate \(P_2\)
Calculating the above expression:
\[ P_2 = \frac{1.013 \times 919 \times 288}{273 \times 1000} \]
Calculating the numerator:
\[ 1.013 \times 919 \times 288 \approx 253,163.344 \]
Calculating the denominator:
\[ 273 \times 1000 = 273,000 \]
Now divide:
\[ P_2 \approx \frac{253,163.344}{273,000} \approx 0.927 \text{ bar} \]

### Step 8: Calculate the Vapor Pressure of Water
The pressure of the moist gas is given as 0.987 bar. The vapor pressure of water (\(P_{H_2O}\)) can be calculated as:
\[ P_{H_2O} = P_{\text{moist}} - P_{\text{dry}} \]
Substituting the values:
\[ P_{H_2O} = 0.987 \text{ bar} - 0.927 \text{ bar} = 0.060 \text{ bar} \]

### Final Answer
The vapor pressure of water at 15°C is approximately **0.060 bar**.

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