A dry gas occupies `127 mL` at `N.T.P.` If the same mass of the gas is collected over water at `23^(@)C` and a total pressure of `0.98 "bar"`, what volume will it occupy ? The vapour pressure of water at `23^(@)C` is `0.028 "bar"`.

Explanation

To solve the problem, we will use the Ideal Gas Law and the concept of gas behavior under different conditions. Here are the steps to find the volume of the gas collected over water at 23°C and a total pressure of 0.98 bar.

### Step 1: Identify the given data
- Volume of gas at NTP (V1) = 127 mL
- Pressure at NTP (P1) = 1.013 bar
- Temperature at NTP (T1) = 273 K
- Total pressure when gas is collected over water (P_total) = 0.98 bar
- Vapor pressure of water at 23°C (P_water) = 0.028 bar

### Step 2: Calculate the pressure of the dry gas (P2)
The pressure of the dry gas can be found by subtracting the vapor pressure of water from the total pressure:
\[ P2 = P_{total} - P_{water} \]
\[ P2 = 0.98 \, \text{bar} - 0.028 \, \text{bar} = 0.952 \, \text{bar} \]

### Step 3: Convert the temperature to Kelvin
The temperature at which the gas is collected is given as 23°C. We need to convert this to Kelvin:
\[ T2 = 23 + 273 = 296 \, \text{K} \]

### Step 4: Use the Ideal Gas Law
We can use the combined gas law, which relates the initial and final states of the gas:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]
Rearranging this equation to solve for V2 gives:
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 5: Substitute the known values
Now we can substitute the values we have into the equation:
- \( P1 = 1.013 \, \text{bar} \)
- \( V1 = 127 \, \text{mL} \)
- \( T1 = 273 \, \text{K} \)
- \( T2 = 296 \, \text{K} \)
- \( P2 = 0.952 \, \text{bar} \)

Substituting these values:
\[ V2 = \frac{1.013 \, \text{bar} \cdot 127 \, \text{mL} \cdot 296 \, \text{K}}{0.952 \, \text{bar} \cdot 273 \, \text{K}} \]

### Step 6: Calculate V2
Now we can perform the calculation:
\[ V2 = \frac{1.013 \cdot 127 \cdot 296}{0.952 \cdot 273} \]
Calculating the numerator:
\[ 1.013 \cdot 127 \cdot 296 \approx 39654.596 \]
Calculating the denominator:
\[ 0.952 \cdot 273 \approx 259.896 \]
Now, dividing the two results:
\[ V2 \approx \frac{39654.596}{259.896} \approx 152.5 \, \text{mL} \]

### Final Answer
The volume of the gas collected over water at 23°C and a total pressure of 0.98 bar is approximately **152.5 mL**.

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