At N.T.P the volume of a gas is found to be 250mL. What will be the volume of this gas at 600mm Hg and `273^o`C?

Options

A. 589.7
B. 458.9
C. 645.7
D. 147.5

Explanation

To solve the problem of finding the volume of a gas at different conditions, we can use the Ideal Gas Law, which states that:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \( P_1 \) and \( V_1 \) are the initial pressure and volume.
- \( T_1 \) is the initial temperature in Kelvin.
- \( P_2 \) and \( V_2 \) are the final pressure and volume.
- \( T_2 \) is the final temperature in Kelvin.

### Step 1: Identify the known values
- Initial volume, \( V_1 = 250 \, \text{mL} \)
- Initial pressure, \( P_1 = 1 \, \text{atm} \) (NTP)
- Initial temperature, \( T_1 = 273 \, \text{°C} = 273 + 273 = 546 \, \text{K} \)
- Final pressure, \( P_2 = 600 \, \text{mm Hg} \) (Convert to atm)
- Final temperature, \( T_2 = 273 \, \text{°C} = 273 + 273 = 546 \, \text{K} \)

### Step 2: Convert final pressure to atm
To convert mm Hg to atm, we use the conversion factor:
\[ 1 \, \text{atm} = 760 \, \text{mm Hg} \]
Thus,
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.789 \, \text{atm} \]

### Step 3: Rearrange the Ideal Gas Law equation
We need to find \( V_2 \), so we rearrange the equation:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ V_2 = \frac{(1 \, \text{atm}) \times (250 \, \text{mL}) \times (546 \, \text{K})}{(0.789 \, \text{atm}) \times (546 \, \text{K})} \]

### Step 5: Calculate \( V_2 \)
The \( 546 \, \text{K} \) cancels out:
\[ V_2 = \frac{250 \, \text{mL}}{0.789} \]
Calculating this gives:
\[ V_2 \approx 316.5 \, \text{mL} \]

### Final Answer
Thus, the volume of the gas at 600 mm Hg and 273 °C is approximately **316.5 mL**.

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