At N.T.P the volume of a gas is found to be 257mL. What will be the volume of this gas at 600mm Hg and `257^o`C?

Options

A. 588.45
B. 400
C. 478.75
D. 200

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 \(P_2\) are the initial and final pressures,
- \(V_1\) and \(V_2\) are the initial and final volumes,
- \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Initial volume (\(V_1\)) = 257 mL
- Initial pressure (\(P_1\)) = 1 atm (NTP)
- Initial temperature (\(T_1\)) = 293 K (NTP)
- Final pressure (\(P_2\)) = 600 mm Hg
- Final temperature (\(T_2\)) = 257 °C

2. **Convert Final Pressure to atm:**
\[ P_2 = \frac{600 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.7895 \text{ atm} \approx 0.79 \text{ atm} \]

3. **Convert Final Temperature to Kelvin:**
\[ T_2 = 257 °C + 273 = 530 \text{ K} \]

4. **Substitute the Values into the Ideal Gas Law:**
Using the formula:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substitute the known values:
\[ \frac{1 \text{ atm} \times 257 \text{ mL}}{293 \text{ K}} = \frac{0.79 \text{ atm} \times V_2}{530 \text{ K}} \]

5. **Cross-Multiply to Solve for \(V_2\):**
\[ 1 \times 257 \times 530 = 0.79 \times V_2 \times 293 \]
\[ 136810 = 231.47 V_2 \]

6. **Solve for \(V_2\):**
\[ V_2 = \frac{136810}{231.47} \approx 590.45 \text{ mL} \]

### Final Answer:
The volume of the gas at 600 mm Hg and 257 °C is approximately **590.45 mL**.