At N.T.P the volume of a gas is found to be 273mL. What will be the volume of this gas at 600mm Hg and `257^o`C?

Explanation

To solve the problem, we will use the Ideal Gas Law, which relates pressure, volume, and temperature of a gas. The formula we will use is:

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

Where:
- \(P_1\) = initial pressure
- \(V_1\) = initial volume
- \(T_1\) = initial temperature
- \(P_2\) = final pressure
- \(V_2\) = final volume
- \(T_2\) = final temperature

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial volume, \(V_1 = 273 \, \text{mL}\)
- Initial pressure, \(P_1 = 1 \, \text{atm}\) (at NTP)
- Initial temperature, \(T_1 = 273 \, \text{K}\) (NTP is 0°C or 273 K)
- Final pressure, \(P_2 = 600 \, \text{mm Hg}\)
- Final temperature, \(T_2 = 257 \, \text{°C}\)

2. **Convert Units:**
- Convert \(P_2\) from mm Hg to atm:
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.789 \, \text{atm} \]
- Convert \(T_2\) from Celsius to Kelvin:
\[ T_2 = 257 + 273 = 530 \, \text{K} \]

3. **Substitute Values into the Ideal Gas Law:**
\[ \frac{1 \, \text{atm} \times 273 \, \text{mL}}{273 \, \text{K}} = \frac{0.789 \, \text{atm} \times V_2}{530 \, \text{K}} \]

4. **Cross-Multiply to Solve for \(V_2\):**
\[ 1 \times 273 \times 530 = 0.789 \times V_2 \times 273 \]
\[ 144390 = 0.789 \times V_2 \times 273 \]

5. **Isolate \(V_2\):**
\[ V_2 = \frac{144390}{0.789 \times 273} \]

6. **Calculate \(V_2\):**
\[ V_2 \approx \frac{144390}{215.697} \approx 668.6 \, \text{mL} \]

### Final Answer:
The volume of the gas at 600 mm Hg and 257°C is approximately **668.6 mL**.