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 `273^o`C?

Explanation

To solve the problem of finding the volume of a gas at a different pressure and temperature, we can use the Ideal Gas Law, which relates pressure, volume, and temperature. The equation we will use is derived from the Ideal Gas Law:

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

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 the Given Values**:
- Initial volume (\(V_1\)) = 273 mL
- Initial pressure (\(P_1\)) = 1 atm (NTP)
- Initial temperature (\(T_1\)) = 273°C = 273 + 273.15 = 546.15 K
- Final pressure (\(P_2\)) = 600 mm Hg
- Final temperature (\(T_2\)) = 273°C = 273 + 273.15 = 546.15 K

2. **Convert Final Pressure to atm**:
- Since \(1 \text{ atm} = 760 \text{ mm Hg}\),
\[ P_2 = \frac{600 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.7895 \text{ atm} \approx 0.79 \text{ atm} \]

3. **Substitute Values into the Ideal Gas Equation**:
\[ \frac{(1 \text{ atm})(273 \text{ mL})}{546.15 \text{ K}} = \frac{(0.79 \text{ atm})(V_2)}{546.15 \text{ K}} \]

4. **Cancel Out the Temperature**:
Since \(T_1\) and \(T_2\) are the same, we can cancel them out:
\[ 1 \text{ atm} \times 273 \text{ mL} = 0.79 \text{ atm} \times V_2 \]

5. **Solve for \(V_2\)**:
\[ V_2 = \frac{(1 \text{ atm} \times 273 \text{ mL})}{0.79 \text{ atm}} = \frac{273}{0.79} \approx 345.57 \text{ mL} \]

6. **Final Calculation**:
\[ V_2 \approx 345.57 \text{ mL} \]

### Final Answer:
The volume of the gas at 600 mm Hg and 273°C is approximately **345.57 mL**.