At N.T.P the volume of a gas is found to be 240mL. What will be the volume of this gas at 600mm Hg and `173^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 in the form of the combined gas equation:

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume at NTP (Standard conditions): \( V_1 = 240 \, \text{mL} \)
- Pressure at NTP: \( P_1 = 1 \, \text{atm} \)
- Temperature at NTP: \( T_1 = 293 \, \text{K} \) (which is \( 0^\circ C \))
- New Pressure: \( P_2 = 600 \, \text{mm Hg} \)
- New Temperature: \( T_2 = 173^\circ C \)

2. **Convert the New Pressure to Atmospheres:**
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.789 \, \text{atm} \]

3. **Convert the New Temperature to Kelvin:**
\[ T_2 = 173 + 273 = 446 \, \text{K} \]

4. **Substitute the Values into the Combined Gas Equation:**
\[ \frac{1 \, \text{atm} \cdot 240 \, \text{mL}}{293 \, \text{K}} = \frac{0.789 \, \text{atm} \cdot V_2}{446 \, \text{K}} \]

5. **Rearranging the Equation to Solve for \( V_2 \):**
\[ V_2 = \frac{240 \, \text{mL} \cdot 0.789 \, \text{atm} \cdot 446 \, \text{K}}{1 \, \text{atm} \cdot 293 \, \text{K}} \]

6. **Calculate \( V_2 \):**
\[ V_2 = \frac{240 \cdot 0.789 \cdot 446}{293} \]
\[ V_2 \approx \frac{105,000.24}{293} \approx 358.51 \, \text{mL} \]

### Final Answer:
The volume of the gas at 600 mm Hg and 173°C is approximately **358.51 mL**.