A sealed tube which can withstand a pressure of 3 atm sphere is filled with air at `27^(@)C and 760` mm pressure. The temperature above which the tube will burst will be:

Options

A. `900^(@)C`
B. `627^(@)C`
C. `727^(@)C`
D. `1173^(@)C`

Answer

To solve the problem step by step, we will use the ideal gas law and the relationship between pressure, volume, and temperature.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial Pressure (P1) = 760 mmHg = 1 atm (since 760 mmHg = 1 atm)
- Maximum Pressure (P2) = 3 atm
- Initial Temperature (T1) = 27°C = 27 + 273 = 300 K (convert to Kelvin)

2. **Use the Ideal Gas Law:**
The ideal gas law can be expressed as:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Since the volume of the sealed tube does not change (V1 = V2 = V), we can simplify the equation to:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

3. **Rearranging the Equation:**
Rearranging the equation to solve for T2 gives:
\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ T_2 = \frac{3 \, \text{atm} \cdot 300 \, \text{K}}{1 \, \text{atm}} = 900 \, \text{K} \]

5. **Convert Temperature to Celsius:**
To convert the temperature back to Celsius:
\[ T_2 = 900 \, \text{K} - 273 = 627 \, \text{°C} \]

6. **Conclusion:**
The temperature above which the tube will burst is **627°C**.

### Final Answer:
The temperature above which the tube will burst is **627°C**.