Steam is present in a pressure cooker at 1.5 atm pressure and `150^(@)C`. If the pressure cooker can withstand a maximum of 4 atm pressure and there is no provision of a weight atop the cooker then at what temperature will the cooker explode?

Options

A. `1128^(@)C`
B. 564K
C. `564^(@)C`
D. 1128K

Answer

To solve the problem of determining the temperature at which the pressure cooker will explode, we can use the ideal gas law in the form of the combined gas law, which relates pressure and temperature. The formula we will use is:

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial pressure, \( P_1 = 1.5 \, \text{atm} \)
- Final pressure (maximum pressure before explosion), \( P_2 = 4 \, \text{atm} \)
- Initial temperature, \( T_1 = 150^\circ C \)

2. **Convert Initial Temperature to Kelvin:**
- The temperature in Kelvin is calculated as:
\[ T_1 = 150 + 273 = 423 \, \text{K} \]

3. **Set Up the Equation:**
- Using the combined gas law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

4. **Rearrange to Solve for \( T_2 \):**
- Rearranging gives:
\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]

5. **Substitute the Known Values:**
- Plugging in the values:
\[ T_2 = \frac{4 \, \text{atm} \cdot 423 \, \text{K}}{1.5 \, \text{atm}} \]

6. **Calculate \( T_2 \):**
- Performing the calculation:
\[ T_2 = \frac{1692}{1.5} = 1128 \, \text{K} \]

7. **Convert \( T_2 \) Back to Celsius:**
- To convert Kelvin back to Celsius:
\[ T_2 = 1128 - 273 = 855^\circ C \]

### Final Answer:
The temperature at which the cooker will explode is \( 855^\circ C \).