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, we will use the ideal gas law in the form of the combined gas law, which states that for a fixed volume and number of moles, the ratio of pressure to temperature is constant. The formula we will use is:

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

### Step-by-Step Solution:

**Step 1: Identify the known values.**
- Initial pressure, \( P_1 = 1.5 \, \text{atm} \)
- Initial temperature, \( T_1 = 150^\circ C \)
- Maximum pressure before explosion, \( P_2 = 4 \, \text{atm} \)

**Step 2: Convert the initial temperature from Celsius to Kelvin.**
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_1 = 150 + 273 = 423 \, \text{K} \]

**Step 3: Set up the equation using the combined gas law.**
Using the formula:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
We can rearrange this to find \( T_2 \):
\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]

**Step 4: Substitute the known values into the equation.**
\[ T_2 = \frac{4 \, \text{atm} \cdot 423 \, \text{K}}{1.5 \, \text{atm}} \]

**Step 5: Calculate \( T_2 \).**
\[ T_2 = \frac{1692}{1.5} = 1128 \, \text{K} \]

**Step 6: Convert \( T_2 \) back to Celsius if needed.**
To convert back to Celsius:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T_2(°C) = 1128 - 273 = 855 \, °C \]

### Final Answer:
The cooker will explode at a temperature of \( 1128 \, \text{K} \) or \( 855 \, °C \).

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