A glass container encloses gas at a pressure `4 xx10^5` Pa and 300 K temperature. The container walls can bear a maximum pressure of `8 xx10^(5)` Pa. If the temperature of container is gradually increased find temperature at which container will break.

Options

A. `600 K`
B. `150 K`
C. `1200 K`
D. `375 K`

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure and temperature when the volume is constant.

### Step-by-step Solution:

1. **Identify Given Values**:
- Initial Pressure, \( P_1 = 4 \times 10^5 \) Pa
- Initial Temperature, \( T_1 = 300 \) K
- Maximum Pressure, \( P_2 = 8 \times 10^5 \) Pa

2. **Use the Ideal Gas Law**:
Since the volume of the container is constant, we can use the relationship derived from the ideal gas law which states that for a fixed volume, the ratio of pressure to temperature remains constant:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

3. **Substitute the Known Values**:
Plugging in the known values into the equation:
\[ \frac{4 \times 10^5}{300} = \frac{8 \times 10^5}{T_2} \]

4. **Cross-Multiply to Solve for \( T_2 \)**:
Cross-multiplying gives:
\[ 4 \times 10^5 \cdot T_2 = 8 \times 10^5 \cdot 300 \]

5. **Calculate the Right Side**:
Calculate \( 8 \times 10^5 \cdot 300 \):
\[ 8 \times 10^5 \cdot 300 = 240 \times 10^5 = 2.4 \times 10^7 \]

6. **Isolate \( T_2 \)**:
Now, divide both sides by \( 4 \times 10^5 \):
\[ T_2 = \frac{2.4 \times 10^7}{4 \times 10^5} \]

7. **Perform the Division**:
\[ T_2 = \frac{2.4}{4} \times 10^{7-5} = 0.6 \times 10^2 = 60 \]
Therefore, \( T_2 = 600 \) K.

### Final Answer:
The temperature at which the container will break is \( T_2 = 600 \) K.

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