A cylinder contains gas at `2 xx 10^(5) Pa`and `47^(@)C`. The cylinder is steadily heated. Neglecting any change in its volume, find the temperature at which the cylinder will break if the walls of cylinder can bear a maximum pressure of `8 xx 10^(5) Pa`

Explanation

To solve the problem, we will use the relationship between pressure and temperature for a gas at constant volume, as described by the ideal gas law. Here are the steps to find the temperature at which the cylinder will break:

### Step 1: Identify the given values
- Initial Pressure, \( P_1 = 2 \times 10^5 \, \text{Pa} \)
- Initial Temperature, \( T_1 = 47^\circ C = 47 + 273 = 320 \, \text{K} \)
- Maximum Pressure, \( P_2 = 8 \times 10^5 \, \text{Pa} \)

### Step 2: Use the relationship between pressure and temperature
Since the volume is constant, we can use the relationship:
\[ \frac{P_1}{P_2} = \frac{T_1}{T_2} \]

### Step 3: Rearrange the equation to solve for \( T_2 \)
From the equation, we can rearrange it to find \( T_2 \):
\[ T_2 = T_1 \times \frac{P_2}{P_1} \]

### Step 4: Substitute the known values into the equation
Now we substitute the values we have:
\[ T_2 = 320 \, \text{K} \times \frac{8 \times 10^5 \, \text{Pa}}{2 \times 10^5 \, \text{Pa}} \]

### Step 5: Calculate \( T_2 \)
First, calculate the fraction:
\[ \frac{8 \times 10^5}{2 \times 10^5} = 4 \]
Now substitute this back into the equation:
\[ T_2 = 320 \, \text{K} \times 4 = 1280 \, \text{K} \]

### Step 6: Convert \( T_2 \) to Celsius
To convert from Kelvin to Celsius, we use:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 \]
Thus,
\[ T_2 = 1280 \, \text{K} - 273 = 1007 \, \text{°C} \]

### Final Answer
The temperature at which the cylinder will break is \( 1007 \, \text{°C} \).

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