A tyre pumped to a pressure of `6 atmosphere` bursts suddenly. Calculate the temperature of escaping air. Given initial room temperature is `15^(@)C` and gamma for air is 1.4`.

Explanation

To solve the problem of calculating the temperature of the escaping air when a tyre bursts, we will follow these steps:

### Step 1: Understand the Given Data
- Initial pressure of the tyre, \( P_1 = 6 \, \text{atm} \)
- Final pressure after bursting, \( P_2 = 1 \, \text{atm} \) (atmospheric pressure)
- Initial temperature, \( T_1 = 15^\circ C = 15 + 273 = 288 \, \text{K} \)
- Value of gamma for air, \( \gamma = 1.4 \)

### Step 2: Identify the Process
Since the tyre bursts suddenly, we can assume that the process is adiabatic (no heat exchange with the surroundings).

### Step 3: Use the Adiabatic Relation
For an adiabatic process, the relationship between pressure and temperature is given by:
\[ \frac{P_1}{P_2} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma - 1}} \]
We can rearrange this equation to find \( T_2 \):
\[ T_2 = T_1 \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}} \]

### Step 4: Substitute the Values
Now we can substitute the known values into the equation:
\[ T_2 = 288 \left(\frac{6}{1}\right)^{\frac{1.4 - 1}{1.4}} \]
\[ T_2 = 288 \times 6^{\frac{0.4}{1.4}} \]

### Step 5: Calculate the Exponent
First, calculate \( \frac{0.4}{1.4} \):
\[ \frac{0.4}{1.4} = \frac{4}{14} = \frac{2}{7} \approx 0.2857 \]

### Step 6: Calculate \( 6^{\frac{2}{7}} \)
Now we need to calculate \( 6^{\frac{2}{7}} \):
Using a calculator or logarithmic tables, we find:
\[ 6^{\frac{2}{7}} \approx 1.724 \]

### Step 7: Final Calculation of \( T_2 \)
Now substitute this back into the equation for \( T_2 \):
\[ T_2 = 288 \times 1.724 \approx 496.032 \, \text{K} \]

### Step 8: Convert to Celsius
Finally, convert the temperature back to Celsius:
\[ T_2 = 496.032 - 273 \approx 223.032^\circ C \]

### Final Answer
The temperature of the escaping air when the tyre bursts is approximately \( 223.03^\circ C \).

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