A tyre filled with air,`(27^(@)C`,o and 2 atm ) bursts, then what is temperature of air `(lambda=1.5)`

Options

A. `-33^(@)C`
B. `0^(@)C`
C. `27^(@)C`
D. `240^(@)C`

Answer

To find the temperature of the air after the tyre bursts, we can use the principles of thermodynamics, specifically the adiabatic process equations. Here’s a step-by-step solution:

### Step 1: Identify Initial Conditions
- The initial temperature \( T_1 \) is given as \( 27^\circ C \).
- Convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The initial pressure \( P_1 \) is \( 2 \, atm \).

### Step 2: Identify Final Conditions
- After the tyre bursts, the air expands to atmospheric pressure, which is \( P_2 = 1 \, atm \).

### Step 3: Use the Adiabatic Process Equation
For an adiabatic process, we can use the equation:
\[ T_1 P_1^{\frac{1 - \gamma}{\gamma}} = T_2 P_2^{\frac{1 - \gamma}{\gamma}} \]
Where \( \gamma = 1.5 \).

### Step 4: Rearrange the Equation to Solve for \( T_2 \)
Rearranging the equation for \( T_2 \):
\[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
- \( T_1 = 300 \, K \)
- \( P_1 = 2 \, atm \)
- \( P_2 = 1 \, atm \)
- \( \gamma = 1.5 \)

So,
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{1.5 - 1}{1.5}} \]

### Step 6: Calculate the Exponent
Calculate the exponent:
\[ \frac{1.5 - 1}{1.5} = \frac{0.5}{1.5} = \frac{1}{3} \]

### Step 7: Substitute the Exponent Back
Now substitute back into the equation:
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{1}{3}} \]

### Step 8: Calculate \( \left( \frac{1}{2} \right)^{\frac{1}{3}} \)
Calculating \( \left( \frac{1}{2} \right)^{\frac{1}{3}} \):
\[ \left( \frac{1}{2} \right)^{\frac{1}{3}} \approx 0.7937 \]

### Step 9: Final Calculation of \( T_2 \)
Now calculate \( T_2 \):
\[ T_2 \approx 300 \times 0.7937 \approx 238.11 \, K \]

### Step 10: Convert Back to Celsius
Convert \( T_2 \) back to Celsius:
\[ T_2 = 238.11 - 273 \approx -34.89 \, ^\circ C \]

### Conclusion
The final temperature of the air after the tyre bursts is approximately \( -34.89 \, ^\circ C \).