A tyre pumped to a pressure `3.375 atm "at" 27^(@)C` suddenly bursts. What is the final temperature `(gamma = 1.5)`?

Options

A. `27^(@)C`
B. `-27^(@)C`
C. `0^(@)C`
D. `-73^(@)C`

Explanation

To solve the problem of finding the final temperature of the air in the tire after it bursts, we can use the principles of thermodynamics, specifically the adiabatic process. Here's a step-by-step solution:

### Step 1: Understand the Initial Conditions
The initial pressure \( P_1 \) is given as \( 3.375 \, \text{atm} \) and the initial temperature \( T_1 \) is \( 27^\circ C \). We need to convert the temperature to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]

### Step 2: Identify the Final Pressure
When the tire bursts, the final pressure \( P_2 \) becomes equal to atmospheric pressure, which is:
\[ P_2 = 1 \, \text{atm} \]

### Step 3: Use the Adiabatic Process Relation
For an adiabatic process, the relation between pressure and temperature can be expressed as:
\[ \frac{P_1 V_1^\gamma}{T_1^\gamma} = \frac{P_2 V_2^\gamma}{T_2^\gamma} \]
Since the volume does not change significantly during the burst, we can simplify this to:
\[ \frac{P_1}{T_1^\gamma} = \frac{P_2}{T_2^\gamma} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ T_2^\gamma = T_1^\gamma \cdot \frac{P_2}{P_1} \]
Taking the \(\gamma\)-th root:
\[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{1}{\gamma}} \]

### Step 5: Substitute the Values
Now substitute the known values into the equation:
- \( P_1 = 3.375 \, \text{atm} \)
- \( P_2 = 1 \, \text{atm} \)
- \( T_1 = 300 \, \text{K} \)
- \( \gamma = 1.5 \)

\[ T_2 = 300 \left( \frac{1}{3.375} \right)^{\frac{1}{1.5}} \]

### Step 6: Calculate \( T_2 \)
Calculate the fraction:
\[ \frac{1}{3.375} \approx 0.2963 \]
Now raise this to the power of \( \frac{1}{1.5} \):
\[ (0.2963)^{\frac{1}{1.5}} \approx 0.408 \]
Now multiply by \( 300 \):
\[ T_2 \approx 300 \times 0.408 \approx 122.4 \, \text{K} \]

### Step 7: Convert to Celsius
To convert the final temperature back to Celsius:
\[ T_2 = 122.4 - 273 \approx -150.6^\circ C \]

### Final Answer
Thus, the final temperature after the tire bursts is approximately:
\[ T_2 \approx -150.6^\circ C \]