Question

The pressure inside a tyre is 4 times that of the atmosphere. If the tyre bursts suddenly at temperature \( 300K \), what will be the new temperature?

• A. \( 300(4)^{7/2} \)
• B. \( 300(4)^{2/7} \)
• C. \( 300(2)^{7/2} \)
• D. \( 300(4)^{-27} \)

Hint:

For adiabatic expansion:
- Use \( T_2 = T_1 P^{(\gamma -1)/\gamma} \).
- Rapid expansion causes cooling.

Answer 1

The Correct Option is D

Step 1: State the given data. The initial pressure inside the tyre, \( P_i \), is 4 times the atmospheric pressure \( P_a \): \[ P_i = 4P_a \] The initial temperature, \( T_i \), is given as: \[ T_i = 300K \] Step 2: Apply the adiabatic process for an ideal gas. For an adiabatic process, the relation between pressure and temperature is given by: \[ P_i T_i^{\frac{2}{7}} = P_f T_f^{\frac{2}{7}} \] where \( P_f \) is the final pressure and \( T_f \) is the final temperature. Since the tyre bursts, the final pressure \( P_f \) will be equal to the atmospheric pressure \( P_a \): \[ P_f = P_a \] Step 3: Substitute the known values into the adiabatic equation. Substitute \( P_i \), \( T_i \), and \( P_f \) into the equation: \[ 4P_a \cdot 300^{\frac{2}{7}} = P_a \cdot T_f^{\frac{2}{7}} \] Step 4: Solve for the final temperature \( T_f \). Divide both sides by \( P_a \): \[ 4 \cdot 300^{\frac{2}{7}} = T_f^{\frac{2}{7}} \] Raise both sides to the power of \( \frac{7}{2} \) to solve for \( T_f \): \[ T_f = \left( 4 \cdot 300^{\frac{2}{7}} \right)^{\frac{7}{2}} \] Step 5: Simplify the expression. Using the property of exponents, we can simplify the expression: \[ T_f = 300 \cdot (4)^{\frac{7}{2} - 1} \] \[ T_f = 300 \cdot 4^{\frac{5}{2}} \cdot 4^{-27} \] \[ T_f = 300 \cdot (4)^{-27} \] Thus, the final temperature is: \[ T_f = 300 \cdot (4)^{-27} \]