Question

A balloon contains 2.27 L air and has a pressure of \(1.013 \times 10^5 \, \text{Nm}^{-2}\). The balloon rises to a certain height and expands to a volume of 4540 mL. What is the final pressure of air in the balloon?

• A. \( 2.026 \times 10^2 \, \text{Nm}^{-2} \)
• B. \( 4.540 \times 10^4 \, \text{Nm}^{-2} \)
• C. \( 5.065 \times 10^{-4} \, \text{Nm}^{-2} \)
• D. \( 5.065 \times 10^4 \, \text{Nm}^{-2} \)

Hint:

In Boyle's law, pressure is inversely proportional to volume, so when volume increases, pressure decreases and vice versa.

Answer 1

The Correct Option is D

Step 1: Applying Boyle’s Law.
Boyle’s law states that for a fixed amount of gas at constant temperature, the pressure is inversely proportional to the volume. The formula is given by: \[ P_1 V_1 = P_2 V_2 \] Where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.
Step 2: Converting units.
Initial volume \(V_1 = 2.27 \, \text{L} = 2270 \, \text{mL}\), final volume \(V_2 = 4540 \, \text{mL}\).
Step 3: Calculation.
Using Boyle's law, we have: \[ P_2 = \frac{P_1 V_1}{V_2} \] Substitute the values: \[ P_2 = \frac{(1.013 \times 10^5 \, \text{Nm}^{-2}) \times 2270 \, \text{mL}}{4540 \, \text{mL}} = 5.065 \times 10^4 \, \text{Nm}^{-2} \] Step 4: Conclusion.
The final pressure is \(5.065 \times 10^4 \, \text{Nm}^{-2}\), so the correct answer is (D).