Final answer:
According to Boyle's law, when the pressure of a gas decreases, its volume increases if the temperature remains constant. In this case, the new volume will be greater than 25.0 L when the pressure is 51.2 kPa.
Step-by-step explanation:
According to Boyle's law, when the pressure of a gas decreases, its volume increases if the temperature remains constant. In this case, the initial volume of the balloon is 25.0 L at a pressure of 98.7 kPa, and it is asked to determine if the new volume will be more or less when the pressure is 51.2 kPa.
Since the pressure is decreasing, we can conclude that the new volume will be greater than 25.0 L.
Boyle's Law states that the product of the initial pressure and the initial volume is equal to the product of the final pressure and the final volume, as long as the temperature remains constant. This can be expressed as P1 * V1 = P2 * V2. Given the initial volume (V1) and pressure (P1) and the final pressure (P2), we can rearrange the equation to solve for the final volume (V2).
Using the equation, we have:
(98.7 kPa)(25.0 L) = (51.2 kPa)(V2)
Solving for V2:
V2 = (98.7 kPa)(25.0 L) / (51.2 kPa)
V2 ≈ 48.3 L
Therefore, when the pressure is 51.2 kPa, the new volume of the balloon will be approximately 48.3 L, which is greater than the initial volume of 25.0 L.