Question

You have a balloon with a volume of 3.2 L at standard air pressure of 760 mm Hg. You submerge the balloon in one foot of water that is the same temperature as the room. Underwater, the pressure increases to 783 mm Hg. What is the new volume of the balloon? Explain your thinking and math to receive full credit.

Answer 1

Let's start by setting up the initial conditions:

We have a balloon with a known volume (3.2 L) and the pressure inside the balloon is at atmospheric pressure (760 mm Hg). When submerging in water, the pressure increases to 783 mmHg.

Next, we need to use the ideal gas law (P1 * V1 = P2 * V2) to solve for the new volume of the balloon underwater:

760 * 3.2 = 783 * V2

Now, we can solve for the new volume of the balloon:

V2 = (783 / 760) * 3.2

V2 = 3.39 L

So, the new volume of the balloon is 3.39 L, which is a slight increase in volume due to the increase in pressure underwater.

We can use the given formula (P1 * V1 = P2 * V2) to solve for the volume change in any situation where there is a change in pressure and volume. This is a very common and important equation in chemistry and physics, and it's important to be able to understand and use it effectively.