Chapter 20, Problem 44A

(a)

To determine

To explain: The effect on the size of the balloon for the given changes.

Answer to Problem 44A

The size will decrease

Explanation of Solution

Introduction:

According to Boyle’s law, the product of pressure and volume is a constant quantity.

  $PV=\gamma$

When the inflated balloon sinks inside the water it will be pressurized by the surrounded water and this pressure is more than the pressure on it when it was above the surface of water.

According to Boyle’s law, pressure and volume are inversely proportional to each other. More the pressure exerted, less is the volume.So, when it will be under the water, due to the water pressure,size will be reduced.

(b)

To determine

The new volume of the balloon.

Answer to Problem 44A

Volume will become half, compared to the volume at the surface.

Explanation of Solution

Given:

Balloon is at a depth of 10.3 m below the water surface.

Formula used:

According to Boyle’s law, the product of pressure and volume is a constant quantity.

  $PV=\gamma$

Calculation:

From Boyle’s law

  $P_s{V}_s=P_b{V}_b$

Where P_s and V_s are the pressure and volume of balloon at the water surface whileP_b and V_b are the pressure and volume of balloon 10.3 m below the water surface

Atmospheric pressure at the water surface $P_s=1$ atm

It is known that 1 atm pressure can also be created by the water column of the height of 10 m

Since depth is 10.3 m, $P_{{}_b}\simeq 2$ atm

Substitute $P_{{}_b}\simeq 2$ atm and $P_s=1$ atm,

  $V_s=2V_b$

  $V_b=\frac{1}{2}{V}_s$

Conclusion:

Therefore, the volume will become half as compared to the volume at the surface.