1 Your Turn 1. A scuba diver swimming underwater in the ocean breathes compresse

Question

1 Your Turn 1. A scuba diver swimming underwater in the ocean breathes compressed air at a pressure of 2 atm. If she holds her breath while returning to the surface by how much might the volume of her lungs increase? Solution: 2. A 5.00 liter rubber baloon is submerged 5 meters under ocean water, where its new volume is measured to be 3.38 liter. Show that the pressure at this depth is 1.48 atm Solution: . 3. A perfectly elastic 419 liter balloon is heated from 25°C (298K) to 50C (823K). To what new volume does it expand? Solution: . 4. A hot air balloon 401,000 liters in volume is warmed from 298K to 323K. As the air inside the balloon expands, it is unable to stretch the fabric, which is not very elastic. Instead the expanded air escapes out of a hole placed at the top of the balloon. Show that 135,000 liters of air escapes. Solution: . 5. Air has a density of 1.18 g/L. Show that the hot air balloon in the previous question is now lighter by 159 kg, which helps the hot air balloon to rise. Solution:

Explanation / Answer

Solution

1)

Pressure Underwater, P1 = 2 atm

Volume of lung Underwater, V1 = V1

Pressure at surface, P2 = 1 atm (atmospheric pressure)

Volume of lung at surface, V2 = ?

We know that

P1V1 = P2V2

2 x V1 = 1 x V2

V2 = 2V1

The volume of lungs becomes double

2)

Pressure outside of water, P1 = 1 atm (atmospheric pressure)

Volume of balloon outside of water, V1 = 5.0 L

Pressure at 5 meters under ocean, P2 = ?

Volume at 5 meters under ocean,, V2 = 3.38 L

We know that

P1V1 = P2V2

P2 = (1 x 5)/3.38 = 1.48 atm (ans)

3)

V1 = 419 L

V2=?

T1 = 298 K

T2 = 323 K

We know that

V1/T1 = V2/T2

V2 = (V1xT2)/T1

= (419 x 323)/298 = 454.15 L

V2 = 454.15 L (ans)

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