1. If a scuba diver\'s lungs have a normal capacity of 4.9 L at sea level (1.0 a

Question

1. If a scuba diver's lungs have a normal capacity of 4.9 L at sea level (1.0 atm), what would be the volume of her lungs if the pressure at a depth of 50 ft is 975 mmHg?

0.26 L

2. If a swimmer has a lung volume of 6 L at sea level, what would the volume of her lungs be when she is at the bottom of a pool that is 2m deep? Assume that the temperature and amount of the air in the lungs remain unchanged.

Express your answer to one significant figure and include the appropriate units.

3. A beach ball is filled with 13.0L of air in a hotel room where the temperature is 73?F.

What is the volume of the air in the beach ball out on the beach where the temperature is 33?C?

Express your answer with the appropriate units.

4. Liquid nitrogen is an extremely cold liquid (-196 ?C) used in chemistry "magic" shows to "shrink" balloons (and do other fun things).

If a balloon is inflated to 2 L at 26 ?C and then cooled in liquid nitrogen, what is the final volume of the balloon?

Express your answer to one significant figure and include the appropriate units.

0.26 L

2. If a swimmer has a lung volume of 6 L at sea level, what would the volume of her lungs be when she is at the bottom of a pool that is 2m deep? Assume that the temperature and amount of the air in the lungs remain unchanged.

Express your answer to one significant figure and include the appropriate units.

3. A beach ball is filled with 13.0L of air in a hotel room where the temperature is 73?F.

What is the volume of the air in the beach ball out on the beach where the temperature is 33?C?

Express your answer with the appropriate units.

4. Liquid nitrogen is an extremely cold liquid (-196 ?C) used in chemistry "magic" shows to "shrink" balloons (and do other fun things).

If a balloon is inflated to 2 L at 26 ?C and then cooled in liquid nitrogen, what is the final volume of the balloon?

Express your answer to one significant figure and include the appropriate units.

Explanation / Answer

1) we know that

PV = nRT

here n and T are constant

so

P1V1 = P2V2

so

1 x 4.9 = ( 975 / 760) x V2

V2 = 3.8 L

so

the volume should be 3.8 L

2)

we know that

for every 10 m depth the pressure increases by 1 atm

so

final pressure = 1.2 atm

now

P1V1 = P2V2

1 x 6 = 1.2 x V2

V2 = 5

so

the volume of the lungs at the bottom is 5 L

3) we know that

PV = nRT

in this case , P and n are constant

so

V/T = constant

V1 / T1 = V2 / T2

given

V1 = 13 L

T1 = 73 F = 295.9 K

T2 = 33 C = 33 + 273 = 306

so

13 / 295.9 = V2 / 306

V2 = 13.44 L

so

the volume of the air will be 13.44 L

4)

V1 / T1 = V2 / T2

given

V1 = 2

T1 = 26 C = 299 K

T2 = -196 C = 77 K

so

2 / 299 = V2 / 77

V2 = 0.515

so

the final volume will be 0.515 L

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