### Boyle's LawProblems #16-30

I tried to put the answers in the form of P1V1 = P2V2. They don't have to be in that order, except that the sub ones must be paired on one side of the equals sign and the sub twos must be paired on the other.

For the most part, only set-ups are provided, no answers.

Problem #16: 4.00 L of a gas are under a pressure of 6.00 atm What is the volume of the gas at 2.00 atm?

Solution:

(6.00 atm) (4.00 L) = (2.00 atm) (x)

Problem #17: A gas occupies 25.3 mL at a pressure of 790.5 mmHg. Determine the volume if the pressure is reduced to 0.804 atm

Solution:

(790.5 mmHg) (25.3 mL) = ( 0.804 atm) (x)

This is wrong!! You MUST change one of the pressures units so both are the same. I'll change the mmHg to atm:

(790.5 mmHg / 760.0 mmHg/atm) (25.3 mL) = ( 0.804 atm) (x)

Problem #18: A sample of gas has a volume of 12.0 L and a pressure of 1.00 atm If the pressure of gas is increased to 2.00 atm, what is the new volume of the gas?

Solution:

(1.00 atm) (12.0 L) = (2.00 atm) (x)

Problem #19: A container of oxygen has a volume of 30.0 mL and a pressure of 4.00 atm If the pressure of the oxygen gas is reduced to 2.00 atm and the temperature is kept constant, what is the new volume of the oxygen gas?

Solution:

(4.00 atm) (30.0 mL) = (2.00 atm) (x)

Problem #20: A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mmHg. Find the volume of the nitrogen when its pressure is changed to 400.0 mmHg while the temperature is held constant.

Solution:

(760.0 mmHg) (14.0 L) = (400.0 mmHg) (x)

Problem #21: A 40.0 L tank of ammonia has a pressure of 8.00 atm Calculate the volume of the ammonia if its pressure is changed to 12.0 atm while its temperature remains constant.

Solution:

(8.00 atm) (40.0 L ) = (12.0 atm) (x)

Problem #22: Two hundred liters of helium at 2.00 atm and 28.0 °C is placed into a tank with an internal pressure of 600.0 kPa. Find the volume of the helium after it is compressed into the tank when the temperature of the tank remains 28.0 °C.

Solution:

(2.00 atm) (200.0 L) = (600.0 kPa) (x)

This is wrong. The pressure units must be the same. I'll change the atm to kPa. You could go the other way if you want, the answer would be the same.

(2.00 atm x 101.325 kPa/atm) (200.0 L) = (600.0 kPa) (x)

In fact, here's the problem with the kPa changed to atm:

(2.00 atm) (200.0 L) = (600.0 kPa / 101.325 kPa/atm) (x)

Problem #23: You are now wearing scuba gear and swimming under water at a depth of 66.0 ft. You are breathing air at 3.00 atm and your lung volume is 10.0 L. Your scuba gauge indicates that your air supply is low so, to conserve air, you make a terrible and fatal mistake: you hold your breath while you surface. What happens to your lungs? Why?

Solution:

Your lungs will "explode." As you go up towards the surface, the pressure on your body and lungs becomes less. The air in your lungs expands. What would happen is the alveoli and small capallaries in the lungs would rupture, causing massive bleeding in the lungs. You'd die.

No, your body would not swell up and burst, like a balloon.

Problem #24: Solve Boyle's Law equation for V2.

Solution:

V2 = (P1V1) / P2

Problem #25: Boyle's Law deals with the relationship between what quantities?

(a) pressure/temperature
(b) pressure/volume
(c) volume/temperature
(d) moles/pressure
(e) none of these

Solution:

(b)

Problem #26: A 1.5 liter flask is filled with nitrogen at a pressure of 12 atmospheres. What size flask would be required to hold this gas at a pressure of 2.0 atmospheres?

Solution:

(12 atm) (1.5 liter) = (2.0 atm) (x)

Problem #27: 300 mL of O2 are collected at a pressure of 645 mm of mercury. What volume will this gas have at one atmosphere pressure?

Solution:

(645 mmHg) (300 mL) = (1.00 atm) (x)

This is wrong. I will change atm to mmHg.

(645 mmHg) (300 mL) = (760 mmHg) (x)

Problem #28: How many cubic feet of air at standard conditions (1.00 atm) are required to inflate a bicycle tire of 0.50 cu. ft. to a pressure of 3.00 atmospheres?

Solution:

(1.00 atm) (x) = (3.00 atm) (0.50 cu. ft.)

Problem #29: How much will the volume of 75.0 mL of neon change if the pressure is lowered from 50.0 torr to 8.00 torr?

Solution:

(50.0 torr) (75.0 mL) = (8.00 torr) (x)

Problem #30: A tank of helium has a volume of 50.0 liters and is under a pressure of 2000.0 p.s.i. This gas is allowed to flow into a blimp until the pressure in the tank drops to 40.00 p.s.i. and the pressure in the blimp is 30.00 p.s.i. What will be the volume of the blimp?

Solution:

(1960.0 p.s.i.) (50.0 liters) = (30.00 p.s.i.) (x)

Important point: 2000 psi − 40 psi = 1960 psi flowed out of tank. 2000 is not used in calculation.

Bonus Problem: A spherical weather balloon is constructed so that the gas inside can expand as the balloon ascends to higher altitudes where the pressure is lower. If the radius of the spherical balloon is 2.5 m at sea level where the pressure is 1.004 x 105 Pa, what will be the radius at an altitude of about 10 km where the pressure of the gas is 2.799 x 104 Pa? For simplicity, assume the temperature has not changed.

Solution:

I used P1V1 = P2V2 for this:

[(4/3)(3.14159)(2.5 m)3] (1.004 x 105 Pa) = [(4/3)(3.14159)(x)3] (2.799 x 104 Pa)

However, notice something about the two volumes shown. Each is done as the formula for the volume of a sphere: V = (4/3)(π)(r3). Please notice that the (4/3)(π) portion will cancel out:

(2.5 m)3 (1.004 x 105 Pa) = (x)3 (2.799 x 104 Pa)

x3 = [(15.625 m3) (1.004 x 105 Pa)] / 2.799 x 104 Pa

x3 = 56.05 m3

x = 3.8 m (to two significant figures)