Problems #1-15

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Problems 16-30 | Return to KMT & Gas Laws Menu |

**Problem #1:** A gas occupies 12.3 liters at a pressure of 40.0 mmHg. What is the volume when the pressure is increased to 60.0 mmHg?

(40.0 mmHg) (12.3 liters) = (60.0 mmHg) (x)x = 8.20 L

Note three significant figures.

**Problem #2:** If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

(1.00 atm) ( 3.60 liters) = (2.50 atm) (x)x = 1.44 L

**Problem #3:** To what pressure must a gas be compressed in order to get into a 3.00 cubic foot tank the entire weight of a gas that occupies 400.0 cu. ft. at standard pressure?

(400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot)x = 133 atm

It doesn't matter what the volume units are. It just matters that they be the same on each side.

**Problem #4:** A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm?

(1.56 L) (1.00 atm) = (3.00 atm) (x)0.520 L

**Problem #5:** A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L?

(11.2 liters) (0.860 atm) = (x) (15.0 L)x = 0.642 atm

**Problem #6:** 500.0 mL of a gas is collected at 745.0 mmHg. What will the volume be at standard pressure?

(745.0 mmHg) (500.0 mL) = (760.0 mmHg) (x)x = 490.1 mL

**Problem #7:** Convert 350.0 mL at 740.0 mmHg to its new volume at standard pressure.

(740.0 mmHg) (350.0 mL) = (760.0 mmHg) (x)

**Problem #8:** Convert 338 L at 63.0 atm to its new volume at standard pressure.

(63.0 atm) (338 L) = (1.00 atm) (x)

**Problem #9:** Convert 273.15 mL at 166.0 kPa to its new volume at standard pressure.

(166.0 kPa) (273.15 mL) = (101.325 kPa) (x)

**Problem #10:** Convert 77.0 L at 18.0 mmHg to its new volume at standard pressure.

(18.0 mmHg) (77.0 L) = (760.0 mmHg) (x)

**Problem #11:** When the pressure on a gas increases, will the volume increase or decrease?

Volume will decrease.

**Problem #12:** If the pressure on a gas is decreased by one-half, how large will the volume change be?

It will double in size.

**Problem #13:** A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm.

(0.755 atm) (4.31 liters) = (1.25 atm) (x)

**Problem #14:** 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00 atm?

(8.00 atm) (600.0 mL) = (2.00 atm) (x)

**Problem #15:** 400.0 mL of a gas are under a pressure of 800.0 torr. What would the volume of the gas be at a pressure of 1000.0 torr?

(800.0 torr) (400.0 mL) = (1000.0 torr) (x)

**Bonus Example #1:** A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 liters. If the balloon is filled with 2.0 liters of helium at sea level (101.3 kPa), and rises to an altitude at which the boiling temperature of water is only 88 degrees Celsius, will the balloon burst?

**Solution:**

Comment: These is no way of determining the starting temperature of the gas. However, we know something not in the problem: at sea level, the boiling point of water is 100 °C. So:

1) Let us use a ratio and proportion to estimate the pressure required for water to boil at 88 °C:

100 °C is to 101.3 kPa as 88 °C is to xx = 89.144 kPa

2) Now, we can solve the problem using Boyle's Law:

P_{1}V_{1}= P_{2}V_{2}(101.3) (2.0) = (88.144) (x)

x = 2.27 L

The balloon will not burst.

Comment: Boyle's Law assumes that the temperature and amount of gas are constant. Since we never knew the starting temperature, we will assume it never changed as the balloon rose. If the temperature actually did change, but by some unknown value, then we cannot solve the problem.

**Bonus Example #2:** Two bulbs of different volumes are separated by a valve. The valve between the 2.00 L bulb, in which the gas pressure is 1.00 atm, and the 3.00 L bulb, in which the gas pressure is 1.50 atm, is opened. What is the final pressure in the two bulbs, the temperature being constant and the same in both bulbs?

**Solution using Boyle's Law:**

1) P_{1}V_{1} = P_{2}V_{2} twice

(1.00 atm) (2.00 L) = (x) (5.00 L)x = 0.400 atm

(1.50 atm) (3.00 L) = (y) (5.00 L)

y = 0.900 atm

2) Add 'em up!

0.400 atm + 0.900 atm = 1.30 atm

**Solution using the Ideal Gas Law:**

1) PV = nRT twice:

(1.00) (2.00) = n_{1}RTin the first bulb moles gas = n

_{1}= 2.00/RT(1.50) (3.00) = n

_{2}RTin the second bulb moles gas = n

_{2}= 4.50/RT

2) PV = nRT for a third time

total volume = 2.00 + 3.00 = 5.00(P

_{3}) (5.00) = (n_{1}+ n_{2})RT(P

_{3}) (5.00) = (2.00/RT + 4.50/RT)RT(P

_{3}) (5.00) = 6.50P

_{3}= 6.50 / 5.00 = 1.30 atm

Ten examples A list of all examples and problems (no solutions) Problems 16-30 Return to KMT & Gas Laws Menu