Chemistry Law

Boyle’s Law

PROBLEM

1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg?

2. If a gas at 25.0 occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

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?

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?

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

Boyle’s Law

ANSWER

1. (40.0 mm Hg) (12.3 liters) = (60.0 mm Hg) (x); x = 8.20 L, note three significant figures!!

2. If a gas at 25.0 occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

3. ( 400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot); x = 133 atm

4. (1.56 L) (1.00 atm) = (3.00 atm) (x); 0.520 L

5. (11.2 liters) (0.860 atm) = (x) (15.0 L); x = 0.642 atm

Charles’ Law

PROBLEM

1. A gas is collected and found to fill 2.85 L at 25.0. What will be its volume at standard temperature?

2. 3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?

3. Calculate the decrease in temperature when 2.00 L at 20.0 is compressed to 1.00 L.

4. 600.0 mL of air is at 20.0. What is the volume at 60.0?

5. A gas occupies 900.0 mL at a temperature of 27.0. What is the volume at 132.0?

Charles’ Law

ANSWER

1. convert 25.0 to Kelvin and you get 298 K. Standard temperature is 273 K.

2. 3.50 liters / 1000.0 K = x / 426.0 K

3. (2.00 L) / 293.0 K) = (1.00 L) / (x); x = 146.5 K

4. (600.0 mL) / (293.0) = (x) / (333.0 K); x = 682 mL

5. (900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL

Combined Gas Law

PROBLEM

1. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 to 30.0 and a pressure change from 760.0 mm Hg to 360.0 mm Hg?

2. 500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 . The gas is placed into a tank under high pressure. When the tank cools to 20.0 , the pressure of the gas is 30.0 atm. What is the volume of the gas?

3. A gas has a volume of 800.0 mL at minus 23.00 and 300.0 torr. What would the volume of the gas be at 227.0 and 600.0 torr of pressure?

4. 690.0 mL of oxygen are collected over water at 26.0 and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 and 800.0 mm pressure?

5. 45.0 mL of wet argon gas is collected at 729.3 mm Hg and 25.0. What would be the volume of this dry gas at standard conditions?

Combined Gas Law

ANSWER

1. x = [ (760 mm Hg) (400 mL) (303 K) ] / [ (295 K) (360 mm Hg) ]; x = 867.3 mL

2. x = [ (700/760) (500) (293) ] / [ (473) (30) ]; x = 9.51 L

3. x = [ (300 torr) (800 mL) (500 K) ] / [ (250 K) (600 torr) ]; x = 800.0 mL

4. Pdry = 725.0 - 25.2 = 699.8 mmHg

5. Pdry = 729.3 - 23.8 = 705.5

Gay-Lussac’s Law

PROBLEM

1. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 to 30.0.

2. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0, what would the final temperature of the gas be?

3. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 is placed inside an oven whose temperature is 50.0. The pressure inside the container at 20.0 was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?

4. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from -100.0 to 1.00 x 103. What is the final pressure inside the tank?

5. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg?

Gay-Lussac’s Law

ANSWER

1. atm / 293 K = x / 303 K; x = 1.03 atm

2. 15.0 atm / 298 K = 16.0 atm / x

3. 3.00 atm. / 293 K = x / 323

4. 30.0 kPa / 173 K = x / 1273

5. 750.0 mm Hg / 323.0 K = x / 273.15 K

Ideal Gas Law

PROBLEM

1. How many moles of gas are contained in a 50.0 L cylinder at a pressure of 100.0 atm and a temperature of 35.0 ?

2. Determine the number of moles of Krypton contained in a 3.25 liter gas tank at 5.80 atm and 25.5 . If the gas is Oxygen instead of Krypton, will the answer be the same? Why or why not?

3. Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm and minus 50.5 . Determine the number of grams of oxygen that the same container will contain under the same temperature and pressure.

4. Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.

5. A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample

Ideal Gas Law

ANSWER

1. n = PV / RT

n = [ (100.0 atm) (5.00 L) ] / [ (0.08206 L atm mol 1 K 1) (308.0 K) ]

2. n = PV / RT

n = [ (5.80 atm) (3.25 L) ] / [ (0.08206 L atm mol 1 K 1) (298.5 K) ]

The moles of gas would be the same if the gas was switched to oxygen. Since the temperature and pressure would be the same, the same volume willcontain the same number of molecules of gas, i.e. moles of gas. This is Avogadro's Hypothesis.

3. n = PV / RT

n = [ (0.4506 atm) (1.80 L) ] / [ (0.08206 L atm mol 1 K 1) (222.5 K) ]

This calculates the number of moles of . Multiply the moles by the molecular weight of to get the grams. Under the same set of conditions, the moles of oxygen would be the same, so multiply the calculated moles by the molecular weight of to get the grams.

4. Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.

V = nRT / P

V = [ (2.34 g / 44.0 g mol 1) (0.08206 L atm mol 1 K 1) (273.0 K) ] / 1.00 atm

5. n = PV / RT

n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol 1 K 1) (273.0 K) ] Multiply the moles by the atomic weight of Ar to get the grams.