Gases

Properties of gases and gas laws.Chapters 18 and 19.

Kinetic Molecular Theory

1. All matter is composed of tiny, discrete particles (atoms, ions, or molecules).

2. These particles are in rapid, random, constant straight line motion. This motion can be described by well-defined and established laws of motion.

3. All collisions between particles are perfectly elastic, meaning that there is no change in the total kinetic energy of two particles before and after their collision.

Physical Characteristics of Gases

Gases have high kinetic energy and:

a. have no definite volume or no shape (compressible).

b. assume the shape and volume of their containers.

c. there is much space between molecules.

Standard Temperature and Pressure of Gases

The volume of a gas(V), the number of gas particles in that volume(n), the pressure of the gas(P), and the temperature of the gas(T) are variables that depend on one another.

The average kinetic energy of all the molecules is proportional to the temperature.

Pressure is the force of the collisions between the gas particles and the sides of the container.

Standard atmospheric pressure =

101.3 kilopascals, 1 atm, 760 mmHgStandard temperature =

0°CelsiusWe indicate that a gas has been measured

at standard conditions by the capital letters STP (standard temp and atm. pressure)

Converting Between Units of Pressure

There are three different units of pressure used in chemistry. This is an unfortunate situation, but we cannot change it. You must be able to use all three. Here they are:

1 atmospheres (symbol = atm) 2 millimeters of mercury (symbol = mmHg),

also referred to as torr.3 kiloPascals (symbol = kPa)

Converting Between Units of Pressure

Conversion bridges are:

1 atm = 101.3 kPa = 760 mmHg (torr)

*ignore STP values for sig figs.

EX: Convert 253.4 kPa to mmHg

253.4 kPa ( 760 mmHg ) = 1901 mmHg

1 ( 101.3 kPa )

Convert 78.5 cmHg into atm.

1. Move decimal based on metrics

785 mmHg

2. Convert

785 mmHg ( 1 atm ) = 1.03 atm

1 (760 mmHg)

Converting Between Units of Temperature

– EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVINS, NOT DEGREES CELSIUS.

K = °C + 273EX: Convert 25°C to K

K = 25 + 273K = 298 K

*ignore 273 for sig figs. Add/sub=least decimal

Boyle’s Law

Robert Boyle, a British chemist, examined the relationship between volume and pressures of gases.

His work has been supported for over 300 years.

Boyle’s Law

If the amount and the temperature of a gas remain constant, the pressure exerted by the gas varies inversely as the volume.

Using Boyle’s Law, we can derive the following equation to calculate the volume of gases at different pressures.

P1V1 = P2V2

Applying Boyle’s Law

EX: A gas is collected and found to fill 2.85 L at 245 kPa. What will be its volume at standard pressure?

Step 1: Standard pressure is 101.3 kPa.

Step 2: Isolate your unknown variable (V2).

Step 3: Plug in your known values.

Step 4: Solve for your unknown.

Isolate V2: P1V1 = P2V2

P1V1 = P2V2

P2 P2

P1V1 = V2

P2

Now plug in your numeric values:

245 kPa x 2.85L =

101.3 kPaSolve: 6.89 L

More Boyle’s Problems

A 5.00 L canister of O2 gas has a pressure of 1110 mmHg. What volume would this gas occupy at 1435 mmHg?

If a 3.50 L bottle of N2 has a pressure of 450 kPa, what would be the new pressure if the volume were reduced to 2.25 L?

10.0 L of Helium gas has a pressure of 1500.0 torr, what is the pressure of 7.50L?

Charles’s Law

Jacques Charles, a French physicist, noted the relationship between the volume of a gas and temperature. His work led to the Kelvin scale.

Charles’s Law

The volume of a quantity of a gas, held at a constant pressure, varies directly with the Kelvin temperature.

Using Charles’s Law, we can derive the following equation:

V1T2 = V2T1

Applying Charles’s Law

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

Step 1: Convert Celsius to Kelvin.

273 + 25.0 = 298.0 K

Step 2: Isolate your unknown variable (V2).

Step 3: Plug in your known values.

Step 4: Solve for your unknown.

Isolate unknown variable:

V1T2 = V2

T1

2.85 L x 273 K =

298.0 K 2.61L

More Charles’s Problems

4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?

5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?

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

Combined Gas Law

We can combine Boyle’s and Charles’s Laws to derive:

P1V1T2 = P2V2T1

We use the combined gas law equation when pressure, volume, and temperature are all changing.

Using Combined Gas Law

2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP?

Step 1: Make a table for known values:

P1 = 745.0 mmHg P2 = 760 mmHg

V1 = 2.00 L V2 = X

T2 = 273 K T1 = 298.0 K

Isolate V2

V2 = P1V1T2

P2T1

V2 = 745.0 mmHg x 2.00 L x 273 K

760 mmHg x 298.0 KV2 = 1.80 L

Combined Gas Law Problems

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

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

The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.0 mL to 350.0 mL. What would the final temperature be if the original temperature was 90.0 °C?

Avogadro’s Law

The volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas.

Using Avogadro’s Law, we can derive the following equation:

V1n2 = V2n1

Using Avogadro’s Law

EX: A balloon is filled with 2.0 moles of Helium gas, occupying 44.8 L at constant temp. How many liters will be occupied if the number of moles is reduced to 1.5?

Isolate unknown:

V1n2 = V2n1

n1 n1

44.8 L x 1.5 mol = 2.0 mol

34 L

The Ideal Gas Equation

When we combine Boyle’s, Charles’s, and Avogadro’s Laws, we can derive the ideal gas equation:

PV = nRT*P = pressure, V = volume, n = moles, T =

temperature, and R is the gas constant0.0821 L x atm

mol x K

Your textbook uses

8.31 dm3 x kPa

mol x K

*whatever you select as your gas constant, remember all units must agree when solving problems.

Solving Problems Using The Ideal Gas Equation

EX: Calculate the volume of 1.000 mol of an ideal gas at 1.000 atm and 0.00°C.

*all units must agree with the units of the gas constant. (L, atm, mol, and K)

*convert 0.00°C to K

0.00 + 273 = 273.00 K

*isolate volume from PV=nRT

V = nRT

P

*solve:

V = 1.000 mol x 0.0821 Latm/molK x 273.00 K =

1.000atm

22.41 L

Hints to using PV=nRT:

1. There are no changing variables. No initials or finals of any P, V, or T.

2. n= moles, may see grams or molecules in word problem that would need to be converted to moles. Show work on side.

More Ideal Gas Problems

EX: A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a 0.250 L flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temp of 31°C. How many moles of CO2 were generated?

EX: A flashbulb contains 2.4 x 10-4 moles of O2 gas at a pressure of 2.01 kPa and a temperature of 19°C. What is the volume?

EX: A 5.20 liter container containing 25 grams of carbon dioxide gas is at a temperature of 27.0ºC. What is the pressure of the gas?

EX: 50. grams of nitrogen tetroxide gas is in a 50. liter container at a pressure of 890 torr. What is the temperature of the sample in Kelvin and Celsius?

Calculating Density and Molar Mass

We can use the ideal gas equation to calculate gas density.

Density has the units m/v so we rearrange the equation to:

n = _P_

V RTNow the units are moles/L so we can multiply

each side my molar mass, units g/mole.

nM = PM

V RTMoles cancel leaving units of g/L (density).

Thus, d (density) = PM

RT

Also, M (molar mass) = dRT

P

Hints:

1. When calculating d, make sure to have correct molar masses and they do count for sig figs.

2. Remember the “Big 7” gases.

3. When asked for name or identity of gas, use molar mass formula and look for molar masses on periodic table to find what gas.

4. If calculating M, density may not be given directly, may have to divide a given grams by a given liters in the place of the d.

Density and Molar Mass Problems

What is the density of carbon tetrachloride vapor at 714 torr and 125°C?

What is the density of Sodium hydroxide at 450.0mmHg and 75°C?

What is the name of the diatomic gas if at 450.0mmHg and 75°C its density is 5.20g/L?

Dalton’s Law of Partial Pressure

Our calculations so far have been for pure gases. John Dalton formed a hypothesis about pressure exerted by a mixture of gases.

Dalton’s Law of Partial Pressure: The total pressure in a container is the sum of the partial pressures of all the gases in the container.

Ptotal = P1 + P2 + P3 …

Gases in a single container are all the same temperature and have the same volume, therefore, the difference in their partial pressures is due only to the difference in the numbers of molecules present.

Partial Pressure of Air

Air is an example of a mixture of gases.

-nitrogen is 78.084 %

-oxygen is 20.948 %

-argon is 0.934 %

-carbon dioxide is 0.0315

-neon, helium, krypton, and xenon are among the other trace gases.

The total pressure of the atmosphere at STP is 101.3 kPa. If 78% of air is nitrogen, then 78% of pressure is due to nitrogen molecules.

0.78 x 101.3 = 79 kPa21% of air is oxygen so 21% of pressure

is due to oxygen molecules.

0.21 x 101.3 = 22 kPa

79 kPa + 22 kPa = 101 kPa

Collecting Gases by Water Displacement

One method to collect gases is by water displacement.

Gases must be insoluble in water.When collection is complete, water vapor

is present in the collection container and must be accounted for in the partial pressures of gases.

Volume of a dry gas:

P dry gas = Ptotal from problem – Pwater (table)

EX: A quantity of gas is collected over water at 8°C in a 0.353 L vessel at 84.5 kPa.What volume would the dry gas occupy at standard atmospheric pressure and 8°C?

Find the pressure of the dry gas:

Pgas = Ptotal – Pwater

Obtain Pwater from table. P = 84.5 kPa – 1.1 kPa

83.4 kPaThe remainder of this problem is a

pressure and volume comparison.Boyle’s equation will now be used.

P1V1 = P2V2

P1V1 = V2

P2

83.4 kPa x 0.353 L = .291 L

101.3 kPa

*Hint: adjust initial pressure, P1 or P only!

Look for dry gas, gas by water displacement.

Partial Pressure Practice Problems

A gas is collected over water and occupies a volume of 596 cm3 at 40°C. The atmospheric pressure is 101.1 kPa. What volume will the dry gas occupy at 40°C and standard atmospheric pressure? The vapor pressure of water at 40°C is 8.6 kPa.

825 mL of oxygen gas was collected over water with a temperature of 25.5°C and a pressure of 835 mmHg. What is the volume of the dry gas at standard conditions?

1.50 moles of a gas was collected over water at 25°C and 95.5 kPa of pressure. Calculate the volume of the dry gas.

Dalton’s Law of Partial Pressure and Mole Fraction

Find the partial pressure of 5.00 moles of oxygen gas in a mixture with 10.0 moles of nitrogen gas and 15.0 moles of argon gas that has a total pressure of 825 mmHg.

Use equation: mole gas x PT = Pgas

total moles

Mole Fraction

Total Pressure

Pressure of Gas

Step 1: Add total number of moles together:

5.00 mol + 10.0 mol + 15.0 mol = 30.0 moles

Step 2: Place values into equation and solve.

5.00 mole O2 x 825 mmHg = 138 mmHg

30.0 mole Total

Practice:1. Find the mole fraction of each gas if 0.25

moles of N2 is mixed with 0.50 moles of Ar.2. What is the mole fraction of each gas if 25.0

grams of CO2, 40.0 grams of Ne, and 35.0 grams of H2 are in a mixture?

3. Based on the gases from problem 2, what would be the total pressure if hydrogen gas had a pressure of 120.0 kPa?

4. A mixture of Ar,and NH3 contains 25.0 grams of each gas.

a. What is the total number of moles of gas?

b. What is the mole fraction for each gas?

c. If the total pressure was 2.05 atm, what is the partial pressure of each gas?

d. If ammonia has a pressure of 950 torr, what is the total pressure of the mixture and what is the partial pressure of each of the other gas?

Diffusion and Graham’s Law

Kinetic theory states that molecules travel in straight lines. Molecules often collide with other molecules which alters its path an sends it on another straight path.

This is the basics of diffusion.As gas molecules diffuse, they become

more and more evenly distributed throughout their container.

Graham’s Law

The relative rates at which two gases under identical conditions of temp and pressure will diffuse vary inversely as the square roots of the molecular masses of the gases.

Non polar molecules diffuse faster than polar molecules.

Molecules of small mass diffuse faster than molecules of large mass because they travel faster.

Smaller molecules can also pass through more substances with greater ease.

Using Graham’s Law, we can derive:

v1 = _m2_

v2 m1

*Molecule with lower molar mass will be m1

Using Graham’s Law

EX: Find the relative rate of diffusion for the gases of krypton and bromine.

vKr = mBr2

vBr2 mKr

vKr = 159.80

vBr2 83.80

The square root of 1.907 = 1.381.Since krypton is the lighter gas, it will

diffuse 1.381 times as fast as bromine.

EX: Compute the relative rate of diffusion of helium to argon.

EX: Compute the relative rate of diffusion of argon to radon.

Deviations from Ideal Gas Behavior

When using the ideal gas equation, two assumptions were made.1. Gases have no volume.2. Gases have no attractive forces between them.*3. Gases occupy 22.4 liters at STP.

We will now examine how real gases can deviate from these assumptions.

1. Gases occupy no volume. At low pressures, both ideal and real gases

are far apart separated by empty space. As pressure is applied, real gases begin to

take up more of the empty space. Ideal gases are still far apart.

An ideal gas can have zero volume, but a real gas will become a liquid under increased pressure.

2. Gases have no attractive forces between them.

If gases are made up of polar molecules such as water, the attractive forces are large and the behavior of this real gas is markedly different from an ideal gas.

There are even weak attractive forces (dispersion forces) between noble gases.

For most gases, the ideal gas laws are accurate to about 1%.

Gas Stoichiometry

Recall the method of determining the amount of substances consumed, reacted, produced, and left over for chemical reactions using the relationship between moles of a balanced equation and molar masses of substances.

Gases demonstrate a relationship as well. 1 mole of a gas is equivalent to 1 mole of another gas, and the molar masses of gases can be calculated. (same as basic stoic)

Gases are generally measured in units of volume rather than mass.

We have 1 more conversion bridge that can be used for converting between gases:

1 mol = 22.4 L

EX: Given the equation:

N2(g) + H2(g) NH3(g)Calculate how many liters of ammonia gas can be produced from 55 liters of hydrogen gas and excess nitrogen gas.

Step 1: Start with a balanced equation.Step 2: Identify given and unknown substances.Step 3: Place your given substance over 1.Step 4: Convert from liters of given to moles of

given.

Step 5: Convert from moles of given to moles of unknown using coefficients from balanced equation.

Step 6: Convert from moles of unknown to liters of unknown.

Step 7: Perform calculation and express answer in correct sig figs, unit, and substance.

N2(g) + 3H2(g) 2NH3(g)

Given is 55 liters of hydrogen and unknown is liters of ammonia.

55 L H2 ( 1 mol) ( 2 mol NH3) ( 22.4 L )

1 ( 22.4 L) ( 3 mol H2 ) ( 1 mol ) =

37 L NH3

N2(g) + 3H2(g) 2NH3(g)

Using the same equation, calculate liters of N2 needed to react with 35 grams of H2.

Given is 35 g of H2 and unknown is L of N2.

( 35 g H2) ( 1 mol ) (1 mol N2 ) ( 22.4 L )

1 ( 2.02 g) ( 3 mol H2 ) ( 1 mol ) =

130 L N2

Practice Problems:

1. magnesium hydroxide + ammonium sulfate magnesium sulfate + water + ammonia

How much (grams) magnesium hydroxide do you need to use in the above reaction to produce 500. liters of ammonia?

2. How much strontium bromide is needed to add to chlorine gas to produce 75 liters of bromine?

Stoichiometry of Gases and Ideal Gas Law

EX: If 250 grams of ethane gas (C2H6) undergoes combustion, what is the pressure of carbon dioxide gas in a 50. liter container at 27.0ºC?

Step 1: Use basic stoichiometry to convert from grams of ethane to moles of carbon dioxide.

Step 2: Place moles of carbon dioxide into ideal gas equation and solve for unknown variable.

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)

250 g C2H6 ( 1mol ) (4 mol CO2 )

1 (30.08g) (2 mol C2H6) = 17 mol

CO2 produced

PV=nRT

P = nRT = 17mol x .0821Latm/molK x 300.0K

V 50. L

= 8.4 atm

Hints on when to use:

Stoichiometry: When converting from one substance to a completely different substance.

Ideal Gas Equation: When the conditions are not standard (STP).

*If conditions are standard and you are asked to find volume, use basic stoic.

*If conditions are not standard and you are asked to find volume, use stoic to find moles and Ideal to find volume.