Properties of Gases &

Gas Laws

Unit 10

Kinetic Molecular Theory

1. Gases are composed of a large number of particles that behave like small, solid, spherical objects in a state of constant, random motion.

2. These particles move in a straight line until they collide with another particle or the walls of the container (they follow Newtonian Physics).

3. These particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space.

4. There is no force of attraction between gas particles or between the particles and the walls of the container.

5. Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container.

6. The average kinetic energy of a collection of gas particles depends on the temperature of the gas and nothing else.

Kinetic Molecular Theory

Kinetic Molecular theory requires: An ideal gas that perfectly fits all the

assumptions of the kinetic-molecular theory. A real gas is a gas that does not behave

completely according to the assumptions of the kinetic-molecular theory.

Ideal gases aren’t real, but they offer a good approximation at normal temperatures and pressures.

Properties of Gases Gases expand to fill their containers Gases are compressible Gases have relatively low densities Gases “flow” like liquids so both liquids and

gases are termed “fluid”. Gases undergo diffusion which is the

spontaneous mixing of the particles of two substances due to their random motion.

Pressure Pressure (P) is defined as the force per unit

area on a surface. Pressure = force/area

The unit for force is the newton (N). Some common units of pressure include:

mmHg, atm, kPa, mbar, torr The equality related to these:

760 mmHg = 1 atm = 101.325 kPa = 1013 mbar = 760 torr

Familiar Pressure Measurments A barometer is a

device used to measure atmospheric pressure.

Because atmospheric pressure is often measured with a mercury barometer, a common unit of pressure is millimeters of mercury, mmHg.

Standard Temperature/Pressure The volume of a gas varies based on

temperature and pressure. In order to make valid comparisons of

volumes, scientists have agreed on standard conditions for gases. Standard Temperature and Pressure

(STP) is equal to exactly 1 atm pressure and 0º C (273K).

Memorize this!!!

Avogadro’s Principle

Equal volumes of gases under the same conditions have equal numbers of molecules. The Molar Volume (1mole) of any gas at STP is 22.4 L.

Volume Stoichiometry

If a gas is at STP, we can use the 4 steps of stoichiometry along with Avogodro’s Principle to calculate the volume of gas produced in a chemical equation.

Remember: the conversion factor is 1 mole = 22.4 L @ STP.

Sample Problem #1

How many liters of oxygen gas are needed to completely react 150L of hydrogen at STP according to this equation:

H2 + O2 H2O

First, balance the equation and add the given and unknown values.

22150 L X L

Sample Problem #1 (cont.)

Step 2: 150 L H2 x 1 mole = 6.7 moles H2

22.4 L

Step 3: 6.7 moles H2 x 1 mole O2 = 3.4 moles O2

2 mole H2

Step 4: 3.4 moles O2 x 22.4 L = 75L O2

1 mole

Sample Problem #1 (Short Way)

Set up a proportion with the given on top and the unknown on the bottom:

2 moles H2 = 150 L H2

1 mole O2 X L O2

Now, cross multiply and solve for X:

2 X = 150

X = 75 L

Sample Problem #2

How many liters of nitrogen are needed to completely react with 10.0 L of hydrogen at STP according to this equation:

H2 + N2 NH3

Solution:

3 H2 + N2 2 NH310.0 L X L

Sample Problem #2 Solution

3 moles H2 = 10.0 L H2

1 mole N2 X L N2

3 X = 10.0

X = 3.33 L N2

How many grams of nitrogen?

3.33 L N2 x 1 mole = 0.149 mole N2

22.4 L

0.149 mol x 28.0134 g/mol = 4.17 g N2

Practice

Stoichiometry Diagram WS #1 Gas Stoich @ STP WS #2 Gas Stoich @ STP

Boyle’s Law

The volume of a given sample of a gas at constant temperature is inversely proportional to its pressure.

P1 V1 = P2 V2

Boyle’s Law

Boyle’s Law Problem #1

The volume of a gas at 1.0 atm is 75.3 mL. If temperature remains constant, what will the new volume be if the pressure is increased to 2.5 atm?

V1=

P1=

V2=

P2=

75.3 mL

1.0 atm

2.5 atm

X mL

(75.3)(1) = X(2.5)

75.3 = 2.5 X

X = 30. mL

Boyle’s Law Problem #2

At 780. mmHg, a gas has a volume of 27.8 mL. What pressure is needed to increase the volume to 45.5 mL if the temperature remains constant?

V1=

P1=

V2=

P2=

27.8 mL

780 mmHg

X mmHg

45.5 mL

(27.8)(780) = 45.5 X

21684 = 45.5 X

X = 477 mmHg

Practice

WS #3 Boyle’s Law

Kelvin Temperature Scale & Gas Laws When using temperature

for ANY gas law problem, you must always convert Celsius to Kelvins.

To convert to Kelvins: K = C + 273.

K = Kelvin temperature; C = degrees Celsius.

To convert back to Celsius: C = K – 273.

Charles’ Law

The volume of a given sample of a gas is directly proportional to its temperature at constant pressure.

V1 = V2

T1 T2

Charles’ Law Problem #1

A sample of gas at 45.0ºC has a volume of 25.3 mL. At constant pressure, what would the new volume be at 25.0ºC?

V1=

T1=

V2=

T2=

25.3 mL

45.0ºC

25.0ºC

X mL

25.3 = X .318 298

318 X = 7539.4

X = 23.7 mL

+ 273

+ 273

Charles’ Law Problem #2

At 37.0ºC a gas has a volume of 48.8 mL. If the pressure is constant, what Celsius temperature is necessary to reduce the volume to 25.0 mL?

V1=

T1=

V2=

T2=

48.8 mL

37.0ºC

X

25.0 mL

48.8 = 25.0310 X

48.8 X = 7750 X = 158.81 K

+ 273

158.81 – 273 = -114 ºC

Practice

WS #4 Charles’ Law

Combined Gas Law

Combined gas law formula is used when temperature or pressure are not constant.

The formula combines Boyle’s and Charles’ laws:

P1V1 = P2V2

T1 T2

Combined Gas Law Problem #1

At 37.0ºC and 790.0 mmHg, a gas has a volume of 355.0 mL. What would the new pressure be if the temperature was reduced to 25.0ºC and the volume to 250.0 mL?

V1=

P1=

T1=

V2=

P2=

T2=

355.0 mL

37.0ºC

25.0ºC

250.0 mL

(790)(355) = 250X . 310 298

77500 X = 83574100

X = 1078 mmHg

+ 273

+ 273

790.0 mmHg

X mmHg

Combined Gas Law Problem #2

At 28.0ºC and 980.5 kPa, a gas has a volume of 285.0 mL. What would the new volume be at STP?

V1=

P1=

T1=

V2=

P2=

T2=

285.0 mL

28.0ºC

0ºC

X mL

(980.5)(285) = 101.325X . 301 273

30498.825 X = 76287802.5

X = 2501 mL = 2.50 x 103 mL

+ 273

+ 273

980.5 kPa

101.325 kPa

Practice

WS #5 Combined Gas Law

Dalton’s Law of Partial Pressures

Dalton’s law states that the pressure of a system is equal to the sum of the pressures of the gases in the system.

Ptotal = Pgas1 + Pgas2 + … The pressures of gases in our lungs at 37ºC at sea level

would be Ptotal = PN2 + PO2 + PCO2 + PH2O 573 mmHg + 100 mmHg + 40 mmHg + 47 mmHg = 760 mmHg (atmospheric pressure)

Gases Collected by Water Displacement

When a gas is collected over water, water vapor molecules are mixed with the gas.

Just as the gas exerts pressure, so does the water vapor.

To find the pressure of just the collected gas, use Dalton’s Law and subtract the pressure exerted by the water vapor.

1mL=1cm3

Gases Collected by Water Displacement

Dalton’s Law Problem #1

Oxygen gas was collected over water when the barometer read 97.5 kPa, and 20.0ºC. What is the partial pressure of the oxygen collected?

At 20.0ºC, the pressure of water is 2.3 kPa. Ptotal – PH2O = PO2

97.5 – 2.3 = 95.2 kPa.

Dalton’s Law Problem #2 A chemist collects 96.0 mL of gas over water at 27.0 ºC

when the pressure is 1220 kPa. What volume would the dry gas occupy at 70.0 ºC and 1270 kPa?

V1=

P1=

T1=

V2=

P2=

T2=

96.0 mL

27.0ºC

70.0ºC

X mL

(1216.4)(96) = 1270X . 300 343

381000 X = 40053619.2

X = 105 mL

+ 273

+ 273

1220 kPa

1270 kPa

- 3.6

Practice

WS #6 Dalton’s Law—Gas Collection

Ideal Gas Law

The Ideal Gas Law is a mathematical relationship between pressure, volume, temperature and number of moles of gas.

The formula is: PV = nRT

P = Pressure (in atm, mmHg or kPa) V = Volume (must be in Liters) n = number of moles R = Proportionality constant depending on units of

pressure: (8.314 L/kPa, 0.0821 L/atm or 62.4 L/mmHg) T = temperature (in Kelvins)

Ideal Gas Law Problem #1

A 500. g block of dry ice (solid CO2) vaporizes to a gas at room temperature. Calculate the volume of gas produced at 25.0ºC and 975 kPa.

P =

V =

n =

R =

T =

X L500 g/44.0098 = 11.4 mol

25ºC

8.314 L/kPa

+ 273

975 kPa 975 X = (11.4)(8.314)(298)

975 X = 28244.3208

X = 28.97 = 29.0 L

Ideal Gas Law Problem #2

A sample of CO2 with a mass of 0.250 g was placed in a 350. mL container at 127ºC. What is the pressure exerted by the gas? (Solve for kPa).

P =

V =

n =

R =

T =

.350 L.250 g/44.0098 = .00568 mol

127ºC

8.314 L/kPa

+ 273

X kPa.350 X = (.00568)(8.314)(400)

.350 X = 18.889048

X = 54.0 kPa

Practice

WS #7 Ideal Gas Law

Ideal Gas Law & Stoichiometry

If a reaction occurs at something other than STP, Avogadro’s Principle cannot be used. (22.4L is the molar volume at STP only.)

To solve these problems, you must combine stoichiometry with the Ideal Gas Law.

Key to all problems: Always find moles of what you know and convert to moles of what you want to find. This may require stoichiometry first or Ideal Gas

Law first.

Ideal Gas Law & Stoichiometry

How many liters of hydrogen gas are needed to produce 73.3 g of ammonia at 45.0ºC and 3.00 atm?

First, find moles of what you know and then convert to moles of what you want.

N2 + 3H2 2 NH3

73.3 g NH3 = 4.30 mol NH3 x 3 mol H2 = 17.0304 g 2 mol NH3

6.45 mol H2

Ideal Gas Law & Stoichiometry

Now use the Ideal Gas Law to find volume. How many liters of hydrogen gas are needed to produce

73.3 g of ammonia at 45.0ºC and 3.00 atm? (6.45 mol H2)

P =

V =

n =

R =

T =

X L

6.45 mol

45.0ºC

0.0821 L/Atm

+ 273

3.00 Atm 3 X = (6.45)(.0821)(318)

3 X = 168.39531

X = 56.1 L

Ideal Gas Law & Stoichiometry

How many liters of CO2 are produced when 55.6 g of sulfuric acid reacts with sodium bicarbonate at 23.0ºC and 758 mmHg?

First, find moles of what you know and then convert to moles of what you want.

H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2

55.6 g H2SO4 = 0.567 mol H2SO4 x 2 mol CO2 = 98.0794 g 1 mol H2SO4

1.13 mol CO2

Ideal Gas Law & Stoichiometry

How many liters of CO2 are produced when 55.6 g of sulfuric acid reacts with sodium bicarbonate at 23.0ºC and 758 mmHg? (1.13 mol CO2)

P =

V =

n =

R =

T =

X L

1.13 mol

23.0ºC

62.4 L/mmHg

+ 273

758 mmHg 758 X = (1.13)(62.4)(296)

758 X = 20871.552

X = 27.5 L

Ideal Gas Law & Stoichiometry

Practice

WS #8 Ideal Gas Law Stoich

Limiting Reactant

How many grams of CO2 are formed if 10.0 g of carbon are burned in 20.0 dm3 of oxygen gas at STP? (what does dm3 represent?)

1. Write a balanced equation:C + O2 CO2

2. Change both quantities to moles:

10.0 g C = .833 mol C x 1 mol CO2 = .833 mol CO2

12.011g/mol 1 mol C

Limiting Reactant (Cont.)

2. Convert to moles to find the Limiting Reactant

20.0 dm3 O2 = .893 mol O2 x 1 mol CO2 = .893

22.4 L 1 mol O2 mol CO2

3. Carbon is the limiting reactant

4. Complete the problem on the basis of the limiting reactant.

.833 mol C x 1 mol CO2 = .833 mol CO2 x 44.0098 =

1 mol C 36.7 g CO2

Limiting Reactant (Cont.)

How many liters of excess reactant remains? You have .893 mol O2

You need .833 mol O2

Excess: .893 - .833 = .060 mol O2

.060 mol O2 x 22.4 L = 1.3 L O2

OR: 0.893 x 22.4L = 20.0 L 0.833 x 22.4 L = 18.7 L 20.0 L – 18.7 L =1.3 L O2

Practice