The Gas Laws

As P (h) increases V decreases

Apparatus for Studying the Relationship

Between Pressure and Volume of a Gas

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperature

Constant amount of gas

Example 1

A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what

will the pressure be at 2.5L? Given:

V1 = 4.0L V2 = 2.5L P1= 210 kPa P2 = ?

P1V1 = P2V2

(210 kPa) × (4.0L) = P2 × (2.5L) P2 = 336 kPa ≈ 340 kPa

Example 2

A sample of neon gas occupies 0.200L at 0.860 atm. What will be its volume at 29.2 kPa

pressure?

Given: V1 = 0.200L V2 = ? P1= 0.860 atm P2 = 29.2 kPa

P1V1 = P2V2

(87.1 kPa) × (0.200L) = (29.2 kPa) × V2 V2 = 0.597L

**Units must match for each variable (doesn’t matter which one is converted)

0.860 atm 101.3 kPa 1 atm

= 87.1 kPa

As T increases V increases

Variation in Gas Volume with

Temperature at Constant Pressure

K = 0C + 273

**Temperature

must be

in Kelvin

Charles’s Law

Constant pressure

Constant amount of gas

Example 1

A gas ample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant?

Given: T1 = 40.0°C = 313K T2 = 75.0°C = 348K V1= 2.32L V2 = ?

V1 V2

T1 T2 =

2.32L V2

313 348 =

V2 = 2.58L

Example 2

A gas ample at 55.0°C occupies a volume of 3.50L. At what new temperature in kelvin will the volume increase to 8.00L?

Given: T1 = 55.0°C = 328K T2 = ? V1= 3.50L V2 = 8.00L

V1 V2

T1 T2 =

3.50L 8.00

328 T2 =

T2 = 750K

Gay-Lussac’s Law

K = 0C + 273

**Temperature

must be

in Kelvin

Constant volume

Constant amount of gas

Example 1

The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank?

Given: P1 = 3.20 atm P2 = ? T1= 22.0°C = 295K T2 = 60.0°C =333K

P1 P2

T1 T2 =

3.20 atm P2

295K 333K =

P2 = 3.61 atm

Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C?

Given: P1 = 665 torr P2 = ? T1= 22.0°C = 295K T2 = 44.6°C =317.6K

P1 P2

T1 T2 =

665 torr P2

295K 317.6K =

P2 = 716 torr

What is the same about these gases? What is different about these gases?

What is the relationship between number of moles and volume?

Avogadro’s Law

Constant temperature

Constant pressure

1 mol of gas at STP = 22.4L

Gas Stoichiometry

When gases are involved, the coefficients in a balanced

chemical equation represent not only molar amounts

(mole ratios) but also relative volumes (volume ratios).

Example 1

5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and

pressure. Given:

V1 = 5.00 L V2 = ? n1= 0.965 mol n2 = 1.80 mol

V1 V2

n1 n2 =

5.00 L V2

0.965 mol 1.80 mol =

V2 = 9.33 L

Example 2

Calculate the volume that 0.881 mol of gas at STP will occupy.

0.881 mol 22.4 L 1 mol

= 19.7 L

Example 3

How many grams of carbon dioxide gas are in a 0.75 L balloon at STP?

0.75 L CO2 1 mol CO2 44.01 g CO2 22.4 L CO2 1 mol CO2

= 1.5 g CO2

Example 4

What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas

(C3H8)? Assume constant temperature and pressure.

= 20.0 L 4.00 L C3H8 5 L O2

1 L C3H8

Gas Law Summary

Gay-Lussac’s Law

Charles’s Law

Combined Gas Law

Combine all 4 to make one master equation:

Example 1

A gas at 110 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0˚C and the pressure increased to 440

kPa, what is the new volume?

Given: P1 = 110 kPa P2 = 440 kPa T1 = 30.0˚C =303K T2 = 80.0˚C = 353K V1 = 2.00L V2 = ?

110 (2.00) 440 (V2)

303 353

=

V2 = 0.583L ≈ 0.58L

Example 2 An unopened bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm and temperature

of 5.00˚C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and

temperature changes to 1.52 atm and 2.90˚C, what will be the volume of gas in the bottle?

Given: V1 = 46.0 mL V2 = ? P1 = 1.30 atm P2 = 1.52 atm T1 = 5.00˚C = 278K T2 = 2.90˚C = 275.9K

1.30 (46.0) 1.52 (V2)

278 275.9

=

V2 = 39.0mL

The Ideal Gas Law

Ideal Gas

Follows all gas laws under

all conditions of temperature

and pressure.

Follows all conditions of the

Kinetic Molecular Theory

(KMT)

An ideal gas does not exist

in real life

Real Gas

Follows some gas laws under

some conditions of temperature

and pressure.

Does not conform to the Kinetic

Molecular Theory. Real gases

have a volume and attractive and

repulsive forces.

• A real gas differs from an ideal

gas the most at low temperature

and high pressure.

Low T gas is moving less High P gas is more attractive

Ideal Gas law

R is the universal gas

constant PV = nRT

R = PV

nT =

(1 atm)(22.4L)

(1 mol)(273K)

Example 1

Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm.

V = 3.00L T = 298K P = 1.50atm n = ? R = 0.0821 L atm mol K

PV=nRT

(1.50)(3.00) = n (0.0821) (298)

n= 0.184 mol

Example 2

What will the pressure (in kPa) be when there are 0.400 mol of gas in a 5.00L container at 17.0˚C?

V = 5.00L T = 17.0˚C = 290K P = ? kPa n = 0.400 mol R = 8.314 L kPa mol K

PV=nRT

P(5.00) = (0.400)(8.314)(290)

P= 193 kPa

Stoichiometry

Example 1

If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of

3.00 atm and 298K, how many grams of ammonia are produced.

0.613 mol N2 2 mol NH3 17.04 g NH3 1 mol N2 1 mol NH3

= 120.9 g NH3

Example 2

How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a

pressure of 2.00 atm and 298K?

CaCO3(s) CO2(g) + CaO(s)

0.552 mol CO2 1 mol CaCO3 100.09 g CaCO3 1 mol CO2 1 mol CaCO3

= 55.2 g CaCO3

Example 3

How many liters of chlorine will be needed to make 95.0 grams of C2H2Cl4 at 3.50 atm and 225K?

Cl2(g) + C2H4(g) C2H2Cl4(l)

95.0 g C2H2Cl4 1 mol C2H2Cl4 2 mol Cl2 167.84 g C2H2Cl4 1 mol C2H2Cl4

= 1.13mol Cl2

37

Density (d) Calculations

n V

= P

RT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT

P M =

d is the density of the gas in g/L

d = m V

d = PM RT

n = m

M

m M V

= P

RT so

Example 1

What is the molar mass of a pure gas that has a density of 1.40g?L at STP?

Example 2

Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.