The Gas Laws

As P (h) increases V decreases

Apparatus for Studying the Relationship

Between Pressure and Volume of a Gas

P ∝ 1/V

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 = 2.5LP1= 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 kPa1 atm

= 87.1 kPa

Day 2• Comparing

volume and temperature

• Keep pressure constant

As T increases V increases

Variation in Gas Volume with

Temperature at Constant Pressure

𝑉 ∝ 𝑇𝑉

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

K = 0C + 273

**Temperature

must be

in Kelvin

Charles’s Law

𝑉1

𝑇1=

𝑉2

𝑇2

Constant pressure

Constant amount of gas

Example 1

A gas sample 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 = 348KV1= 2.32L V2 = ?

V1 V2

T1 T2=

2.32L V2

313 348=

V2 = 2.58L

Example 2

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

Given:T1 = 55.0°C = 328.0K T2 = ?V1= 3.50L V2 = 8.00L

V1 V2

T1 T2=

3.50L 8.00328.0 T2

=

T2 = 750K-273=477°C

Day 3

Gay-Lussac’s Law

𝑃 ∝ 𝑇

𝑃

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

K = 0C + 273

**Temperature

must be

in Kelvin

𝑃1

𝑇1=

𝑃2

𝑇2

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 = 295.0K T2 = 60.0°C =333.0K

P1 P2

T1 T2=

3.20 atm P2

295.0K 333.0K=

P2 = 3.61 atm

Example 2A 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

STOP

Avogadro’s Law and Combine Gas Law

What is the relationship between number of moles and volume?

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

Avogadro’s LawV ∝ n (number of moles)

𝑉

𝑛= constant

Constant temperature

Constant pressure𝑉1

𝑛1=

𝑉2

𝑛2

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 L1 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.

𝐶3𝐻8 + 5 𝑂2 → 3 𝐶𝑂2 + 4 𝐻2𝑂

= 20.0 L4.00 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:

𝐵𝑜𝑦𝑙𝑒′𝑠 𝑙𝑎𝑤: 𝑃 × 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑇

𝐶ℎ𝑎𝑟𝑙𝑒′𝑠 𝑙𝑎𝑤:𝑉

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑃

𝐺𝑎𝑦 − 𝐿𝑢𝑠𝑠𝑎𝑐′𝑠 𝑙𝑎𝑤:𝑃

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑉

𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′𝑠 𝑙𝑎𝑤:𝑉

𝑛= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 𝑎𝑛𝑑 𝑇)

𝑃𝑉

𝑛𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑃1𝑉1𝑛1𝑇1

=𝑃2𝑉2𝑛2𝑇2

𝑃1𝑉1𝑇1

=𝑃2𝑉2𝑇2

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 kPaT1 = 30.0˚C =303K T2 = 80.0˚C = 353KV1 = 2.00L V2 = ?

110 (2.00) 440 (V2)

303 353=

V2 = 0.583L ≈ 0.58L

𝑃1𝑉1𝑇1

=𝑃2𝑉2𝑇2

Example 2An 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 atmT1 = 5.00˚C = 278K T2 = 2.90˚C = 275.9K

1.30 (46.0) 1.52 (V2)

278 275.9=

V2 = 39.0mL

𝑃1𝑉1𝑇1

=𝑃2𝑉2𝑇2

STOP

Warm Up

A sample of gas of unknown pressure occupies 3.2 L at a temperature of 302 K. The sample of gas is then tested under known condition and has a pressure of 22.6 kPa and occupies 1.2 L at 310 K. What was the original pressure of the gas?

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,

high pressure, and high

concentration.

o Low T gas is moving less

High P gas is more attractiveHigh Conc has more collisions

Ideal Gas law

𝑅 =0.0821𝐿 𝑎𝑡𝑚

𝑚𝑜𝑙 𝐾=8.314𝐿 𝑘𝑃𝑎

𝑚𝑜𝑙 𝐾=

62.4𝐿 𝑚𝑚𝐻𝑔

𝑚𝑜𝑙 𝐾=

62.4𝐿 𝑡𝑜𝑟𝑟

𝑚𝑜𝑙 𝐾

𝑃𝑉

𝑛𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (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 = 3.00LT = 298KP = 1.50atmn = ?R = 0.0821 L atm

mol K

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

n= 0.184 mol

PV=nRT

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 = 5.00LT = 17.0˚C = 290KP = ? kPan = 0.400 molR = 8.314 L kPa

mol K

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

P= 193 kPa

PV=nRT

STOP

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.

𝑁2 + 3𝐻2 → 2𝑁𝐻3 PV=nRT

(3.00 atm)(5.00 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚

𝑚𝑜𝑙 𝑘)(298 K)

n= 0.613 mol N2

0.613 mol N2 2 mol NH3 17.04 g NH3

1 mol N2 1 mol NH3

= 20.9 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)PV=nRT

(2.00 atm)(6.75 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚

𝑚𝑜𝑙 𝑘)(298 K)

n= 0.552 mol CO2

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?

2Cl2(g) + C2H2(g) C2H2Cl4(l)

95.0 g C2H2Cl4 1 mol C2H2Cl4 2 mol Cl2

167.84 g C2H2Cl4 1 mol C2H2Cl4= 1.13mol Cl2

PV=nRT

(3.50 atm)(V) = (1.13 mol)(0.0821 𝐿 𝑎𝑡𝑚

𝑚𝑜𝑙 𝑘)(225 K)

V= 5.97 L Cl2

Graphic Organizer for PV=nRT and Stoichiometry

47

Density (d) Calculations

nV

=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

PM =

d is the density of the gas in g/L

d = mV

d = PMRT

n = m

M

mM V=

P

RTso

Example 1

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

d = PMRT

Example 2

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

Test Review1. Using the principles behind Graham’s Law, knowing

that M H2 = 2.02 g/mol and M N2 = 28.02 g/mol which gas would diffuse faster in a closed container?

2. Why do real gases deviate from ideal gas behavior at high pressure, low temperature and high

concentration ?

3. When can you use 22.4 L = 1 mole and when can you use volume to volume ratios?

Hydrogen

High P-increased collisions, increase temp Low T-slow particles down,

less collisions and pressure

High C- particles collisions increase and so doespressure

At STP = 1 atm and 273 K

Test Review

B. What volume of NH3 gas is produced at 55.5oC and 670

torr when 0.6789 g of H2 reacts with N2?

4. Given the two problems below, which problem starts with stoichiometry and then uses PV=nRt, and which problem starts with PV=nRt and then uses stoich?

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

A. What is the mass of NH3 gas is produced at 55.5oC and 670 torr, when 1L of H2 reacts with an excess of N2?

PV=nRT, then Stoich

Stoich, then PV=nRT