Chapter 5 Gases Gas – neither definite shape or volume

Chapter 5Chapter 5

GasesGases

Gas – neither definite Gas – neither definite shape shape oror volume volume

OxygenOxygen

Joseph PriestleyJoseph Priestley

(1733 -1804)(1733 -1804)

21% of atmosphere21% of atmosphere Necessary for life (respiration Necessary for life (respiration

photosynthesis)photosynthesis)

Combines with Si, Al, Fe, Ca, Mg, etc. to Combines with Si, Al, Fe, Ca, Mg, etc. to form rocks and mineralsform rocks and minerals

Lewis structureLewis structure

Colorless, odorless, tasteless gasColorless, odorless, tasteless gas reactivereactive

Supports combustionSupports combustion

Ozone (OOzone (O33))

Poisonous blue gas, pungent odorPoisonous blue gas, pungent odor

HydrogenHydrogen

Most abundant elementMost abundant element Rocket fuel (liquid)Rocket fuel (liquid) Possible fuel of the futurePossible fuel of the future

Colorless, odorless, tasteless gasColorless, odorless, tasteless gas reactivereactive

NitrogenNitrogen

78% of atmosphere78% of atmosphere Non reactiveNon reactive Important to life Important to life

on earthon earth Nitrogen cycle Nitrogen cycle

Carbon DioxideCarbon Dioxide Colorless gasColorless gas Important to lifeImportant to life

Tests for common gases Tests for common gases

Characteristics of GasesCharacteristics of Gases

1.1. ExpansionExpansion

2.2. FluidityFluidity

(cont.)(cont.)

3. Low density3. Low density

(cont.)(cont.)

4.4. CompressibilityCompressibility

5.5. DiffusionDiffusion

Kinetic theoryKinetic theory

Particles of matter are always in motionParticles of matter are always in motion

Ideal gasIdeal gas

Imaginary gas, conforms perfectly to Imaginary gas, conforms perfectly to kinetic theorykinetic theory

Kinetic theory applies only to an ideal Kinetic theory applies only to an ideal gas, but many gases are close to ‘Ideal’gas, but many gases are close to ‘Ideal’

Kinetic Theory of GasesKinetic Theory of Gases1.1. Gases consist of a lg. # of tiny particles, Gases consist of a lg. # of tiny particles,

occupy a volume 1000x the volume of occupy a volume 1000x the volume of liquid or solid w/ the same # of particlesliquid or solid w/ the same # of particles

2.2. Constant motionConstant motion

3.3. Elastic collisions of particles (no net Elastic collisions of particles (no net loss of kinetic E)loss of kinetic E)

(cont.)(cont.)

4.4. No forces of attraction or repulsionNo forces of attraction or repulsion

5.5. Avg. KE is proportional to Kelvin temp. Avg. KE is proportional to Kelvin temp. of gasof gas

4 measurable quantities of 4 measurable quantities of gasesgases

1.1. Volume (V)Volume (V)

2.2. Pressure (P)Pressure (P)

3.3. Temperature (T, Temperature (T, always in Kalways in K))

4.4. # of molecules (moles)# of molecules (moles)

Temperature ScalesTemperature Scales

Absolute zero = Absolute zero = --273.15273.15ooC (rounded C (rounded

to to --273273ooC) C)

= 0 K (= 0 K (not 0not 0ooKK))

Therefore K = 273 + Therefore K = 273 + ooCC 0 0 ooC = 273 KC = 273 K

Temp. conversionsTemp. conversions

K = K = ooC + 273C + 273

25.0 25.0 ooC = ? KC = ? K

273 + 25.0 = 298273 + 25.0 = 298

= 298K= 298K

Units of PressureUnits of Pressure

Pressure at sea level & 0 Pressure at sea level & 0 ooC =760 C =760 mmHg or 1 atmmmHg or 1 atm

Standard temperature and Standard temperature and

pressure (STP) pressure (STP)

= 1 atm & 0= 1 atm & 0ooCC

Convert a pressure of 0.830 Convert a pressure of 0.830 atm to mmHgatm to mmHg

0.830 atm x 760 mmHg/ 1 atm0.830 atm x 760 mmHg/ 1 atm

= 631 mmHg= 631 mmHg

Write an equation for the Write an equation for the relationship between P, V, and T relationship between P, V, and T (assume a constant # moles)(assume a constant # moles)

P x V = a constantP x V = a constant

P/ T = a constantP/ T = a constant

V/ T = a constantV/ T = a constant

PV/ T = a constantPV/ T = a constant

PP11VV11/ T/ T11 = P = P22VV22/ T/ T22

Boyle’s LawBoyle’s Law

Volume off a fixed Volume off a fixed mass of gas varies mass of gas varies inversely with inversely with pressure at a pressure at a constant temperatureconstant temperature

Robert Boyle (1627-1691)Robert Boyle (1627-1691)

Born at Lismore Castle, Munster, Ireland, Born at Lismore Castle, Munster, Ireland, the 14th child of the Earl of Cork. the 14th child of the Earl of Cork.

1662, delineated the quantitative 1662, delineated the quantitative relationship that the volume of a gas relationship that the volume of a gas varies inverselyvaries inversely with pressure. with pressure.

Boyle’s LawBoyle’s Law PP11VV11 = k and P = k and P22VV22 = k = k

Therefore PTherefore P11VV11 = P = P22VV22

A sample of oxygen gas A sample of oxygen gas occupies a vol. of 150 mL at a occupies a vol. of 150 mL at a pressure of 720 mmHg. What pressure of 720 mmHg. What would the volume be at 750 would the volume be at 750 mmHg press.?mmHg press.?

PP11VV11 = P = P22VV22 or V or V22 = P = P11VV11/ P/ P22

= (720 mmHg)(150 mL)/ 750 mmHg= (720 mmHg)(150 mL)/ 750 mmHg

= 144 mL= 144 mL

= 140 mL oxygen gas= 140 mL oxygen gas

France, early 1800’sFrance, early 1800’s

Hot air balloons were extremely popularHot air balloons were extremely popular Scientists were eager to improve the Scientists were eager to improve the

performance of their balloons. Two of the performance of their balloons. Two of the prominent French scientists were prominent French scientists were Jacques Charles and Joseph-Louis Gay-Jacques Charles and Joseph-Louis Gay-Lussac, Lussac,

Charles’ LawCharles’ Law

The volume of a The volume of a fixed mass of gas fixed mass of gas varies directly with varies directly with the Kelvin the Kelvin temperature at temperature at constant pressureconstant pressure

Jacques CharlesJacques Charles

Charles’ LawCharles’ Law

VV11/ T/ T11 = V = V22/ T/ T22

A sample of Ne gas has a vol. of A sample of Ne gas has a vol. of 752 mL at 25.0 deg. C. What is the 752 mL at 25.0 deg. C. What is the vol. at 50.0 deg. C?vol. at 50.0 deg. C?

TT11 = 25.0 deg. C = 298 K = 25.0 deg. C = 298 K

TT22 = 50.0 deg C = 323 K = 50.0 deg C = 323 K

VV11 = 752 mL = 752 mL

VV11/ T/ T11 = V = V22/ T/ T22 oror V V22 = V = V11TT22/ T/ T11

= 752 mL x 323 K/ 298K= 752 mL x 323 K/ 298K

= 815 mL= 815 mL

Gay-Lussac’s LawGay-Lussac’s Law

The pressure of a The pressure of a fixed mass of gas fixed mass of gas varies directly with varies directly with the Kelvin temp. at the Kelvin temp. at constant volumeconstant volume

PP11/ T/ T11 = P = P22/ T/ T22

PP11/ T/ T11 = P = P22/ T/ T22

A sample of N gas is at 3.00 atm of A sample of N gas is at 3.00 atm of pressure at 25 deg C what would pressure at 25 deg C what would the pressure be at 52 deg C?the pressure be at 52 deg C?

PP11 = 3.00 atm = 3.00 atm

TT11 = 25 deg C = 298 K = 25 deg C = 298 K

TT22 = 52 deg C = 325 K = 52 deg C = 325 K

PP22 = ? = ?

PP11/ T/ T11 = P = P22/ T/ T22 oror P P22 = P = P11TT22/ T/ T11

(cont.)(cont.)

= (3.00 atm)(325 K)/ 298K= (3.00 atm)(325 K)/ 298K

= 3.27 atm= 3.27 atm

= 3.3 atm= 3.3 atm

Combined Gas LawCombined Gas Law

PP11VV11/ T/ T11 = P = P22VV22/ T/ T22

A helium filled balloon has a vol. of A helium filled balloon has a vol. of 50.0 L at 25 deg C and 820. mmHg 50.0 L at 25 deg C and 820. mmHg of pressure. What would the vol. of pressure. What would the vol. be at 650 mmHg pressure and 10. be at 650 mmHg pressure and 10. deg C?deg C?

PP11VV11/ T/ T11 = P = P22VV22/ T/ T2 2 oror V V22 = P = P11VV11TT22/ P/ P22TT11

VV11 = 50.0 L = 50.0 L

VV22 = ? = ?

TT11 = 25 deg C = 298 K = 25 deg C = 298 K

TT22 = 10. deg C = 283 K = 10. deg C = 283 K

PP1 1 = 820. mmHg= 820. mmHg

PP22 = 650 mmHg = 650 mmHg

(cont)(cont)

VV22 = (820. mmHg)(50.0 L)(283K) = (820. mmHg)(50.0 L)(283K)//

(650. mmHg(650. mmHg)(298 K))(298 K)

=59.9 L=59.9 L

= 60. L= 60. L

Molar Volume of a GasMolar Volume of a Gas One mole of gas (6.02 x 10One mole of gas (6.02 x 102323 molecules) molecules)

has the same volume at STP (0 deg C, 1 has the same volume at STP (0 deg C, 1 atm) as any other gasatm) as any other gas

Standard Molar Volume of a Standard Molar Volume of a GasGas

= 22.4 L / mol= 22.4 L / mol

therefore 1 mol gas = 22.4 L (at therefore 1 mol gas = 22.4 L (at STP)STP)

What volume would 0.0680 mol What volume would 0.0680 mol of oxygen gas occupy at STP?of oxygen gas occupy at STP?

0.0680 mol x 22.4 L/ 1 mol0.0680 mol x 22.4 L/ 1 mol

= 1.52 L O= 1.52 L O22

Ideal Gas LawIdeal Gas Law

PV = nRTPV = nRT n = number of molesn = number of moles

R = ideal gas constantR = ideal gas constant

= 0.0821 L = 0.0821 L .. atm/ mol atm/ mol .. K K

Volume must be in LVolume must be in L

Pressure must be in atmPressure must be in atm

Temp must be in KTemp must be in K

What is the P (in atm) exerted by a 0.500 mol What is the P (in atm) exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container sample of nitrogen gas in a 10.0 L container

at 298 K?at 298 K?

PV = nRT or P= nRT/ VPV = nRT or P= nRT/ V

= (0.500 mol)(0.0821 L = (0.500 mol)(0.0821 L ..atm/mol atm/mol ..K)(298K)K)(298K)

10.0 L10.0 L

= 1.22 atm= 1.22 atm

Gas StoichiometryGas Stoichiometry

volume-volume calculationsvolume-volume calculations

e.g. L e.g. L L, use like mole ratios L, use like mole ratios

CC33HH88(g) + 5O(g) + 5O22(g) (g) 3CO 3CO22(g) + (g) +

4H4H22O(gO(g)) How many L of oxygen are required for the How many L of oxygen are required for the

complete combustion of 0.250 L of complete combustion of 0.250 L of propane?propane?

5 L O5 L O2 2 : : 1 L C1 L C33HH88

0.250 L C0.250 L C33HH88 xx 5 L O 5 L O22

1 L C1 L C33HH88

= 1.25 L O= 1.25 L O22

CaCOCaCO33(s) (s) CaO(s) + CaO(s) + COCO22(g)(g)

How many grams of calcium carbonate must How many grams of calcium carbonate must be decomposed to produce 2.00 L of CObe decomposed to produce 2.00 L of CO22 at at

STP?STP?

2.00 L CO2.00 L CO22 xx 1 mol CO 1 mol CO22 xx 1 mol CaCO 1 mol CaCO33

22.4 L CO22.4 L CO22 1 mol CO 1 mol CO22

xx 100.086 g CaCO 100.086 g CaCO33

1 mol CaCO1 mol CaCO33

= 8.94 g CaCO= 8.94 g CaCO33

Graham’s Law of effusion or Graham’s Law of effusion or diffusiondiffusion

Graham’s Law of Effusion or Graham’s Law of Effusion or DiffusionDiffusion

The rate of effusion or diffusion is inversely The rate of effusion or diffusion is inversely proportional to the square roots of their proportional to the square roots of their molar massesmolar masses

Rate A Rate A == M MBB

Rate B MRate B MAA

Compare the rate of effusion of Compare the rate of effusion of hydrogen and nitrogen gashydrogen and nitrogen gas

rate Hrate H22 == 28 = 14 = 3.7 28 = 14 = 3.7

rate Nrate N22 2.0 2.0

therefore hydrogen effuses 3.7 times faster therefore hydrogen effuses 3.7 times faster than nitrogenthan nitrogen