Ch.12: Ch.12:

Study GuideStudy Guide(answers)(answers)

small

insignificant

compressibilityforces

expandrandom

elasticKinetic energyKelvin

AT

AT

AT

NT

NT

c

d

ea

b

1) Gas volume is insignificant compared to the space between the particles. This explains why all gases take up the same space at a given P and T. 2) There is no significant attractive or repulsive forces between gas particles. This explains why gases spread out to fill their container.

3) Gases move constantly in random straight lines and collide with each other and the walls without losing energy. This explains why gases don’t slow down and become a liquid or solid.

doubles

half

decreasing

doubles

temperature

doubles

AT

AT

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a

b

a

b

This means that if the Kelvin temperature increases the pressure also increases or if Kelvin temperature decrease so does the pressure.

Kelvin temperature is directly related to pressure.

The driver can increase the pressure or decrease temperature (refrigerated trailer) so they can get more gas in each load.

inversely

increases

Boyle’s

amount

Kelvin

Charles

Gay-Lussac’s

directly

Combined

none

NT

AT

ST

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c

be

a

d

P1 = 55 kPaT1 = -100.0 °C

Change Temp to KELVIN!!+ 273

173 KP2 = _____ kPa

T2 = 200 °C+ 273 = 473 K

P1

T1

=P2

T2

55

173=

P2

473

P2 =(473)(55)

173

(473)(473)

= 150 kPa

V1 = _____T1 = 0 °C = 273KP1 = 101.3 kPa

V2 = 75.0 mL

P2 = 91 kPa

T2 = 30 °C + 273 = 303 K

V1

273=

(91)

303

(75)(101.3)P1

T1

=P2

T2

V1V2

V1 =91x 75

(303 x 101.3)

x 273 = 60.7 mL

mL

number of moles

PV = nRT

n

ideal gas constant

8.31

ideal

no

forces

volume

kPa·Lmol·K

AT

AT

NT

NT

AT

dc

ba

n = _____ mol O2

V = 12.5 L

R = 8.31

P = 25,325 kPa

T = 22 °C+ 273 = 295 K

(12.5)

(8.31) (295)

(25,325)

R=

P

Tn

V

n = 129 mol O2

kPa·Lmol·K

PV = n

=(8.31 x 295)

RT

n = _____ mol NO2

V = 275 mL

R = 8.31

P = 240.0 kPa

T = 28 °C+ 273 = 301 K

(0.275)

(8.31) (301)

(240)

R=

P

Tn

V

n =0.0264 mol NO2

kPa·Lmol·K

PV = n

=(8.31 x 301)

RT You can convert moles to grams for any formula

Moles or grams indicates that you are using the Ideal gas law.

1 L 1000 mL

= 0.275 L

46.01 g NO2 1 mol NO2

x

x = 1.21 g NO2

Avogadro’s

Temperaturepressure

1 mole22.4 L

suminversely

molar mass

Graham’s law of effusion

total

NT

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AT

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d

ab

c

Kinetic Energy of Gas Kinetic Energy of Gas ParticlesParticles

At the same conditions of temperature, all gases have the same average kinetic energy.

2

2

1mvKE

m = mass

v = velocity

KE = ½ mv2

A BBoth objects at the same temperature.

Same temp. means same KEave

KEA = KEB

½ mAvA2 = ½ mBvB

2

multiply both sides by 2

2• 2•

mAvA2 = mBvB

2

divide both sides by mA and VB2

mAvB2 mAvB

2 vA

2 mB

mAvB2 =

Take the square root of both sides.

A

B

B

A

m

m

v

v

g

i

jf

dec

ahb

83.73g - 83.32g = 0.41g

83.82g - 83.39g = 0.43g

267 mL x 1 L1000 mL

= 0.267 L

99.0 + 273 = 372.0 K

99.0 + 273 = 372.0 K

773.5mm Hgx101.3 kPa 760 mm Hg

=103.1 kPa

812.0mm Hgx101.3 kPa

760 mm Hg=108.2 kPa

267 mL x 1 L1000 mL

= 0.267 L

99.0 + 273 = 372.0 K

99.0 + 273 = 372.0 K

773.5mm Hgx101.3 kPa 760 mm Hg

=103.1 kPa

812.0mm Hgx101.3 kPa

760 mm Hg=108.2 kPa

PV = nRTn =PVRT

(103.1)(0.267)(8.31) (372.0)=0.00890 moles=

n =(108.2)(0.267)(8.31) (372.0)=0.00935 moles

267 mL x 1 L1000 mL

= 0.267 L

99.0 + 273 = 372.0 K

99.0 + 273 = 372.0 K

773.5mm Hgx101.3 kPa 760 mm Hg

=103.1 kPa

812.0mm Hgx101.3 kPa

760 mm Hg=108.2 kPa

PV = nRT

n =PVRT

(103.1)(0.267)(8.31) (372.0)=0.00890 moles=

n =(108.2)(0.267)(8.31) (372.0)=0.00935 moles

Molar mass = gramsmole

83.73g - 83.32g = 0.41g

83.82g - 83.39g = 0.43g

0.41 g0.00890 moles=46.1 g/mol

0.43 g0.00935 moles=46.0 g/mol

Molar mass = gramsmole

0.41 g0.00890 moles=46.1 g/mol

0.43 g0.00935 moles=46.0 g/mol

(46.1 + 46.0)/2

C = 12.01

H = 1.01

O = 16.00

C

H

H

O-H

H C

H

O-H

H

C

H

H

H

methanol ethanol

= 46.05 g/mol

Which one has a molar mass of 46 g/mol?

12.011.01

1.01

1.01

1.01

16

12.0112.01

161.01

1.011.01

1.01

1.01 1.01

=32.05 g/mol = 46.08 g/mol

Ethanol