ch.12: study guide (answers)
DESCRIPTION
Ch.12: Study Guide (answers). small. insignificant. compressibility. forces. expand. random. elastic. Kinetic energy. Kelvin. AT. AT. NT. NT. AT. c. d. e. a. b. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/1.jpg)
Ch.12: Ch.12:
Study GuideStudy Guide(answers)(answers)
![Page 2: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/2.jpg)
small
insignificant
compressibilityforces
expandrandom
elasticKinetic energyKelvin
![Page 3: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/3.jpg)
AT
AT
AT
NT
NT
![Page 4: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/4.jpg)
c
d
ea
b
![Page 5: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/5.jpg)
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.
![Page 6: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/6.jpg)
doubles
half
decreasing
doubles
temperature
doubles
![Page 7: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/7.jpg)
AT
AT
NT
NT
NT
![Page 8: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/8.jpg)
a
b
a
b
![Page 9: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/9.jpg)
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.
![Page 10: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/10.jpg)
inversely
increases
Boyle’s
amount
Kelvin
Charles
Gay-Lussac’s
directly
Combined
none
![Page 11: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/11.jpg)
NT
AT
ST
NT
NT
AT
![Page 12: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/12.jpg)
c
be
a
d
![Page 13: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/13.jpg)
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
![Page 14: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/14.jpg)
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
![Page 15: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/15.jpg)
number of moles
PV = nRT
n
ideal gas constant
8.31
ideal
no
forces
volume
kPa·Lmol·K
![Page 16: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/16.jpg)
AT
AT
NT
NT
AT
![Page 17: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/17.jpg)
dc
ba
![Page 18: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/18.jpg)
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
![Page 19: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/19.jpg)
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
![Page 20: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/20.jpg)
Avogadro’s
Temperaturepressure
1 mole22.4 L
suminversely
molar mass
Graham’s law of effusion
total
![Page 21: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/21.jpg)
NT
AT
NT
AT
AT
![Page 22: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/22.jpg)
d
ab
c
![Page 23: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/23.jpg)
![Page 24: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/24.jpg)
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
![Page 25: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/25.jpg)
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
![Page 26: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/26.jpg)
g
i
jf
dec
ahb
![Page 27: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/27.jpg)
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
![Page 28: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/28.jpg)
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
![Page 29: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/29.jpg)
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
![Page 30: Ch.12: Study Guide (answers)](https://reader034.vdocuments.net/reader034/viewer/2022042508/568139bc550346895da15faa/html5/thumbnails/30.jpg)
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