Post on 05-Jan-2016
description
transcript
Ch.12: Ch.12:
Study GuideStudy Guide(answers)(answers)
small
insignificant
compressibilityforces
expandrandom
elasticKinetic energyKelvin
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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
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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
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ST
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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
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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
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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