AP Chemistry WS 4.1 Key All gas molecules or atoms are always in

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AP Chemistry
WS 4.1 Key
1.
a) All gas molecules or atoms are always in constant, random motion. KE is proportional to Temp. Note:
assumptions for the KMT also apply to real gases under all but exceptional conditions of low
temperatures and high pressures.
b) Ideal gas molecules have negligible volume in comparison to their container and gas ideal gas molecules
don’t attract or repel each other. NOTE: these postulates are for Ideal Gases only, but again they
apply to real gases under most conditions.
πŸ‘
2. a) π‘²π‘¬π’‚π’—π’ˆ = 𝑹𝑻
πŸ‘π‘Ήπ‘»
π“œ
b) π’–π’“π’Žπ’” = √
3.
4.
5.
6.
𝒓𝒂𝒕𝒆𝒂
𝒓𝒂𝒕𝒆𝒃
= √π“œπ’ƒ
𝒓𝒂𝒕𝒆𝒂
𝒓𝒂𝒕𝒆𝒃
= √π“œπ’ƒ
(32𝑠)
4.67
π“œ
π“œ
molβˆ™K
) (274𝐾) = 3415.41 𝐽/π‘šπ‘œπ‘™ with Sigfigs. 3420 J/mol
(πŸ‘)(πŸ–.πŸ‘πŸ 𝑱/π’Žπ’π’ 𝑲)(πŸπŸ•πŸ’π‘²)
.πŸŽπŸπŸŽπŸπŸ– π’Œπ’ˆ/π’Žπ’π’
=
581.80 π‘š/𝑠 with SigFigs 582 m/s
πŸ‘πŸ
= √πŸ’.𝟎 He effuses 2.8 times faster than O2
𝒓𝒂𝒕𝒆 𝒐𝒇 π‘―πŸ
𝒓𝒂𝒕𝒆 𝒐𝒇 π‘ͺπ‘ΆπŸ
𝒂
J
𝟐
π’–π’“π’Žπ’” = √
𝒓𝒂𝒕𝒆 𝒐𝒇 𝑯𝒆
𝒓𝒂𝒕𝒆 𝒐𝒇 π‘ΆπŸ
𝒂
πŸ‘
π‘²π‘¬π’‚π’—π’ˆ = (8.31
𝟐
πŸ’πŸ’.𝟎𝟏
H2 effuses 4.67 times faster than CO2
= √ 𝟐.𝟎𝟐
= 6.85𝑠
𝒓𝒂𝒕𝒆𝒂
𝒓𝒂𝒕𝒆𝒃
π“œ
𝒓𝒂𝒕𝒆 𝒐𝒇 𝑯𝒆
𝒓𝒂𝒕𝒆 𝒐𝒇 π‘΅π‘―πŸ‘
= √π“œπ’ƒ
𝒂
πŸπŸ•.πŸŽπŸ’
= √πŸ’.πŸŽπŸŽπ’ˆ He diffuses 2.06 times faster than NH3 ; therefore, the reciprocal
will be the diffusion rate of NH3 compared to He: 0.49 times
7. (
8.
20𝑠
.49
) = πŸ’πŸ 𝒔𝒆𝒄
𝒓𝒂𝒕𝒆𝑯𝒆
π’“π’‚π’•π’†π’–π’π’Œπ’π’π’˜π’
π“œ
= √πŸ’.πŸŽπŸŽπ’ƒ
𝟏
𝟎.πŸπŸ“
π“œ
= √πŸ’.πŸŽπŸŽπ’ƒ = πŸ”πŸ’ π’ˆ/π’Žπ’π’
πŸ‘π‘Ήπ‘»
π“œ
9. a) A is O2 and B is He. O2 is heavier and would have slower speeds compared to He. π’–π’“π’Žπ’” = √
πŸ‘π‘Ήπ‘»
π“œ
b) B is higher temperature. Hotter gas has greater KE, therefore greater speed than cooler gas. π’–π’“π’Žπ’” = √
10. 1.22 π‘Žπ‘‘π‘š ×
1.22π‘Žπ‘‘π‘š ×
1.22π‘Žπ‘‘π‘š ×
760π‘šπ‘šπ»π‘”
1π‘Žπ‘‘π‘š
101.325π‘˜π‘ƒπ‘Ž
1π‘Žπ‘‘π‘š
760π‘‡π‘œπ‘Ÿπ‘Ÿ
1π‘Žπ‘‘π‘š
= 927.2 π‘šπ‘šπ»π‘” (927𝑠𝑖𝑔𝑓𝑖𝑔)
= 123.62 π‘˜π‘ƒπ‘Ž (124𝑠𝑖𝑔𝑓𝑖𝑔)
= 927.2 π‘‡π‘œπ‘Ÿπ‘Ÿ (927𝑠𝑖𝑔𝑓𝑖𝑔)
11. 520π‘šπ‘šπ»π‘”
749π‘šπ‘šπ»π‘” − 670π‘šπ‘šπ»π‘” = 79π‘šπ‘šπ»π‘”
749π‘šπ‘šπ»π‘” + 103π‘šπ‘šπ»π‘” = 852π‘šπ‘šπ»π‘”
12. a) The piston would move upward to double the volume because doubling the number of molecules exerts twice
𝐹
the pressure. More collisons ∴ more force and more pressure. 𝑃 =
π‘Ž
b) The piston would move upward to double the volume because the faster moving molecules exert twice the
πŸ‘
pressure. π‘²π‘¬π’‚π’—π’ˆ = 𝑹𝑻
𝟐
c) Double pressure ∴ volume decreases by ½ . Molecules are more crowded, increased # of collisions, greater
force, greater pressure.
13.
P
V
n
T
2.00 atm
1.00 L
1.500 mol
16.2 K
30.3 kPa
1.250 L
0.0152 mol
27oC
650 torr
11.2 L
0.333 mol
350 K
585 mL
0.250 mol
295 K
10.4 atm
14. a) 𝑃𝑉 = 𝑛𝑅𝑇
𝑛=
b) 0.382π‘šπ‘œπ‘™π‘‚2 ×
15.
π‘·πŸ π‘½πŸ
=
π’πŸ π‘»πŸ
𝑇2 =
16.
π‘š
𝑉
=
32.0 𝑔 𝑂2
1 π‘šπ‘œπ‘™
(94.6π‘˜π‘ƒπ‘Ž)(10.0𝐿)
(8.31)(298𝐾)
(πŸ•πŸ“πŸŽπ’•π’π’“π’“)(πŸ‘πŸ“πŸŽπ’Žπ‘³)
π’πŸ π‘»πŸ
π’πŸ (πŸπŸ–πŸ–π‘²)
(750π‘‘π‘œπ‘Ÿπ‘Ÿ)(450π‘šπΏ)(288𝐾)(𝑛1 )
(750π‘‘π‘œπ‘Ÿπ‘Ÿ)(350π‘šπΏ)(𝑛2 )
π‘ƒβˆ™β„³
given
π‘…βˆ™π‘‡
π‘š
π‘ƒβˆ™β„³
= π‘…βˆ™π‘‡
𝑉
π‘ƒβˆ™β„³
π‘…βˆ™π‘‡
18.
=
= 𝟎. πŸ‘πŸ–πŸ π’Žπ’π’
= 𝟏𝟐. 𝟐 π’ˆ π‘ΆπŸ
π‘·πŸ π‘½πŸ
=𝑑=
17. 𝑑 =
𝑃𝑉
𝑅𝑇
=
(πŸ•πŸ“πŸŽπ’•π’π’“π’“)(πŸ’πŸ“πŸŽπ’Žπ‘³)
π’πŸ π‘»πŸ
rearrange and solve for T2
= πŸ‘πŸ•πŸŽ. πŸ‘ 𝑲 This is really just Charles Law
the information, rearrange solving for molar mass(M)
π‘šπ‘…π‘‡
𝑃
=β„³
(1π‘Žπ‘‘π‘š)(17.04𝑔/π‘šπ‘œπ‘™)
(.0821)(273)
=
(4.93𝑔)(.0821)(400𝐾)
1.05π‘Žπ‘‘π‘š
= πŸπŸ“πŸ’. πŸπ’ˆ/π’Žπ’π’
= .76 𝑔/𝐿
Total Pressure
Partial Pressure of He
Density
Average Kinetic energy per molecule
1
1
2
1
19. a. the number of moles of each gas.
n = PV/RT = (265/760)(1.0)/(0.0821)(298)
n = 0.014 mol
n = PV/RT = (800/760)(1.0)/(0.0821)(298)
n = 0.043 mol
n = PV/RT = (532/760)(0.5)/(0.0821)(298)
n = 0.014 mol
N2
Ne
H2
b.
the total pressure after all valves are opened.
P = nRT/V = (0.071)(0.0821)(298)/(2.5)
P = 0.695 atm (528 torr)
c.
the partial pressure of each gas.
PN2 = XN2Ptot = (0.014/0.071)(528 torr)
N2
PN2 = 104 torr
PNe = XNePtot = (0.043/0.071)(528 torr)
Ne
PNe = 320 torr
PH2 = XH2Ptot = (0.014/0.071)(528 torr)
H2
PH2 = 104 torr
740 = PH2 + 32torr  PH2 = 708 torr
20. a) Ptot = PH2 + PH2O
b) 𝑃𝑉 = 𝑛𝑅𝑇 𝑛 =
𝑃𝑉
𝑅𝑇
=
(0.932π‘Žπ‘‘π‘š)(2.0𝐿)
(.0821)(303𝐾)
= πŸ•. πŸ“ × πŸπŸŽ−𝟐 π’Žπ’π’
21.
a)
𝑃𝐻2 = 745 π‘šπ‘šπ»π‘” − 23.8π‘šπ‘šπ»π‘” = πŸ•πŸπŸ. 𝟐 π’Žπ’Ž π‘―π’ˆ
𝑛=
𝑃𝑉
𝑅𝑇
=
(.949π‘Žπ‘‘π‘š)(.090 𝐿)
𝑅(298𝐾)
= πŸ‘. πŸ’πŸ— × πŸπŸŽ−πŸ‘ π’Žπ’π’π’†π’” π‘―πŸ
3
2
3
1
1
3
1
1
Both I and III have the same
# of molecules, therefore
same pressure
b)
First convert 23.8 mm Hg to atm
𝑃𝑉 (0.0313π‘Žπ‘‘π‘š)(.090 𝐿)
6.02 × 1023
𝑛=
=
= 𝟏. πŸπŸ“ × πŸπŸŽ−πŸ’ π’Žπ’π’π’†π’” π‘―πŸ 𝑢 ×
= πŸ”. πŸ—πŸ‘ × πŸπŸŽπŸπŸ— π’Žπ’π’π’†π’„π’–π’π’†π’” π‘―πŸ 𝑢
1 π‘šπ‘œπ‘™π‘’
𝑅𝑇
𝑅(298𝐾)
c)
π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ 𝐻2
π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ 𝐻2𝑂
=
√18.02
√2.02
= 𝟐. πŸ—πŸ– π’•π’Šπ’Žπ’†π’”
d)
H2O because it is a polar molecule and in fact attracts to other water molecules due to hydrogen IMF bonding. The stronger the
IMF, the more the molecule or atom will deviate from ideal gas behavior. At low T and high P conditions molecules with
stronger IMF will attract more to each other and collide less often with the container walls.
a)
𝑷𝑻 =
22.
b)
𝑛𝑅𝑇
𝑉
=
(0.60π‘šπ‘œπ‘™)𝑅(300𝐾)
20.0 𝐿
π‘·π‘ΆπŸ = πŒπ‘ΆπŸ βˆ™ 𝑷𝑻 =
(
𝟎.πŸ”π’Žπ’π’
1π‘šπ‘œπ‘™
23. 24.2 g He × 4.00𝑔 = 6.05 π‘šπ‘œπ‘™ 𝐻𝑒
𝑃𝑇 =
(6.185π‘šπ‘œπ‘™)𝑅(298𝐾)
1.5 𝐿
= 𝟎. πŸ•πŸ‘πŸ—π’‚π’•π’Ž 𝒐𝒓 πŸ•πŸ’. πŸ–πŸ“ π’Œπ‘·π’‚
𝟎.πŸπ’Žπ’π’
= 12.1π‘Žπ‘‘π‘š
) βˆ™ πŸ•πŸ’. πŸ–πŸ“π’Œπ‘·π’‚ = πŸπŸ’. πŸ—πŸ“π’Œπ‘·π’‚
4.32 g 𝑂2 ×
𝑃𝐻𝑒 = (
1π‘šπ‘œπ‘™
32.00𝑔
= 0.135 π‘šπ‘œπ‘™ 𝑂2
6.05 π‘šπ‘œπ‘™
6.185 π‘šπ‘œπ‘™
) βˆ™ 12.1π‘Žπ‘‘π‘š = 𝟏𝟏. πŸ–π’‚π’•π’Ž
𝑃𝑂2 = (
0.135 π‘šπ‘œπ‘™
6.185 π‘šπ‘œπ‘™
) βˆ™ 12.1π‘Žπ‘‘π‘š = 𝟎. πŸπŸ” π’‚π’•π’Ž
24.
a)
b)
All gases have the same KE avg. because temperature is the same and temperature is a measure of KE.
He because it has the lowest MM. At the same temperature, the square of the velocity of any gas varies inversely with its molar
πŸ‘π‘Ήπ‘»
mass. π’–π’“π’Žπ’” = √
c)
d)
e)
f)
π“œ
He because partial pressure is proportional to the number of moles of gas present and the # of moles increases with decreasing
molar mass. 3.00 g of each gas, so if you convert to moles of each, you will see that the lower MM will have the most moles.
SO2 is the largest, most complex and they are the most polar. SO2 uses dipole-dipole bonding which is the strongest of the
IMF’s, the other gases only use LDF.
SO2 because again it uses dipole-dipole intermolecular bonding which is stronger than the LDF, therefore higher BP.
decreasing T and increasing P.
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