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Notes by Aaron Hui based on Professor Mack’s lectures
...
27
...
7 psi
○ Standard atmospheric pressure (1 atm)
■ The pressure that support a column of mercury exactly 760 mm
high at 0℃ at sea level
...

■ Constant amount of gas
...

○ Ex: A sample of chlorine gas occupies a volume of 946 mL at a pressure
of 726 mmHg
...
46 * 103 mmHg

● Charle’s Law
○ V α T (K)
■ Fixed amount of gas
■ Constant pressure
○ T↑V↑
○

V1
T1

=

V2
T2

○ Temperature MUST be in Kelvin
■ T(K)= t(℃) +273
...
20 L at 125℃
...
54 L if the pressure
remains constant
...
15)K = 398 K

●

3
...
54 L
T2

3
...
54 L)
T 2 = 192 K
T 2 = (192 − 273
...
0℃
● Another form of Charles’ Law is Gay Lussac’s Law
○ P α T (K) - constant n, V
○

P1
T1

=

P2
T2

● Avogadro's Law - Relationship between Volume and Amount
○ V α number of moles (n) - constant T and P
○

V1
n1

=

V2
n2

○ Equal volumes of any 2 gases contain equal # of atoms/molecules and
equal # of moles
● Ideal Gases
○ Molecules do not interact with one another
...

○ Ideal gases occur at low pressures and high temperatures
...
414 L - ​standard molar
volume​
...
08206 mol
*K

3

● Ex: An aerosol spray deodorant can with a volume of 350
...
2 g of
propane gas (C​3​H​8​) as propellant
...
mL *

L
1000 mL

n = 3
...
350 L

mol C 3 H 8
44
...
0726 mol

T = (20
...
15)K = 293 K
nRT
V

P =

=

L*atm )(293 K)
(0
...
08206 mol
*K
0
...
0 atm

● Using Standard Molar Volume in Calculations
○ 22
...
40 gof NH​3​ ​at STP?
■

7
...
03 g N H 3

*

22
...
74 L N H 3

○ What volume (in Liters) is occupied by 49
...
8 g HCl *

mol HCl
36
...
414 HCl
mol HCl

= 30
...

○ Ex: A gas initially at 4
...
2 atm, and 66℃ undergoes a change so that
its final volume and temperature are 1
...
What is the final
pressure? Assume number of moles remains constant
...
15)K = 339 K
T 2 = (42 + 273
...
2 atm)(4
...
7 L)
315 K

P 2 = 2
...
A certain lightbulb containing argon at 1
...
What is the final pressure of argon in
the lightbulb (in atm)?
P 1V 1
n1 T 1

■

=

P 2V 2
n2 T 2
P1
T1

●

=

P2
T2

T 1 = (18 + 273
...
15)K = 359 K
1
...
48 atm
Lecture 5
...
17
● Density (d) Calculations
○

d=

m
v

=

PM
RT

m- mass , M- molar mass

○ Expressed in g/L not g/mL
● Molar mass of a gaseous substance
○ Can determine molar mass if formula is unknown
○

M=

dRT
P

d- density (g/L)

● What is the density (in g/L) of uranium hexafluoride (UF​6​) at 779 mmHg and
62℃?
○

D=

m
V

P V = nRT

T = (63 + 273
...
025 atm

Assume 1
...
00 mol)*(0
...
025 atm

mm (molar mass) U F 6 = 352
...
0 g
26
...
8 L

g
mol

= 13
...
10 L vessel contains 4
...
00 atm and 27
...
What is the
molar mass of the gas?
○

molar mass =

g
mol

P V = nRT
n=

PV
RT

=

T = (27
...
15)K = 300
...
00 atm)(2
...
15 K)
(0
...
65 g
0
...
0853 mol

= 54
...
414 L standard molar volume
...
She measures the
volume as 1
...
54 g at STP
...
15 L gas *

mol gas
22
...
0513 mol

1
...
0513 mol

= 30
...

44
...
414 L

= 1
...
14% B and 21
...
At 27℃, 74
...
12 atm
...
0934 g, what is the
molecular formula?
○

78
...
81 g B

21
...
008 g H

= 7
...
69 mol H

B 7
...
69 = B H 3
7
...
228

P V = nRT
n=

PV
RT

=

T = (27 + 273
...
15 K

(1
...
0743 L)
L*atm )(300
...
08206 mol
*K

= 0
...
6 g
13
...
0934 g
0
...
6 g/mol

=2

= (BH 3 )2
= B2H 6
● What is the volume of CO​2​ produced at 37
...
00 atm when 5
...
60 g C 6 H 12 O6 *

mol C 6 H 12 O6
180
...
186 mol CO2

T = (37
...
15)K = 310
...
15 K)
(0
...
08206 mol
*K
1
...
73 L CO2

● If 34
...
mmHg, how many grams of NaN​3
were used in the reaction:
2N aN 3 (s) → 2N a (s) + 3N 2 (g)
L N 2 ⇒ (P V = nRT ) ⇒ mol N 2 ⇒ mol N aN 3 ⇒ g N aN 3
n=

PV
RT

=

(1
...
6 L)
L*atm )(75+273
...
0826 mol
*K

1
...
21 mol N 2

65
...
4 g N aN 3

● Review: At STP (0℃ and 1 atm), 1 mole of any ideal gas occupies 22
...

○ What is the volume of N​2​ produced at STP when 52
...
3 g N aN 3 ⇒ mol N aN 3 ⇒ mol N 2 ⇒ L N 2
mol N aN 3
65
...
3 g N aN 3 *

*

3 mol N 2
2 mol N aN 3

*

22
...
0 L N 2

● At any constant T and P, coefficients relate the volume as well as mol (only for
gases)
○ Calculate the volume of O​2​ in L for the complete combustion of 14
...

2C 4 H 10 (g) + 13O2 (g) → 8CO2 (g) + 10H 2 O (g)
14
...
9 L O2

○ Calculate the liters of CO​2​ produced during the complete combustion of
27
...

2C 8 H 18 (g) + 25O2 (g) → 16CO2 (g) + 18H 2 O (g)
27
...
14 L sample of HCl gas at 2
...
Calculate the molarity of the acid
solution assuming no change in volume
...
61 atm)(2
...
15 K)
(0
...
226 mol HCl
0
...
15)K = 301
...
226 mol HCl

= 0
...
4 x 10​5​ L and a
pressure of 7
...
A solution of LiOH with negligible volume
was introduced
...
2 x 10​-3​ atm
...
9*10

62
...
2*10 atm)(2
...
08206 mol
*K

mol Li2 CO3
mol CO2

*

73
...
8 mol CO2
= 4
...
0 L of C​4​H​10​ and 25
...
25 atm?
2C 4 H 10 (g) + 13O2 (g) → 8CO2 (g) + 10H 2 O (g)
○ Method 1:
5
...
5 L O2 needed

O2 is the limiting reagent because need 32
...
0 L O2 *

8 L CO2
13 L O2

= 15
...
0 L C 4 H 10 *
25
...
L CO2

= 15
...
5 g of C​4​H​10​ and 14
...
5 g C 4 H 10 *
14
...
12 g C 4 H

mol O2
32
...
172 mol CO2

= 0
...
172 mol CO2 *

22
...
9 L CO2

● Mixtures of Gases
○ Each particle of a gas acts independently
9

○ Partial pressure​ - the pressure exerted by a gas in a mixture
○ Air

P​T​ = 1
...
78(1
...
78 atm
P O2 = 0
...
00 atm) = 0
...
01(1
...
01 atm
P total = P N 2 + P O2 + P other = 1
...
How much oxygen (in g) is produced?
2M gO (s) → 2M g (s) + O2 (g)
P O2 = P T − P H 2 O
P O2 = (757 − 23
...
964 atm
T = (25 + 273
...
964 atm)(0
...
08206 mol
*K

1
...
00 g O2
mol O2

= 1
...
045 g O2

Lecture 5
...
17
● Mixture of A and B
○

PA =

nA RT
V

PB =

nB RT
V

10

(nA +nB )RT
V
n
( X i) = n i
T

PT = PA + PB =
○ Mole fraction

XA =

nA
nA +nB

XA + XB = 1

P i = X iP T
○ A sample of natural gas contains 8
...
421 moles of C​2​H​6
and 0
...
If the total pressure of the gases is 1
...
116 mol
(8
...
42+0
...
0132

= (0
...
37 atm) = 0
...
0% oxygen, 78
...
9% argon by volume has a total pressure of 790
...
What is the
partial pressure of each gas?
■

P O2 = 0
...
mmHg) = 166 mmHg
P N 2 = 0
...
mmHg) = 616 mmHg
P Ar = 0
...
mmHg) = 7 mmHg

● Hydrogen gas is generated when Ca reacts with water
...
The volume of gas
collected was 641 mL
...
82 mmHg at 30℃
...
82)mmHg = 956 mmHg
956 mmHg *

1 atm
760 mmHg

= 1
...
15)K = 303 K
11

nH 2 =

PV
RT

=

(1
...
641 L)
L*atm )(303 K)
(0
...
0325 mol H 2 *

2
...
0325 mol H 2 *

mol Ca
mol H 2

= 0
...
0655 g H 2

Part 2:

●

*

40
...
30 g Ca

Unequal Pressures
○ Density Hg = 13
...
00 g/mL
○ 1 mmHg = 13
...
7 mmHg?

P gas = P atm − P H 2 O − P height dif f erence
P h = 154 mmH 2 O *

1 mmHg
13
...
3 mmHg

P gas = (743
...
5 − 11
...
9 mmHg
● Kinetic molecular theory of gases - explains the behavior of gases that act
ideally
...
V gas particle << (much less than) space between particles (particles have
negligible volume)
2
...
Collisions are perfectly elastic
(the total kinetic energy remains constant)
3
...

4
...
K E = 12 m(↑)v 2 (↓) α T
Example: 1 mol He and 1 mol Ne at 273 K
→ same KE
12

→ He is faster
● Maxwell Speed Distribution Curves
○ At higher temperatures, more molecules are moving faster
●

Diffusion - the gradual mixing of molecules of one gas with molecules of another

by virtue of their kinetic properties
...
5 rHCl or

■

rHCl =

2
3

3
2

= 1
...
5
rHCl

rN H 3

● Graham’s law of effusion
○ Effusion - gas under pressure escapes into another area by passing
through a small opening
...
37 = 1
...
4 rO2
● If CO gas effuses at a rate that is 1
...
48 rX
rCO
rX

√

= (1
...
48)2 =

MX 2
)
M CO

MX
28
...
4 g/mol
13

● Real Gases
○ 1 mol of ideal gas
■ PV = nRT
n=

PV
RT

= 1
...

● Modified Ideal Gas Equation
○ Van der Waals equation - nonideal gas
■

(P +

an2
)(V
V2

− nb) = nRT

● Corrected P due to attractive forces
● Corrected Volume
● (equation wont be tested)
●

Note that on SA-6 #13 only Zn reacts with HCl not Cu

INTERMOLECULAR FORCES AND LIQUIDS AND
SOLIDS
Lecture 5
...
17
● For Molecular Compounds:
○ Intermolecular ​forces - attractive forces ​between​ molecules
■ Determines boiling point, melting point, state at room temp, etc
...
Covalent
bonds)
●

For Ionic Compounds:
○ “Intermolecular forces” are the ionic bonds (ion-ion forces)

●

Intermolecular vs Intramolecular
○ 41 kJ to vaporize 1 mole of water (inter)
14

○ 930 kJ to break all O-H bonds in 1 mol of water (intra)
○ Generally, ​inter​molecular forces are much weaker than ​intra​molecular
forces
...


■
■ Constant motion of electrons within molecules can create
instantaneous dipole​
...


●

C H 3 OCH 3 no, H is not attached to FON
...


○

Al3+ + H 2 O
18

■ Dispersion, Ion-Dipole force
●

Summary of Intermolecular Forces

○
● Due Tuesday
○ SA7 and Exp 9-Voicethread

Lecture 5
...
17

● Surface Tension - the resistance of a liquid to spread out and increase its surface
are
...

●

Cohesion - the intermolecular attraction between like molecules
...


○
●

Properties of Liquids - Viscosity
○ Viscosity - a measure of a fluid’s resistance to flow
■ Strong intermolecular forces → high viscosity
19

●

Water is a unique substance
○ Ice is less dense than water

●

Phase Changes from liquid to gas

○
●

Evaporation (Vaporization)
○ Liquids are constantly evaporating even when the temperature is less than
the boiling point
...

●

Boiling point - vapor pressure = atm pressure

●

To figure which substance has stronger IMF,
○ Compare vapor pressures (at the same temperature)
○ Compare boiling points (at the same vapor pressure)

20

○
●

Volatile liquid ○ Easily forms gas molecules due to weak intermolecular forces
...
Why?
■ A higher vapor pressure has weaker IMF
● HF - ​Hydrogen bonds, dipole dipole, dispersion
● HCl - ​dipole dipole, dispersion
■ HCl is weaker, so it has higher vapor pressure
...

■ All dispersion forces (if only C and H, the comp is nonpolar so only
disp)
●
C 3H 8
Low VP
Strong IMF

< C 2H 6

< CH 4
→

High VP
Weaker IMF
21

Higher molar
Mass

Lowest MM

● Since they are all nonpolar and dispersion forces, compare
molar masses
...

●

Boiling
○ Boiling point- temperature at which the (equilibrium) vapor pressure of a
liquid is = to the external pressure
...


●
○ Higher molar weight = higher boiling point b/c stronger IMFs
●

Molar heat of vaporization ( ΔH vap )
○ Energy required to vaporize 1 mol of liquid

●

Dispersion Forces and Boiling Point
○ Which substance will have a larger boiling point:
■ F​2​ (38
...
46 g/mol)
● F has dispersion, HCl has dipole dipole AND dispersion
○ HCl is stronger IMF so higher BP
■ Xe or Ar
● Both dispersion,​ but Xe has higher MM so it is stronger
IMF with higher BP
■ NaCl or H​2​O
● NaCl has ion-ion
22

● Water has H-bond, d-d, disp
○ NaCl is a solid (with a higher melting point) and
ion-ion is significantly stronger than h-bonds
● Critical Temperature and Pressure
○ Critical Temperature ( T c ) - the temperature above which the gas cannot
liquefy, no matter what the pressure
...

○ Stronger IMF → High T c
●

Phase Changes for Solids to Liquids

○
○ Melting point​ of a solid or ​freezing point​ of a liquid - the temperature at
which the solid and liquid phases coexist in equilibrium
○ Normal Melting point - ​the melting point at 1 atm
...

○

ΔH f us < ΔH vap always

●

Supercooling - a liquid is temporarily cooled to below its freezing point
...

● Calculate the amount of energy (in kJ) released when 346 g of steam at 182℃ is
converted to liquid water at 0℃
...
Mix 87
...
40 M ammonium sulfate + 225
...
20 M potassium
hydroxide
...
5 mL)(0
...
0 mL)(0
...


C hange
E end
E (M )

25
312
...


− 45

+ 45

0

0

25
...
11M


...
080M

0

45
312
...
14M

25

V olumef = 87
...
0 mL = 312
...
18
...


■
■ ↑P - condensation
■ ↓T - gas to solid or deposition
■ For water, since the slope of the line between solid and liquid is
negative, then the density of the solid is less than the liquid
...


○
●

Types of Solids
○ Crystalline solid - ordered
26

○ Amorphous solid - no order
●

X-ray diffraction by crystals
○ X-ray diffraction - scattering of x-rays by the units of a crystalline solid
○ Max von Laue - crystals diffract x-rays since the λ of x-rays is similar to the
distance between lattice points
...
154 nm are diffracted from a crystal at an
angle of 14
...
Assuming that n=1, what is the distance (in pm) between
tlayers in the crystal?
■ 1000 pm = 1 nm
■

2 d sinθ = nλ
●

●

d=

nλ
2sinθ

103 pm

=

(1)(0
...
17)

= 315 pm

Types of Crystals
○ Ionic Crystals
■ Basic units: cations and anions
■ Forces: ionic (ion-ion)
■ High melting point
■ Poor conductor of heat and electricity unless melted or dissolved in
water
■ Hard, brittle
○ Covalent Network
27

■ Basic units: atoms
■ Forces: covalent bonds
■ No individual molecules
■ High melting point
■ Poor conductor of heat and electricity
■ Hard
■ Examples: C (diamond, graphite), SiO​2​ (Quartz)
○ Molecular Crystals
■ Basic units: molecules
■ Forces: intermolecular forces
■ Low melting point
■ Poor conductor of heat and electricity
■ Soft
■ Larger IMF = higher the melting point
○ Metallic Crystals
■ Basic units: metal atoms
■ Forces: metallic bonds
■ Soft to hard, range of melting points
■ Good conductors of heat and electricity
○ Atomic Solid
■ Solid containing atoms from a ​nonmetallic element​ (ex: solid form
of Ne and other noble gases)
■ Single atms held in place by weak dispersion forces
● Low melting points
○ Amorphous solids
■ Amorphous solid - no order
■ Glass
●
Type Solid

Basic Unit

IMF

MP

Example
28

Ionic

Ions

Ion ion

High

NaCl

Covalent
Network

Atoms

Covalent
bonds

High

C - (diamond/grahite)
SiO​2​ - (quartz)

Molecular

Molecules

H-bonds,
d-d,and/or
dispersion

Low

H​2​O, F​2

Metallic

Metal atoms

Metallic
bonding

High

Cu

Atomic

Nonmetal
atoms

dispersion

Low

Ne

●
●

Practice:
○ Classify the solid and what type of intermolecular forces are present:
■ SO​2​ - Molecular solid, dipole dipole, dispersion
■ KI - Ionic solids, ion ion
■ Zn - Metallic solid, metallic bonding
■ NaBr - Ionic solid, ion ion
■ CH​4 -​ Molecular solid, dipole dipole, dispersion
○ Arrange the following in order of increasing melting point
...


●

Hydration - an ion is surrounded by water molecules

●

“Like dissolves like”
○ Substances with similar ​INTER​molecular forces are soluble
■ Nonpolar dissolve in nonpolar
● CCl​4​ in C​6​H​6
■ Polar​ and ​ionic ​compounds dissolve in polar solvents
● C​2​H​5​OH​ in H​2​O
● NaCl ​in H​2​O or NH​3
○ Miscible​ - 2 liquids that are completely soluble in one another in all
proportions

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