Chap13_Part2

Chap13_Part2 - 1
 Entropy
(S)
 2


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Unformatted text preview: 1
 Entropy
(S)
 2
 The
entropy
of
liquid
water
 is
greater
than
the
 entropy
of
solid
water
 (ice)
at
0˚
C.
 Order
of
Entropy:
Gas
>
Liquid
>
Solids
 3
 The
Change
in
the
Gibbs
Free
Energy
Indicates
Whether

 or
Not
a
Given
Process
Will
Occur
Spontaneously
 This
result
comes
from
combining
the
First
&
Second
Laws
of
Thermodynamics.
 ΔG = ΔH − TΔS where 
ΔG
=
change
in
Gibbs
free
energy
 
ΔH
=
change
in
enthalpy
[ΔE+Δ(PV)]
 
ΔS
=
change
in
entropy
 
T
=
absolute
temperature
(constant) 

 
 
 
 
If
ΔG
<
0,
the
process
can
occur
spontaneously.
 
If
ΔG
>
0,
the
process
cannot
occur
spontaneously.
 
If
ΔG
=
0,
the
system
is
in
equilibrium.
 ReducTons
in
the
enthalpy
(ΔH
<
0)
and

increases
in
the
entropy
(ΔS
>
0)

 contribute
to
favorable
processes.
[More
on
ΔG
in
Chapter
19.]
 4
 OZen,
but
not
always,
the
magnitude
of
ΔH
is
larger
than
the
magnitude
of
TΔS,

 and
the
sign
of
the
change
in
the
Gibbs
free
energy
is
determined
by
the
sign
of
ΔH.
 ΔG = ΔH − TΔS If
ΔG
<
0,
the
process
can
occur
spontaneously.
 If
ΔG
>
0,
the
process
cannot
occur
spontaneously.
 If
ΔG
=
0,
the
system
is
in
equilibrium.
 1)
SoluTons
that
are
exothermic
(ΔH
<
0;
release
heat)
occur
spontaneously.
 Dissolving
MgSO4
(magnesium
sulfate)
in
water
has
a
net
ΔH
=
‐91.2
kJ/mol.

The
entropy

 also
increases,
so
ΔG
<
0.
 2)
SoluTons
that
are
endothermic
(ΔH
>
0;
absorb
heat)
may
or
may
not
occur
spontaneously.
 
2a)
Dissolving
NH4NO3
(ammonium
nitrate)
in
water
has
a
net
ΔH
=
+26.4
kJ/mol.


 
But
the
entropy
increases
enough
so
that
ΔG
<
0
and
the
process
occurs
spontaneously.
 
2b)
Water
does
not
dissolve
in
nonpolar
solvents
such
as
octane
(C8H18)
because
ΔS
 
is
not
large
enough
to
overcome
the
very
posiTve
ΔH.

The
hydrogen
bonding
of
water
is
 
too
strong.
 5
 The
enthalpy
of
soluTon
of
KBr
in
water
is
+198
kJ/mol.
Nonetheless,
 the
solubility
of
KBr
in
water
is
high.

Why

does
the
KBr
readily
 dissolve
in
water
even
though
the
process
is
endothermic?
 6
 The
enthalpy
of
soluTon
of
KBr
in
water
is
+198
kJ/mol.
Nonetheless,
 the
solubility
of
KBr
in
water
is
high.

Why

does
the
KBr
readily
 dissolve
in
water
even
though
the
process
is
endothermic?
 The
contribuTon
to
ΔG
from
the
entropy
change
(‐TΔS)
 is
negaTve
enough
(i.e.,
ΔS
is
posiTve
enough)
to
more
 than
balance
the
posiTve
ΔH,
giving
a
net
negaTve
ΔG.
 7
 Carbon
tetrachloride
and
hexane
are
mixed.
Because

 the
intermolecular
interacTons
between
CCl4,
C6H14

 &
CCl4/C6H14
are
similar
in
magnitude,
the
entropy

 change
dominates
the
Gibbs
free
energy
change,
which

 is
negaTve,
and
the
liquids
readily
dissolve.
 On
the
contrary,
NaCl
does
not
dissolve
in
gasoline,
 because
the
reducTon
in
intermolecular
akracTons
is

 not
counterbalanced
by
entropy.
 FormaTon
of
soluTons
is
favored
by
the
increase
in

 entropy
that
accompanies
mixing.
 8
 Josiah
Willard
Gibbs
(1839‐1903)
 ‐ 
TheoreTcal
Physicist
&
MathemaTcian
 ‐ 
Father
of
Physical
Chemistry
 ‐ 
First
one
to
apply
second
law
of
Thermo
dynamics
 ‐ 
Concept
of
chemical
potenTal
and
free
energy
 ‐ 
Entropy
 ‐ 
Gibbs‐
Helmholtz
equaTon
..
and
many
more!!
 Gibbs
developed
much
of
what
we
know
about
chemical
thermodynamics.
 9
 Entropy
of
Mixing
 The
stopcock
is
opened
and
the
two
gases
thoroughly
mix.
 Does
the
entropy
increase
or
decrease?





 10
 Entropy
of
Mixing
 The
stopcock
is
opened
and
the
two
gases
thoroughly
mix.
 Does
the
entropy
increase
or
decrease?




increase
 11
 Chemical
ReacQons
 The
situaTon
is
different
if
a
chemical
reacTon
occurs.

For
example,
 
 
Ni(s)
+
2
HCl(aq)
→
NiCl2(aq)
+
H2(g)
 Physical
process

 



Vs
 Chemical
ReacQon
 HCl
(aq)
dissolves
Ni(s).

During
the
process,
NiCl2(aq)
is
formed,
 which
is
found
as
NiCl2
6H2O
aZer
evaporaTon.

 12
 Hydrates
 Water
molecules
oZen
remain
in
salts
obtained
by
evaporaTon
of
water
 from
soluTons.


 hydrated
cobalt
chloride
 






CoCl2●6H2O
 anhydrous
cobalt
chloride
 












CoCl2
 13
 Solid
sodium
sulfate
becomes
hydrated
with
water
vapor,
 and
the
water
vapor
is
all
consumed.
 Does
the
entropy
increase
or
decrease?







 14
 Solid
sodium
sulfate
becomes
hydrated
with
water
vapor,
 and
the
water
vapor
is
all
consumed.
 Does
the
entropy
increase
or
decrease?






decrease
 15
 Saturated
SoluQons
 In
a
soluTon
with
excess
solute,

 ions
on
the
surface
of
the
solute
conTnually
move

 into
the
soluTon
as
hydrated
ions
(solvaTon)
while
 hydrated
ions
in
the
soluTon
conTnually
move
onto

 the
surface
of
the
solute
(crystallizaTon).
 A
dynamic
equilibrium
is
present.
 solvation solute + solvent ⇔ solution crystallization 16
 Saturated
SoluQons
 A
soluTon
in
equilibrium
with
excess
(undissolved)
solute
is
called
saturated.
 A
soluTon
which
has
the
capability
of
dissolving
more
solute
is
called
unsaturated.
 The
amount
of
solute
required
to
form
a
saturated
soluTon
in
a
given
solvent
is
 called
the
solubility.
 A
soluTon
can
be
made
to
be
supersaturated
by
saturaTng
it
at
a
high
temperature
 and
then
lowering
the
temperature.
 Try
making
sugar
crystal
from
 powdered
Sugar,
using
the

 concept
of

‘supersaturated
 soluQons’
 17
 Which
of
the
following
is
the
best
representaTon
of
a
saturated
soluTon?
 18
 Which
of
the
following
is
the
best
representaTon
of
a
saturated
soluTon?
 19
 The
solubility
of
MnSO4H2O
in
water
at
20oC
is
70
gm
per
100
mL.
 Is
a
1.22
M
soluTon
unsaturated,
saturated
or
supersaturated?
 (a)
Find
the
formula
weight
in
gm/mol.
 (b)
Find
the
concentraTon
in
units
of
gm
per
100
mL.
 (c)
Determine
the
answer:
 

 20
 The
solubility
of
MnSO4H2O
in
water
at
20oC
is
70
gm
per
100
mL.
 Is
a
1.22
M
soluTon
unsaturated,
saturated
or
supersaturated?
 (a)
Find
the
formula
weight
in
gm/mol.
 1 Mn (55) + 1 S (32) + 5 O (16) + 2 H (1) = 169 gm mol (b)
Find
the
concentraTon
in
units
of
gm
per
100
mL.
 (c)
Determine
the
answer:
 

 21
 The
solubility
of
MnSO4H2O
in
water
at
20oC
is
70
gm
per
100
mL.
 Is
a
1.22
M
soluTon
unsaturated,
saturated
or
supersaturated?
 (a)Find
the
formula
weight
in
gm/mol.
 1 Mn (55) + 1 S (32) + 5 O (16) + 2 H (1) = 169 gm mol (b)
Find
the
concentraTon
in
units
of
gm
per
100
mL.
 (1.22 mol gm gm gm )(169 ) = 206 = 20.6 L mol L 100 mL (c)
Determine
the
answer:
 

 22
 The
solubility
of
MnSO4H2O
in
water
at
20oC
is
70
gm
per
100
mL.
 Is
a
1.22
M
soluTon
unsaturated,
saturated
or
supersaturated?
 (a)
Find
the
formula
weight
in
gm/mol.
 1 Mn (55) + 1 S (32) + 5 O (16) + 2 H (1) = 169 gm mol (b)
Find
the
concentraTon
in
units
of
gm
per
100
mL.
 (1.22 mol gm gm gm )(169 ) = 206 = 20.6 L mol L 100 mL (c)
Determine
the
answer:
 
unsaturated
 23
 Factors
AffecQng
Solubility
 1)
Solute‐Solvent
InteracTons
 2)
Pressure
(for
gases)
 3)
Temperature
 24
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 28
g /mol
 28
g /mol
 32
g /mol
 40
g /mol
 84
g /mol
 The
solubiliTes
of
simple
gases
in
water
increase
with
molecular
 size
and
polarizability
due
to
increased
London
dispersion
forces.
 25
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 Liquids
that
readily
mix
to
form
soluTons
are
called
miscible.
 Liquids
that
do
not
readily
mix
to
form
soluTons
are
immiscible.
 a)
Two
polar
liquids
tend
to
be
miscible
because
of
favorable
dipole‐dipole

 interacTons.
 b)
Polar
liquids
as
well
as
those
capable
of
forming
hydrogen
bonds
tend

 to
be
miscible
in
water.
 c)
Nonpolar
liquids
tend
to
be
immiscible
in
water
because
the
solute‐
 solvent
interacTons
are
not
strong
enough
to
overcome
the
hydrogen
bonding

 in
water.
 d)
Two
nonpolar
liquids
are
usually
miscible.
 26
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 e)
Alcohols
become
increasingly
insoluble
in
the
polar
solvent
water
as
their
 hydrocarbon
components
increase
in
length,
because
the
nonpolar
character
of
the
 alcohol
molecules
begins
to
dominate
the
polar
character.
 f)
Alcohols
become
increasingly
soluble
in
the
nonpolar
solvent
hexane
as
their
 hydrocarbon
components
increase
in
length,
because
the
London
dispersion
 interacTons
increase
in
magnitude.
 27
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 (a)  hydrogen
bonding
between
two
ethanol
molecules
 (b)
hydrogen
bonding
between
ethanol
and
water
molecules
 28
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 Adding
OH
groups
to
a
nonpolar
molecule
increases
the
solubility
in
water.
 29
 Factors
AffecQng
Solubility:


 1)
Solute‐Solvent
InteracQons
 Some
vitamins
like
vitamin
A
(reTnol)
are
primarily
nonpolar.

They
are

 stored
in
faky
Tssue
and
can
build
to
toxic
levels.

Other
vitamins
like

 vitamin
C
(ascorbic
acid)
are
water
soluble,
readily
excreted,
and
must
be
 conTnually
replenished.
 30
 Factors
AffecQng
Solubility:


 2)
Pressure
 The
solubiliTes
of
solids
and
liquids
are
not
significantly
changed
by
pressure.
 The
solubiliTes
of
gases
in
liquids
are
appreciably
affected
by
pressure.
 When
the
pressure
on
a
gas
dissolved
in
a
liquid
is
increased,
the
 concentraTon
of
gas
in
the
liquid
is
increased.
 31
 Henry’s
Law:

RelaTonship
Between
Gas
Solubility
in
Liquids
&
Pressure
 “The
solubility
of
a
gas
in
a
liquidincreases
in
direct
proporTon
to
its
 parTal
pressure
above
the
soluTon”.
 S g = kPg Sg
=
solubility
of
the
dissolved
gas
 Pg
=
parTal
pressure
of
the
gas
over
the
liquid
 k
=
Henry’s
law
constant
 The
constant
k
depends
on
the
idenTty
of
the
gas,
the
idenTty
of
the

 liquid,
and
the
temperature.
 Higher
Pg
means
higher
Sg.
 32
 Carbonated
beverages
are
 bokled
under
pressure.
 When
the
bokle
is
opened,
 the
parTal
pressure
of
CO2
 above
the
liquid
is
reduced.
 The
CO2
solubility
decreases.
 The
CO2
leaves
the
soluTon
 by
forming
bubbles
which
rise
 and
join
the
air.
 33
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 parTal
pressure
of
CO2
(in
air)
is
0.05
atm.
 34
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 S g = kPg = (3.1x10 − 2 M )(10 atm) = 0.31 M atm (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 parTal
pressure
of
CO2
(in
air)
is
0.05
atm.
 35
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 S g = kPg = (3.1x10 − 2 M )(10 atm) = 0.31 M atm (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 parTal
pressure
of
CO2
(in
air)
is
0.05
atm.
 S g = kPg = (3.1x10 −2 M )(0.05 atm) = 0.002 M atm 36
 William
Henry
(English
Chemist,
1775‐1836)
 ‐ 
English
Chemist
 ‐ 
Henry’s
Law
 ‐ 
Worked
on
IlluminaTng
gasses
and
Fire‐
damps
 ‐ 
ComposiTon
of
Hydrochloric
acid
and
Ammonia
 37
 The
parTal
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentraTon
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 (c)
Find
the
parTal
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentraTon
of
O2
in
the
lake.
 38
 The
parTal
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentraTon
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 (c)
Find
the
parTal
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentraTon
of
O2
in
the
lake.
 39
 The
parTal
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentraTon
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
parTal
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentraTon
of
O2
in
the
lake.
 40
 The
parTal
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentraTon
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
parTal
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 P(0 2 ) = 0.21(0.875 atm) = 0.184 atm (d)
Find
the
molar
concentraTon
of
O2
in
the
lake.
 41
 The
parTal
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentraTon
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
parTal
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 P(0 2 ) = 0.21(0.875 atm) = 0.184 atm (d)
Find
the
molar
concentraTon
of
O2
in
the
lake.
 S (0 2 ) = kP(O2 ) = (1.38 x 10 −3 M / atm)(0.184 atm) = 2.54 x10 −4 M 42
 Deep‐Sea
Divers
and
N2
Solubility
in
Water
 Workers
in
caissons
(underwater
construcTon
sites)
breathe
air
at
more
 than
normal
pressure.

If
they
are
returned
to
the
surface
too
rapidly,
N2,
 dissolved
in
their
blood
at
high
pressure,
comes
out
of
soluTon
and
may
 cause
emboli
(gas
bubbles
in
the
bloodstream).

Blood
at
37oC
and
1
atm
 dissolves
1.3
mL
of
pure
N2
in
100
mL
of
blood.

 Calculate
the
volume
of
N2
liberated
from
the
blood
of
a
caisson
worker
 returned
to
1
atm
pressure
aZer
prolonged
exposure
to
pressure
300
m
 below
sea
level
where
the
pressure
is
30
atm.

The
total
blood
volume
of
 an
average
adult
is
3.2
L;
air
contains
78%
N2.
 43
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 44
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 45
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 46
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 S1 5.1x10 −4 M M k= = = 6.5 x10 − 4 P1 0.78(1 atm) atm (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 47
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 S1 5.1x10 −4 M M k= = = 6.5 x10 − 4 P1 0.78(1 atm) atm (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 S 30 = kP30 = (6.5 x10 − 4 M )(0.78)(30 atm) = 0.015 M atm 48
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 49
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 50
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 51
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 n = (0.048 − 0.0016) mol = 0.046 mol (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 52
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
thebody
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 n = (0.048 − 0.0016) mol = 0.046 mol (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 V= nRT = P L − atm )(310.1 K ) mol − K = 1.5 L 0.78(1 atm) (0.046 mol )(0.08206 53
 Factors
AffecQng
Solubility:


 3)
Temperature
 The
solubility
of
most
solids
in
water
increases
with
temperature.
 54
 Factors
AffecQng
Solubility:


 3)
Temperature
 The
solubility
of
most
gases
in
water
decreases
with
temperature.
 Warming
up
cooler
water

bubbles.
 55
 ...
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