Review%20Session-1

Review%20Session-1 - REVIEW
SESSION‐
1
...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: REVIEW
SESSION‐
1
 GAS
LAWS
 Conversion
of
Pressure:
 Arrange
the
pressures
in
ascending
order
(1)
30
cm
of
Hg
(2)
30
lb/in2
 (3)
0.76
torr
 Express
the
following
gas
pressures
in
‘atm’
 1)
30
cm
of
Hg
 2)
30
lb/in2
 3)
0.76
torr
 Conversion
of
Pressure:
 Arrange
the
pressures
in
ascending
order
(1)
30
cm
of
Hg
(2)
30
lb/in2
 (3)
0.76
torr
 Express
the
following
gas
pressures
in
‘atm’
 1)  30
cm
of
Hg:
300
mm
of
Hg
(300
mm)
(
1atm/
760
mm)
 














































=
0.395
atm
 2)
30
lb/in2:
(30
lb/in2)
(1atm/
14.7lb/in2)

 





































































=
2.05
atm
 3)
0.76
torr
:
(0.76
torr)
(1
atm/
760
torr)
=
0.001
atm
 The
ascending
order:
0.76
torr
<
30cm
<
30
lb/in2
 ‐
The
volume
of
the
lungs
is
measured
by
the
volume
of
air
inhaled
 or
exhaled.

If
the

volume
of
the
lungs
is
2.400
L
during
exhalaQon
 and
the
pressure
is
101.70
kPa,
and
the
pressure
during
inhalaQon
is
 100.01
kPa,
what
is
the
volume
of
the
lungs
during
inhalaQon?

 ‐
The
volume
of
the
lungs
is
measured
by
the
volume
of
air
inhaled
 or
exhaled.

If
the

volume
of
the
lungs
is
2.400
L
during
exhalaQon
 and
the
pressure
is
101.70
kPa,
and
the
pressure
during
inhalaQon
is
 100.01
kPa,
what
is
the
volume
of
the
lungs
during
inhalaQon?

 P1
=
101.7
kPa













 V1
=
2.4
L













 P2
=
100.01
kPa
 V2
=
?
 P1V1
=
P2V2
 V2
=
P1V1/P2
 V2
=
(101.7
kPa)
(2.4L)
/
(100.01
kPa)
 V2
=
2.441L
 ‐ Does
the
volume
of
a
fixed
quanQty
of
gas
decreases
to
half
its
original
 
value
when
the
temperature
is
lowered
from
100oC
to
50oC
?
 ‐ Does
the
volume
of
a
fixed
quanQty
of
gas
decreases
to
half
its
original
 
value
when
the
temperature
is
lowered
from
100oC
to
50oC
?
 ‐
NO
 ‐ Does
the
volume
of
a
fixed
quanQty
of
gas
decreases
to
half
its
original
 
value
when
the
temperature
is
lowered
from
100oC
to
50oC
?
 ‐
NO
 V1
/
T1
=

V2
/
T2
 V1
/
V2
=
T1
/
T2
 V1
/
V2
=
(100+273)
/
(50+273)
 V1
/
V2
=
1.154
 V1
=
V2
(1.154)
 V2
=
(0.87)
V1

 ‐ Tennis
balls
are
filled
with
N2
gas
to
higher
pressures
to
increase
their
 
bounce.
If
a
parQcular
tennis
ball
has
a
volume
of
144
cm3
and
contains
 
0.33g
N2
gas,
what
is
the
pressure
inside
the
tennis
ball
at
24oC

 ‐ Tennis
balls
are
filled
with
N2
gas
to
higher
pressures
to
increase
their
 
bounce.
If
a
parQcular
tennis
ball
has
a
volume
of
144
cm3
and
contains
 
0.33g
N2
gas,
what
is
the
pressure
inside
the
tennis
ball
at
24oC

 V=

144
cm3
=
144
mL
=
0.144L
 T
=
24oC=
273.1
+24
=
297.1
K
 n=
Wt/
MWt
=
0.33g/
28
(g /mol)
=
0.0118
mol
 R
=
0.08206
L‐atm/Mol.
K
 P
=
nRT/V
 P=
(0.0118
mol)
(0.08206
L‐atm/
Mol.
K)
297.1
K
/
0.144L
 P
=2
atm


 ‐
6.2
liters
of
an
ideal
gas
are
contained
at
3.0
atm
and
37
°C.
How
 many
moles
of
this
gas
are
present?
 ‐
6.2
liters
of
an
ideal
gas
are
contained
at
3.0
atm
and
37
°C.
How
 many
moles
of
this
gas
are
present?
 Given,
 P
=
3
atm
 V=
6.2
L
 T
=
273.1+37
=
310.1K
 R=
0.08206
L‐atm/mol.K
 ‐
6.2
liters
of
an
ideal
gas
are
contained
at
3.0
atm
and
37
°C.
How
 many
moles
of
this
gas
are
present?
 Given,
 Asking
for
“n”
 P
=
3
atm
 PV=
nRT
 V=
6.2
L
 n
=
PV/RT
 T
=
273.1+37
=
310.1K
 


=
(3
atm)
(6.2
L)
 





(0.08206
L‐atm/mol.
K)
310.1
K
 R=
0.08206
L‐atm/mol.K
 n
=
0.731
mol
 Solid
potassium
chlorate
(KClO3)
decomposes
to
produce
solid
 potassium
chloride
(KCl)
and
oxygen
gas
(O2)
according
to
the
balanced
 chemical
equaQon:
2KClO3
(s)

2KCl
(s)
+
3O2
(g)
 What
volume
of
gas,
measure
at
23.4oC
and
75.4
kPa,
will
be
produced
 when
32.5
g
of
potassium
chlorate
is
decomposed?
 Solid
potassium
chlorate
(KClO3)
decomposes
to
produce
solid
 potassium
chloride
(KCl)
and
oxygen
gas
(O2)
according
to
the
balanced
 chemical
equaQon:
2KClO3
(s)

2KCl
(s)
+
3O2
(g)
 What
volume
of
gas,
measure
at
23.4oC
and
75.4
kPa,
will
be
produced
 when
32.5
g
of
potassium
chlorate
is
decomposed?
 Given,
 P
=
75.4kPa


(75.4kPa)
(1
atm)/
(101.325kPa)
=
0.745
atm
 T
=
23.4oC
=
273.1+23.4
=
296.5
K
 n
=
wt/MWt
=
(32.5g)/
(122.5g/mol)
=
0.265
mol
 R=
0.08206
L.atm/mol.K
 Solid
potassium
chlorate
(KClO3)
decomposes
to
produce
solid
 potassium
chloride
(KCl)
and
oxygen
gas
(O2)
according
to
the
balanced
 chemical
equaQon:
2KClO3
(s)

2KCl
(s)
+
3O2
(g)
 What
volume
of
gas,
measure
at
23.4oC
and
75.4
kPa,
will
be
produced
 when
32.5
g
of
potassium
chlorate
is
decomposed?
 Given,
 P
=
75.4kPa


(75.4kPa)
(1
atm)/
(101.325kPa)
=
0.745
atm
 T
=
23.4oC
=
273.1+23.4
=
296.5
K
 32.5g
of
KClO3
decomposed

?
g
of
oxygen
produced
 2
moles
of
KClO3
produce
3
moles
of
Oxygen
 2(122.55g)
of
KClO3
produce
3(32g)
of
oxygen
 32.5g
of
KClO3
produce
(3X32)
(32.5)/
(2X
122.55)
=
12.73g
of
Oxygen
 n
=
12.73/32
=
0.398
mol
 Solid
potassium
chlorate
(KClO3)
decomposes
to
produce
solid
 potassium
chloride
(KCl)
and
oxygen
gas
(O2)
according
to
the
balanced
 chemical
equaQon:
2KClO3
(s)

2KCl
(s)
+
3O2
(g)
 What
volume
of
gas,
measure
at
23.4oC
and
75.4
kPa,
will
be
produced
 when
32.5
g
of
potassium
chlorate
is
decomposed?
 Given,
 P
=
75.4kPa


(75.4kPa)
(1
atm)/
(101.325kPa)
=
0.745
atm
 T
=
23.4oC
=
273.1+23.4
=
296.5
K
 32.5g
of
KClO3
decomposed

?
g
of
oxygen
produced
 2
moles
of
KClO3
produce
3
moles
of
Oxygen
 2(122.55g)
of
KClO3
produce
3(32g)
of
oxygen
 32.5g
of
KClO3
produce
(3X32)
(32.5)/
(2X
122.55)
=
12.73g
of
Oxygen
 n
=
12.73/32
=
0.398
mol
 V=
(0.398)
(0.08206)
(296.5)
/
0.745
=
13
L
 Nitrogen
and
hydrogen
gases
react
to
form
ammonia
gas.
 
 
 
 
 
N2(g)
+
3
H2(g)
→
2
NH3(g)
 At
a
certain
temperature
and
pressure,
2
L
of
N2
reacts
with
7
L
of
H2.
 What
volume
of
NH3
is
produced?
What
gases
remain
at
the
end
of
the

 reacQon?
 Nitrogen
and
hydrogen
gases
react
to
form
ammonia
gas.
 
 
 
 
 
N2(g)
+
3
H2(g)
→
2
NH3(g)
 At
a
certain
temperature
and
pressure,
2
L
of
N2
reacts
with
7
L
of
H2.
 What
volume
of
NH3
is
produced?
What
gases
remain
at
the
end
of
the

 reacQon?
 
1
mole
of

N2
needs
3
moles
of

H2to
react
completely
&
produce
 2
moles
of
NH3.

 N2
and
H2
react
with
each
other
in
1:3
raQo
 Nitrogen
and
hydrogen
gases
react
to
form
ammonia
gas.
 
 
 
 
 
N2(g)
+
3
H2(g)
→
2
NH3(g)
 At
a
certain
temperature
and
pressure,
2
L
of
N2
reacts
with
7
L
of
H2.
 What
volume
of
NH3
is
produced?
What
gases
remain
at
the
end
of
the

 reacQon?
 
1
mole
of

N2
needs
3
moles
of

H2to
react
completely
&
produce
 2
moles
of
NH3.

 N2
and
H2
react
with
each
other
in
1:3
raQo
 
If
you
have
2
L
of
N2,
you
need
at
least
6
L
of
H2.
 1(2L)
of
N2
+
3
(2L)
H2

2(2L)
NH3
 So,
4L
of
NH3
will
be
produced
and
1L
of
H2
(unreacted)
will
 remain
in
the
container
along
with
4L
of
NH3.
 What
is
the
total
pressure
exerted
by
a
mixture
of
2.0g
of
H2
and
8.0g
of
 N2
at
273K
in
a
10.0
L
vessel?
 What
is
the
total
pressure
exerted
by
a
mixture
of
2.0g
of
H2
and
8.0g
of
 N2
at
273K
in
a
10.0
L
vessel?
 Given,
 T
=
273
K
 V
=
10
L
 R
=
0.08206
L.atm/mol.K
 n
=
?

and
P
=?
 What
is
the
total
pressure
exerted
by
a
mixture
of
2.0g
of
H2
and
8.0g
of
 N2
at
273K
in
a
10.0
L
vessel?
 Given,
 T
=
273
K
 V
=
10
L
 R
=
0.08206
L.atm/mol.K
 n
=
?

and
P
=?
 n
=
Number
of
moles=
nH2
+nN2
 nH2
=
wt/Mwt
=
(2.0g)/(2.0g/mol)
=
1
mol
 nN2
=
wt/Mwt
=
(8.0g)/(28.0g/mol)=
0.286
mol
 n
=
1+
0.286
=1.286
mol
 What
is
the
total
pressure
exerted
by
a
mixture
of
2.0g
of
H2
and
8.0g
of
 N2
at
273K
in
a
10.0
L
vessel?
 Given,
 T
=
273
K
 V
=
10
L
 R
=
0.08206
L.atm/mol.K
 n
=
Number
of
moles=
nH2
+nN2
 nH2
=
wt/Mwt
=
(2.0g)/(2.0g/mol)
=
1
mol
 nN2
=
wt/Mwt
=
(8.0g)/(28.0g/mol)=
0.286
mol
 n
=
1+
0.286
=1.286
mol
 n
=
?

and
P
=?
 P
=nRT/V
 


=
1.286
X
0.08206
X
273
/10
 

P
=
2.88
atm
 In
the
first
step
of
making
nitric
acid,
ammonia
and
oxygen
reacts

to
 form
nitric
oxide
as
shown
in
the
equaQon:
 
 
 
4NH3(g)
+
5O2(g)

4NO(g)
+
6
H2O(g)
 How
many
liters
of
NH3(g)
at
850oC
and
5atm
are
required
to
react
with

 1.0
mol
of
O2(g)
in
this
reacQon?
 In
the
first
step
of
making
nitric
acid,
ammonia
and
oxygen
reacts

to
 form
nitric
oxide
as
shown
in
the
equaQon:
 
 
 
4NH3(g)
+
5O2(g)

4NO(g)
+
6
H2O(g)
 How
many
liters
of
NH3(g)
at
850oC
and
5atm
are
required
to
react
with

 1.0
mol
of
O2(g)
in
this
reacQon?
 First,
we
have
to
calculate
the
volume
occupied
by
Oxygen
under
the
 given
condiQons.
 P
=
5atm
 T=
850+273
=1123K
 n=
1,
R=
0.08206
L.atm/mol.K
 V
=
nRT/P
=
18.43L
 In
the
first
step
of
making
nitric
acid,
ammonia
and
oxygen
reacts

to
 form
nitric
oxide
as
shown
in
the
equaQon:
 
 
 
4NH3(g)
+
5O2(g)

4NO(g)
+
6
H2O(g)
 How
many
liters
of
NH3(g)
at
850oC
and
5atm
are
required
to
react
with

 1.0
mol
of
O2(g)
in
this
reacQon?
 First,
we
have
to
calculate
the
volume
occupied
by
Oxygen
under
the
 given
condiQons.
 
 
 
 
 
 
 
 
 
Now,
look
at
the
equaQon
 P
=
5atm 
 
 
 
 
 
 
5
moles
of
Oxygen
required
to
react
 
 
 
 
 
 
 
 
 
 
with
4
moles
of
Ammonia. 
 
 

 T=
850+273
=1123K 
 
 
 

 
 
 
 
 
 
 
 
 
means,
5
(22.4
L)
of
Oxygen
requires

 n=
1,
R=
0.08206
L.atm/mol.K





4(22.4L)
of
Nitrogen. 
 

 
 
 
 
 
 
 
 
 
means,
112L
of
Oxygen
requires 

 V
=
nRT/P
=
18.43L 
 
 
 
89.6
L
of
ammonia.
 In
the
first
step
of
making
nitric
acid,
ammonia
and
oxygen
reacts

to
 form
nitric
oxide
as
shown
in
the
equaQon:
 
 
 
4NH3(g)
+
5O2(g)

4NO(g)
+
6
H2O(g)
 How
many
liters
of
NH3(g)
at
850oC
and
5atm
are
required
to
react
with

 1.0
mol
of
O2(g)
in
this
reacQon?
 First,
we
have
to
calculate
the
volume
occupied
by
Oxygen
under
the
 given
condiQons.
 
 
 
 
 
 
 
 
 
Now,
look
at
the
equaQon
 P
=
5atm 
 
 
 
 
 
 
5
moles
of
Oxygen
required
to
react
 
 
 
 
 
 
 
 
 
 
with
4
moles
of
Ammonia. 
 
 

 T=
850+273
=1123K 
 
 
 

 
 
 
 
 
 
 
 
 
means,
5
(22.4
L)
of
Oxygen
requires

 n=
1,
R=
0.08206
L.atm/mol.K





4(22.4L)
of
Nitrogen. 
 

 
 
 
 
 
 
 
 
 
means,
112L
of
Oxygen
requires 

 V
=
nRT/P
=
18.43L 
 
 
 
89.6
L
of
ammonia.
 So
how
18.43
liters
of
O2
requires
‐‐‐‐‐
Liters
of
ammonia?
14.75L
 A
container
has
1g
of
H2,
4g
of
O2
and
7g
of
N2
gases
at
12
atm
pressure.
 Calculate
the
parQal
pressure
exerted
by
individual
gases?
 A
container
has
1g
of
H2,
4g
of
O2
and
7g
of
N2
gases
at
12
atm
pressure.
 Calculate
the
parQal
pressure
exerted
by
individual
gases?
 First,
Calculate
the
mole
fracQons

need
number
of
moles
of
each
gas
 Number
of
moles
 
nH2
=
wt/Mwt
=
1/2
=
0.5
mols
 nO2
=
wt/Mwt
=
4/32
=
0.125
mols
 nN2
=
wt/Mwt
=
7/28
=
0.25
mols
 n
=
0.5+0.125+0.25
=
0.875
mols
 A
container
has
1g
of
H2,
4g
of
O2
and
7g
of
N2
gases
at
12
atm
pressure.
 Calculate
the
parQal
pressure
exerted
by
individual
gases?
 First,
Calculate
the
mole
fracQons

need
number
of
moles
of
each
gas
 Number
of
moles
 
nH2
=
wt/Mwt
=
1/2
=
0.5
mols
 Mole
fracQons
 
XH2
=
0.5/0.875
=
0.571
 nO2
=
wt/Mwt
=
4/32
=
0.125
mols
 XO2
=
0.125/0.875=
0.143
 nN2
=
wt/Mwt
=
7/28
=
0.25
mols
 XN2
=
0.25/0.875
=
0.286
 n
=
0.5+0.125+0.25
=
0.875
mols
 X
=
0.571+0.143+0.286
=1.0

 A
container
has
1g
of
H2,
4g
of
O2
and
7g
of
N2
gases
at
12
atm
pressure.
 Calculate
the
parQal
pressure
exerted
by
individual
gases?
 First,
Calculate
the
mole
fracQons

need
number
of
moles
of
each
gas
 Number
of
moles
 
nH2
=
wt/Mwt
=
1/2
=
0.5
mols
 Mole
fracQons
 
XH2
=
0.5/0.875
=
0.571
 ParQal
Pressures
 PH2
=
XH2P
=
0.571X12
=
6.84atm
 nO2
=
wt/Mwt
=
4/32
=
0.125
mols
 XO2
=
0.125/0.875=
0.143
 PO2
=
XO2P
=
0.143X12
=
1.72atm
 nN2
=
wt/Mwt
=
7/28
=
0.25
mols
 XN2
=
0.25/0.875
=
0.286
 PN2
=
XN2P
=
0.286X12
=
3.44atm
 n
=
0.5+0.125+0.25
=
0.875
mols
 X
=
0.571+0.143+0.286
=1.0

 ...
View Full Document

Ask a homework question - tutors are online