equal volumes of gas have the same number of molecules (atoms)
•
V
∝
n
where n = moles
See practice problems in book.
Fig 5.10.
Volume relationships of gases.
Volume relationship is the same as the
mole relationship.
Chapter 5
Chemistry 1303
7
5.4
Ideal Gas Equation
Ideal gas

v
olume of molecules is negligible; applies in most temperature
and pressure ranges.
•
combination of gas laws and Avogadro's Principle
•
Standard Molar Volume (
STP
)
•
Standard Temperature (
0°C or 273 K
)
•
Standard Pressure (
1 atm
)
•
1 mole of any gas occupies 22.4 L
•
or
22.4 L
=
1 mole
but ONLY at STP
PV
=
nRT
where
R
=
"universal gas constant"
=
0.0821 L
⋅
atm / mole
⋅
K
Chapter 5
Chemistry 1303
8
Where does R come from?
R = PV/nT
(Ideal gas law rearranged)
R = (1 atm)(22.414 L) / (1 mol)(273.15K) = 0.082057 L
⋅
atm / mole
⋅
K
{ Useful in many different kinds of calculations involving gases!!! }
Example:
At
STP
, the density of a certain gas is 4.29 g/L.
What is the
molecular mass of the gas?
4.29 g
x
22.4 L
= 96.0 g / mole
L
mole
Example
:
What is the density (g/L) of UF
6
at 779 mm Hg and 62
°
C?
Note:
alternate way shown in book means memorizing
yet another equation.
Just think density  mass /volume
Chapter 5
Chemistry 1303
9
5.5
Gas Stoichiometry
Example:
Acetylene (welding gas), C
2
H
2
, is produced by hydrolysis of
calcium carbide.
CaC
2 (s)
+
2 H
2
O
→
Ca(OH)
2 (s)
+
C
2
H
2 (g)
Starting with 50.0 g of CaC
2
, what is the theoretical yield of
acetylene in liters, collected at 24°C and a pressure of 745
Torr?
CaC
2 (s)
+
2 H
2
O
→
Ca(OH)
2 (s)
+
C
2
H
2 (g)
1st find yield in
moles
:
50.0 g CaC
2
x
x
=
0.780 mole C
2
H
2
1 mole C
2
H
2
1 mole CaC
2
64.10 g CaC
2
1 mole CaC
2
2
now use ideal gas law to find
volume
of C
2
H
2
:
PV = nRT
P
V
=
nRT
V
=
(0.780 mole) x (0.0821 L atm / mole K) x (297 K)
(745 torr) x (1 atm / 760 torr)
=
19.4 L
Chapter 5
Chemistry 1303
10
3 A
+
2 B
→
4
C
+
D
Chapter 5
Chemistry 1303
11
5.6
Dalton's Law of Partial Pressures
A.
For a mixture of gases:
P
T
=
P
a
+
P
b
+
P
c
+
. . . . .
P
T
=
P
a
+
P
b
= n
a
RT / V
+ n
b
RT / V
P
T
= (n
a
+
n
b
) (RT / V)
P
a
=
n
a
RT / V
=
n
a
= X
a
P
t
(n
a
+ n
b
)RT / V
n
a
+ n
b
X
a
= mole fraction
( dimensionless quantity)
X
a
=
moles of one component / total moles present
P
i
= X
i
.
P
T
Chapter 5
Chemistry 1303
12
Example:
A sample of natural gas contains 8.24 moles of CH
4
, 0.421 moles of
ethane, 0.116 moles of C
3
H
8
.
If the total pressure of the gas is 1.37
atm., what is the partial pressure of each gas?
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 Fall '08
 Maguire
 Chemistry