Chapter 7
Gases
The Combined Gas Law
Volume and Moles
(Avogadro’s Law)
Partial Pressures
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Combined Gas Law
P 1V 1
T1
=
P 2V 2
T2
Rearrange the combined gas law to solve for
V2
P 1V 1T 2
V2
=
=
P 2V 2T 1
P 1 V 1T 2
P 2T 1
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Combined Gas Law
P 1V 1
T1
=
P 2V 2
T2
Isolate V2
P 1V 1T 2
V2
=
=
P 2V 2T 1
P 1 V 1T 2
P 2T 1
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Learning Check C1
Solve the combined gas laws for T2.
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Solution C1
Solve the combined gas law for T2.
(Hint: cross-multiply first.)
P 1V 1
T1
=
P 2V 2
T2
P 1V 1T 2 = P 2V 2T 1
T2
= P 2V 2T 1
P 1V 1
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Combined Gas Law Problem
A sample of helium gas has a volume of
0.180 L, a pressure of 0.800 atm and a
temperature of 29°C. What is the new
temperature(°C) of the gas at a volume of
90.0 mL and a pressure of 3.20 atm?
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Data Table
Set up Data Table
P1 = 0.800 atm
V1 = 0.180 L
T1 = 302 K
P2 = 3.20 atm
V2= 90.0 mL
T2 = ??
??
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Solution
Solve for T2
Enter data
T2 = 302 K x
T2 =
atm x
atm
K - 273 =
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mL =
mL
K
°C
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Calculation
Solve for T2
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
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Learning Check C2
A gas has a volume of 675 mL at 35°C and
0.850 atm pressure. What is the
temperature in °C when the gas has a
volume of 0.315 L and a pressure of 802
mm Hg?
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Solution G9
T1 = 308 K
T2 = ?
V1 = 675 mL
V2 = 0.315 L = 315 mL
P1 = 0.850 atm
= 646 mm Hg
P2 = 802 mm Hg
T2 = 308 K x 802 mm Hg
646 mm Hg
P inc, T inc
= 178 K - 273
= - Timberlake
95°C
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x
315 mL
675 mL
V dec, T dec
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Volume and Moles
How does adding more molecules of a gas
change the volume of the air in a tire?
If a tire has a leak, how does the loss of air
(gas) molecules change the volume?
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Learning Check C3
True (1) or False(2)
1.___The P exerted by a gas at constant V is not
affected by the T of the gas.
2.___ At constant P, the V of a gas is directly
proportional to the absolute T
3.___ At constant T, doubling the P will cause the
V of the gas sample to decrease to one-half its
original V.
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Solution C3
True (1) or False(2)
1. (2)The P exerted by a gas at constant V is not
affected by the T of the gas.
2. (1) At constant P, the V of a gas is directly
proportional to the absolute T
3. (1) At constant T, doubling the P will cause the
V of the gas sample to decrease to one-half its
original V.
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Avogadro’s Law
When a gas is at constant T and P, the V is
directly proportional to the number of
moles (n) of gas
V1
n1
initial
=
V2
n2
final
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STP
The volumes of gases can be compared
when they have the same temperature and
pressure (STP).
Standard temperature
0°C or 273 K
Standard pressure
1 atm (760 mm Hg)
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Learning Check C4
A sample of neon gas used in a neon sign has
a volume of 15 L at STP. What is the volume
(L) of the neon gas at 2.0 atm and –25°C?
P1 =
P2 =
V2 = 15 L x
V1 =
V2 = ??
atm
atm
T1 =
T2 =
x
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K
K
K
K
= 6.8 L
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Solution C4
P1 = 1.0 atm
P2 = 2.0 atm
V1 = 15 L
V2 = ??
V2 = 15 L x 1.0 atm
2.0 atm
x
T1 = 273 K
T2 = 248 K
248 K = 6.8 L
273 K
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Molar Volume
At STP
4.0 g He
1 mole
(STP)
V = 22.4 L
16.0 g CH4
1 mole
(STP)
44.0 g CO2
1mole
(STP)
V = 22.4 L
V = 22.4 L
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Molar Volume Factor
1 mole of a gas at STP = 22.4 L
22.4 L
1 mole
and
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1 mole
22.4 L
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Learning Check C5
A.What is the volume at STP of 4.00 g of
CH4?
1) 5.60 L
2) 11.2 L
3) 44.8 L
B. How many grams of He are present in
8.0 L of gas at STP?
1) 25.6 g
2) 0.357 g
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3) 1.43 g
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Solution C5
A.What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4
1 mole CH4
B. How many grams of He are present in 8.0 L of gas
at STP?
8.00 L x
1 mole He x
22.4 He
4.00 g He = 1.43 g He
1 mole He
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Daltons’ Law of Partial
Pressures
Partial Pressure
Pressure each gas in a mixture would exert
if it were the only gas in the container
Dalton's Law of Partial Pressures
The total pressure exerted by a gas mixture
is the sum of the partial pressures of the
gases in that mixture.
PT = P1 +
P2 + P3 + .....
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Gases in the Air
The % of gases in air
Partial pressure (STP)
78.08% N2
593.4 mmHg
20.95% O2
159.2 mmHg
0.94% Ar
7.1 mmHg
0.03% CO2
0.2 mmHg
PAIR = PN + PO + PAr + PCO = 760 mmHg
2
2
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Learning Check C6
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
1) 35.6
2) 156
3) 760
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
2) 9.14
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3) 0.109
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Solution C6
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
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Partial Pressures
The total pressure of a gas mixture depends
on the total number of gas particles, not on
the types of particles.
P = 1.00 atm
P = 1.00 atm
1 mole H2
0.5 mole O2
+ 0.3 mole He
+ 0.2 mole Ar
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Health Note
When a scuba diver is several hundred feet
under water, the high pressures cause N2 from
the tank air to dissolve in the blood. If the
diver rises too fast, the dissolved N2 will form
bubbles in the blood, a dangerous and painful
condition called "the bends". Helium, which is
inert, less dense, and does not dissolve in the
blood, is mixed with O2 in scuba tanks used for
deep descents.
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Learning Check C7
A 5.00 L scuba tank contains 1.05 mole of
O2 and 0.418 mole He at 25°C. What is the
partial pressure of each gas, and what is
the total pressure in the tank?
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Solution C7
P = nRT
V
PT
=
PT = PO + PHe
2
1.47 mol x 0.0821 L-atm x 298 K
=
5.00 L
7.19 atm
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(K mol)
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