Transcript The Nature of Gases

Chapter 12
Gases
The Combined Gas Law
Volume and Moles
(Avogadro’s Law)
Partial Pressures
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Combined Gas Law
Combines Boyles Law,
Charles’ Law and Gay Lussac’s Law
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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
P2 = 3.20 atm
V1 = 0.180 L
V2= 90.0 mL
T1 = 302 K
T2 = ??
Fill in with increase or decrease
P change will _____________ T.
V change will _____________ T.
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Solution
Enter data for T and P, then T and V
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
(inc. T)
(dec. T)
T2 = 604 K - 273 = 331 °C
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Learning Check C1
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 C1
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 C2
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 C2
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 C3
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 C3
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 C4
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 C4
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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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 goes under water, the
high pressure of the water causes more N2 (g)
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 C6
A scuba tank contains O2 with a pressure of
0.540 atm and He at 855 mm Hg? What is
the total pressure in mm Hg in the tank?
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Solution C6
0.450 atm x 760 mm Hg =
1 atm
342 mm Hg
342 mm Hg + 855 mm Hg = 1197 mm Hg
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