Transcript Gas Law Notes

Gases
Chapter 5
1
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Relating temperature, pressure,
volume and amount of gases
Heat an aerosol can
Blow up a balloon and place it in a freezer
Adding 5 mols of nitrogen to a tank that already contains
10 mols of nitrogen
Exploding tires on a hot summer day
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Elements that exist as gases at 250C and 1 atmosphere
3
NO2 gas
4
Physical Characteristics of Gases
solid
liquid
gas
•
Gases assume the volume and shape of their containers.
•
Gases are the most compressible state of matter.
•
Gases will mix evenly and completely when confined to
the same container.
•
Gases have much lower densities than liquids and solids.
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Atmospheric Pressure
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
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Force
Pressure = Area
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Barometer
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The pressure outside of a jet plane flying at high altitude falls
considerably below standard atmospheric pressure. Therefore
the air inside of the cabin must be pressurized to protect the
passengers. What is the pressure in the cabin if the barometer
reading is 749 mmHg?
1 atm = 760 mm Hg
P = 749 mmHg x
1 atm
760 mmHg
= 0.986 atm
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Problem 5.13
Convert 562 mmHg to atm.
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The atmospheric pressure in New York City on a given day is
744 mmHg. What was the pressure in kPa?
1 atm = 760 mm Hg = 101.325 kPa
P = 744 mmHg x
101.325 kPa
760 mmHg
= 99.2 kPa
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Problem 5.14
The atmospheric pressure at the summit
of Mt. McKinley is 606 mmHg on a certain
day.
What is the pressure in atm? In kPa?
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Manometers Used to Measure Gas Pressures
closed-tube
open-tube
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Forces Affecting Gases
•
•
•
•
.
.
.
.
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The Gas Laws
• Boyle’s Law
– Robert Boyle (1627-1691)
– Relates pressure and volume at constant
temperature
– For a given mass of a gas at constant
temperature, the volume of the gas varied
inversely with pressure.
Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P (h) increases
V decreases
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Boyle’s Law
P a 1/V
PxV=k
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
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A sample of chlorine gas occupies a volume of 946 mL at a
pressure of 726 mmHg. What is the pressure of the gas (in
mmHg) if the volume is reduced at constant temperature to 154
mL?
P x V = constant
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
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The Gas Laws
• Charles's Law
– Jacques Charles (1746-1823)
– Relates temperature and volume at
constant pressure
– For a given mass of a gas at constant V = b x T
pressure, the volume of the gas varied V = V
1
2
directly with temperature.
T1 T2
Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases
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Variation of Gas Volume with Temperature
at Constant Pressure
Charles’ Law
VaT
V = constant x T
Temperature must be
in Kelvin
V1/T1 = V2 /T2
T (K) = t (0C) + 273.15
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A sample of carbon monoxide gas occupies 3.20 L at 125 0C.
At what temperature will the gas occupy a volume of 1.54 L if
the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
T1 = 125 (0C) + 273.15 (K) = 398.15 K
T2 =
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
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Problem 5.24
Under constant pressure conditions a sample of
hydrogen gas initially at 88ºC and 9.6 L is
cooled until its final volume is 3.4 L. What is the
final temperature (in K)?
V=bxT
V1 = V2
T1 T2
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The Gas Laws
• Gay-Lussac's Law
– Joseph Gay-Lussac (1778-1850)
• Gay-Lussac studied the effect of temperature
on the pressure of a gas at constant volume
– For a given mass of a gas at constant
volume, the pressure of the gas varied
directly with temperature.
P1 = P2
T1 T2
A sample of nitrogen gas has a pressure of 2.50 kPa at 25 0C.
At what temperature will the gas have a pressure of 1.54 kPa
if the volume remains constant?
P1 /T1 = P2 /T2
P1 = 2.50 kPa
T1 = 298.15 K
P2 = 1.54 kPa
T2 = ?
T1 = 25 (0C) + 273.15 (K) = 298.15 K
T2 =
P2 x T1
P1
1.54 kPa x 298.15 K
=
= 184 K
2.50 kPa
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Problem 5.20
At 46ºC a sample of ammonia gas exerts a
pressure of 5.3 atm. What is the temperature
when the pressure of the gas is reduced to onetenth of the original pressure at the same
volume?
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Problem 5.22
A sample of air occupies 3.8 L when the pressure is 1.2
atm. (a) What volume (in L) does it occupy at
5010mmHg if the temperature remains constant?
(b) At the original pressure and volume the temperature
was 101ºC. What temperature is necessary to increase
the pressure to 5010mmHg, but keep the volume
constant?
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Explain the following. State
the gas law.
• Helium balloon pops at a party on a
summer day
• Predict how the size of exhaled bubbles
change as a scuba diver approaches
the surface of the water
• Aerosol cans have a warning not to put
near a flame.
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Describe the relationships
of…
1. Boyle’s Law
2. Charles’s Law
3. Gay-Lussac’s Law
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Avogadro’s Law
V a number of moles (n)
Constant temperature
Constant pressure
V=axn
V1 / n1 = V2 / n2
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Ammonia burns in oxygen to form nitric oxide (NO) and water
vapor. How many volumes of NO are obtained from one volume
of ammonia at the same temperature and pressure?
4NH3 + 5O2
1 mole NH3
4NO + 6H2O
1 mole NO
At constant T and P
1 volume NH3
1 volume NO
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Summary of Gas Laws
Boyle’s Law
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Charles Law
Gay-Lussac’s Law
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Avogadro’s Law
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Ideal Gas Equation
Boyle’s law: P a 1 (at constant n and T)
V
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
nT
P
V = constant x
nT
P
=R
nT
P
R is the gas constant
PV = nRT
R = 0.082057 L • atm / (mol • K)
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The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
Calculate the volume of 1 mole of an ideal gas at STP
PV = nRT
R = 0.082057 L • atm / (mol • K)
When the moles are constant,
PV = nR
T
Therefore
P1V1 = P2V2
T1
T2
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What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = 273.15 K
P = 1 atm
PV = nRT
nRT
V=
P
1 mol HCl
n = 49.8 g x
= 1.37 mol
36.45 g HCl
1.37 mol x 0.0821
V=
L•atm
mol•K
x 273.15 K
1 atm
V = 30.7 L
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Problem 5.40
Calculate the volume (in L) of 88.4 g of
CO2 at STP.
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Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb containing argon
at 1.20 atm and 18 0C is heated to 85 0C at constant volume.
What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant
nR
= P = constant
T
V
P1
P2
=
T1
T2
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
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Extra Practice: Problem 5.36
The temperature of 2.5 L of a gas initially
at STP is raised to 250ºC at constant
volume. Calculate the final pressure of
the gas in atm.
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Extra Practice: Problem 5.32
Given that 6.9 moles of carbon monoxide
are present in a container of volume 30.4
L, what is the pressure of the gas (in atm)
if the temperature is 62ºC?
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Density (d) Calculations
PM
m
d=
=
V
RT
m is the mass of the gas in g
M is the molar mass of the gas
m
n =M
Molar Mass (M ) of a Gaseous Substance
dRT
M=
P
d is the density of the gas in g/L
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A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0
0C. What is the molar mass of the gas?
dRT
M=
P
M=
g
2.21
L
4.65 g
m
=
= 2.21
d=
2.10
L
V
x 0.0821
L•atm
mol•K
g
L
x 300.15 K
1 atm
M = 54.5 g/mol
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Problem 5.44
At 741 torr and 44ºC 7.10 g of a gas
occupy a volume of 5.40 L. What is the
molar mass of the gas (in g/mol)?
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Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm
when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l)
g C6H12O6
mol C6H12O6
5.60 g C6H12O6 x
6 mol CO2
1 mol C6H12O6
x
180.12 g C6H12O6
1 mol C6H12O6
V=
nRT
=
P
mol CO2
V CO2
L•atm
x 310.15 K
mol•K
1.00 atm
= 0.187 mol CO2
0.187 mol x 0.0821
= 4.76 L
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Problem 5.60
Ethanol (C2H5OH) burns in air:
C2H5OH(l) + O2(g)  CO2(g) + H2O(l)
Determine the volume of air in liters at 35.0ºC and 790
mmHg required to burn 227 g of ethanol. Assume that
air is 21.0 percent O2 by volume.
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Dalton’s Law of Partial Pressures
V and T are constant
P1
P2
Ptotal = P1 + P2
Studying gaseous mixtures shows that each gas
behaves independantly of the others
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Consider a case in which two gases, A and B, are in a
container of volume V.
nART
PA =
V
nA is the number of moles of A
nBRT
PB =
V
nB is the number of moles of B
PT = PA + PB
PA = XA PT
nA
XA =
nA + nB
nB
XB =
nA + nB
PB = XB PT
Pi = Xi PT
mole fraction (Xi ) =
ni
nT
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A sample of natural gas contains 8.24 moles of CH4, 0.421
moles of C2H6, and 0.116 moles of C3H8. If the total pressure
of the gases is 1.37 atm, what is the partial pressure of
propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane =
8.24 + 0.421 + 0.116
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
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Problem 5.63
A mixture of gases contains 0.31 mol
CH4, 0.25 mol C2H6, and 0.29 mol C3H8.
The total pressure is 1.50 atm. Calculate
the partial pressure of each gas.
PCH 4 = _____ PC 2 H 6 = ______ PC 3 H 8 = ______
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Collecting a Gas over Water
2KClO3 (s)
2KCl (s) + 3O2 (g)
PT = PO2 + PH2 O
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Vapor of Water and Temperature
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Problem 5.66
A mixture of helium and neon gasses is collected over water at
28.0ºC and 745 mmHg. If the partial pressure of helium is 368
mmHg, what is the partial pressure of neon (in mmHg)?
(vpH20 = 28.3 mmHg at 28.0ºC)
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Problem 5.68
A sample of zinc metal reacts completely with an excess of
hydrochloric acid:
Zn(s) + 2HCl(aq) 
The hydrogen gas produced is collected over water at 25.0ºC. The
volume of the as is 7.80L, and the pressure is 0.980 atm. Calculate
the amount of zinc metal (in grams) consumed in the reaction.
(vpH20 = 28.3 mmHg at 28.0ºC)
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