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Gases
A Review
Elements that exist as gases at 250C and 1 atmosphere
Physical Characteristics of Gases
•
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.
Properties of Gases
Gas properties defined using
only 4 varibles
• V = volume of the gas (L)
• T = Kelvin temperature (K)
• n = number of moles (mol)
• P = pressure in either
(atm), (mm Hg) or (kPa)
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
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
Atmospheric
pressure
Barometer
Manometer
Boyle’s Law
If n and T are
constant, then
PV = (nRT) = k
This means, for
example, that P
goes up as V goes
down.
Robert Boyle
(1627-1691).
Son of Early of
Cork, Ireland.
Boyle’s Law
A bicycle pump is a good example of
Boyle’s law.
As the volume of the air trapped in
the pump is reduced, its pressure
goes up, and air is forced into the
tire.
Boyle’s Law
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
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?
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
Charles’s
Law
If n and P are
constant, then
V = (nR/P)T = kT
V and T are directly
related.
Jacques Charles (17461823). Isolated boron
and studied gases.
Balloonist.
Charles’s original balloon
Modern long-distance balloon
As T increases
V increases
Variation of gas volume with temperature
at constant pressure.
Charles’ &
Gay-Lussac’s
Law
VaT
V = constant x T
V1/T1 = V2/T2
Temperature must be
in Kelvin
T (K) = t (0C) + 273.15
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
Avogadro’s Law
Equal volumes of gases at the
same T and P have the same
number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many
molecules
Avogadro’s Law
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
Constant temperature
Constant pressure
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
Ideal Gas Equation
Boyle’s law: V a 1 (at constant n and T)
P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
V = constant x
nT
P
=R
nT
P
nT
P
R is the gas constant
PV = nRT
The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.
PV = nRT
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
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=
V = 30.6 L
L•atm
mol•K
1 atm
x 273.15 K
What is the volume (in liters) occupied by 49.8 g of HCl
at STP?
grams of HCl  moles of HCl  liters of HCl at STP
1 mol HCl
22.41 L
V  49.8 g HCl 

 30.6 L
36.45 g HCl 1 mol HCl
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
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
Molar Mass (M ) of a Gaseous Substance
dRT
M=
P
d is the density of the gas in g/L
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm
and 27.00C. What is the molar mass of the gas?
dRT
M=
P
M=
g
2.21
L
M = 54.6 g/mol
4.65 g
m
=
= 2.21
d=
2.10
L
V
x 0.0821
L•atm
mol•K
1 atm
x 300.15 K
g
L
Gas
Stoichiometry
• You can apply the discoveries of Gay -Lussac and
Avogadro to calculate the stoichiometry of chemical
reaction.
• Step 1  Write a balanced chemical equation.
• Step 2  Convert all units to common units
• Step 3  Start with given and Use factor-label skills
(Via MOLES) to get to desired units.
• Step 4 If gas laws are needed use appropriately at
this point or prior to factor- label step.
Gas Stoichiometry
What is the volume of CO2 produced at 370 C 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
= 0.187 mol CO2
180 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 x 0.0821
= 4.76 L
Gas Stoichiometry problems
• 1) Ammonia can react with Oxygen to produce nitrogen and water.
If 1.78 L of O2 reacts, what volume of nitrogen will be produce?
(Temp and pressure remain constant)
Ans.= 1.19 L of N2
• 2) Ethylene gas( C2H4) combust in Air to form H2O and CO2. If 13.8
L of C2H4 measured at 21°C and 1.038 atm burns completely with
O2, calculate the volume of CO2 produced assuming the CO2 is
measured at 44°C and 0.989 atm.
Ans.= 32.6 L of CO2
• 3) When Arsenic (III) sulfide is roasted in air, it reacts with oxygen to
produce arsenic (III) oxide and sulfur dioxide. When 89.5 g As2S3
is roasted with excess O2, what volume of SO2 is produced? The
gaseous product is measured at 20°C and 98.0 kPa.
Ans.= 27.1 L of SO2
Dalton’s Law of Partial Pressures
V and T
are
constant
P1
P2
Ptotal = P1 + P2
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
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
Chemistry in Action:
Scuba Diving and the Gas Laws
P
Depth (ft)
Pressure
(atm)
0
1
33
2
66
3
V
PV = nRT
Kinetic Molecular Theory of Ideal Gases
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be points;
that is, they possess mass but have negligible volume.
2. Gas molecules are in constant motion in random directions.
Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces
on one another.
4. The average kinetic energy of the molecules is proportional
to the temperature of the gas in kelvins. Any two gases at
the same temperature will have the same average kinetic
energy
Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
“12m10an3.mov”
• Charles’ Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
PaT
“12m10an2.mov”
Kinetic theory of gases and …
• Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
Pan
“12m10an1.mov”
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = SPi
Deviations from
Ideal Gas Law
• Real molecules
have volume.
• There are
intermolecular
forces.
–Otherwise a
gas could not
become a
liquid.
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
PV = 1.0
n=
RT
Repulsive Forces
Attractive Forces
Deviations from Ideal Gas Law
Account for volume of molecules
and intermolecular forces with
Van der Waal’s Equation.
Measured V = V(ideal)
Measured P
(
P
n2 a
+ ----V2
)
V
-
nb
nRT
vol. correction
intermol. forces
J. van der Waals,
1837-1923,
Professor of
Physics,
Amsterdam.
Nobel Prize 1910.
Deviations from Ideal Gas Law
Cl2 gas has a = 6.49, b = 0.0562
For 8.0 mol Cl2 in a 4.0 L tank at 27 oC.
P (ideal) = nRT/V = 49.3 atm
P (van der Waals) = 29.5 atm
Van der Waals equation
nonideal gas
2a
n
( P + V2 ) (V – nb) = nRT
}
}
corrected
pressure
corrected
volume
During Calculations plug in
a and b values into the
equation and solve
algebraically