Transcript Document

Gases
Gases
I.
comparison of solids, liquids and Gasses
A.
fluids- things that flow - gases and liquids
B.
condensed states - liquids and solids
C.
Vapor - refers to a gas that is formed by
evaporation of a liquid or the sublimation
of a solid (sublimation - going directly
from a solid to a gas)
II
some common properties of gases
A. Gases can be compressed into smaller volumes;
that is, their densities can be increased by
applying increased pressure.
B.
Gases exert pressure on their surroundings;
in turn, pressure must be exerted to confine
gases.
C.
Gases expand without limits, and so gas
samples completely and uniformly occupy
the volume of any container.
D.
Gases diffuse into each other, and so samples of
gas placed in the same container mix completely.
Conversely,different gases in a mixture do not
separate on standing.
1.
diffusion - going from a high concentration
to a low concentration
E.
expand to fill container
F.
take shape of container
G.
some are invisible
H.
I.
some are combustible
four properties determine the physical
behavior of a gas
1.
amount of gas
2.
3.
volume of the gas
temperature of the gas
4.
pressure of the gas
Elements that exist as gases at 250C and 1 atmosphere
5.1
5.1
J.
exert Pressure
1.
2.
Pressure = force = Newtons = pascal
Area
m2
a.
Named after Blaise Pascal (1625-1662) father of modern Hydraulics
b.
units Newtons/m2 = 1 Pascal (Pa)
c.
1 Pascal is rather small - more comely
used unit is the kilopascal
liquid pressure = g x h x d
Force
Pressure = Area
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mm Hg = 760 torr
= 101,325 Pa
= 14.7 psi
= 29.92 in. Hg
Barometer
5.2
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
5.2
4.
pressure of gases is due to the billion
collision/sec of the gas molecules with
the walls of the container
5.
Barometer - device used to measure gas pressure
a.
invented by Evangelista Torricelli
Boyle’s Law
P α 1/V
This means Pressure and
Volume are INVERSELY
PROPORTIONAL if moles
and temperature are
constant (do not change).
For example, P goes up as
V goes down.
P1V1 = P2 V2
Robert Boyle
(1627-1691).
Son of Earl of
Cork, Ireland.
Gas Law Problems
have 5 steps
e.g. A gas under a pressure of 750. mm of Hg
occupies 25 liters. What will its volume be
under a pressure of 800. mm of Hg?
Temperature remains constant.
P1 = 750. mm of Hg
V1 = 25 L
Step 1
V2 = X
P2 = 800. mm or Hg
P1V1 = P2V2 - base equation
Step 2
P1V1 = P2V2
P2
Divide by P2 on both sides
P2
Make sure your unkown is on the top
P1V1 = V2
P2
- working equation
Step 3
(750. mm of Hg)( 25 L) =
(800. mm of Hg)
Step 4
23 L
Step 5
Charles’s Law
If n and P are constant,
then V α T
V and T are directly
proportional.
V1
V2
=
T1
T2
• If one temperature goes up, the
volume goes up!
Jacques Charles (17461823). Isolated boron
and studied gases.
Balloonist.
There are 3 temperature scales
remember - temperature is a measure of molecular
motions
Fahrenheit
Celsius
Can you have F and C temperatures below 0?
If temperature is a measure of molecular movement How can you have negative temperatures?
We had to come up with a new
temperature scale
Kelvin
K = C + 273
What is 22 C in Kelvin?
K = 22 + 273 = 295 K
Absolute zero
0 K = no molecular motion
0 K = -273 C = -460 F
Really Really Cold
Example of Charles Law
A gas occupies a volume of 2.5 liters at a temperature
of 22 C. What temperature is required for the gas to
occupy 3.0 liters
V1 = 2.5 l
T1 = 22 C + 273 = 295 K
V2 = 3.0 liters
T2 = x
Step 1
V 1 = V2
T1
T2
Step 2
First you must move your unknown to the top
Multiply both sides by T2
T2
V1 = V2 T2
T1
T2
T1 T2 V1 = V2 T1
T1
V1
V1
T2 = V2T1
V1
= (3.0 liters)(295 K)
2.5 L
354 K = 350 K
Step 3
Step 4
Step 5
Gay-Lussac’s Law
If n and V are
constant,
then P α T
P and T are directly
proportional.
P1
P2
=
T1
T2
If one temperature goes up,
the pressure goes up!
Joseph Louis GayLussac (1778-1850)
Combined Gas Law
• The good news is that you don’t
have to remember all three gas
laws! Since they are all related to
each other, we can combine them
into a single equation. BE SURE
YOU KNOW THIS EQUATION!
P1 V1
P2 V2
=
T1
T2
By doing step 1, you will be able to
select the right equations.
If any on the variables are held
constant, disregard them.
e.g. A gas occupies 10.0 L at 700. mm of Hg and
a temperature of 30.0 C. What volume will the gas
Occupy at 800. mm of Hg and 20.0 C
V1=10.0 L
P1= 700. mm of Hg
T1 = 273 +30.0 = 303.0 K
V2 = x
P2 = 800. mm of Hg
T2= 273 + 20.0 C = 293.0 K
Step 1
P1V1 = P2V2
T1
T2
T2P1V1 = V2
Step 2
Step 3
P2T1
(293.0 K)(700. mm of Hg)(10.0 L)
(800. mm of Hg)(303.0 K)
8.46 L = V2
Step 5
Step 4
What temperature is required to change 250 ml of a gas
at 50.0 C and 860.0 mm of Hg to 150 ml at 1200. mm of Hg
V1= 250
P1= 800.0 mm of Hg
T1 = 273 +50.0 = 323.0 K
V2 = 150 ml
P2 = 1200. mm of Hg
T2=x
P1V1 = P2V2
T1
T2
T2 P1V1 = P2V2
T1
T2
T1
P1V1
T2P1V1
T1
T2 = P2V2T1
P1V1
T2
= P2V2 T1
P1V1
Now just put your numbers in
on handout - due question 7-15
7-10 are Charles law
11-15 are combined Gas Laws
Stop here
Standard Pressure =
1 atm (or an
equivalent)
Standard
Temperature = 0 deg
C (273 K)
STP allows us to
compare amounts of
gases between different
pressures and
temperatures
Avogadro’s Law
V a number of moles (n)
V = constant x n
Constant temperature
Constant pressure
V1/n1 = V2/n2
5.3
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
n = number of moles
nT
P
V = constant x
nT
P
=R
nT
P
R is the gas constant
PV = nRT
5.4
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.
22.4 L is standard volume
PV = nRT
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K) or .0821 L Atm
mole K
5.4
When you use the ideal gas equation
Pressure must be in atmospheres
1 atm = 760 mm of Hg
Temperature must be in Kelvin
volume in liters
R is the ideal gas constant = .0821 L Atm
mole K
How many liters will 12.0 g of O2 occupy at
780.0 mm of Hg and 30.0 C
) = 1.026 Atm
P = (780.0 mm of Hg) (1atm
760 mm of Hg
V= X
n=
R=
(12.0 g O2)( 1mole ) = .375 moles
32.0 g O2
.0821 L atm/mole K
T = K = 30.0 C + 273 = 303.0 K
PV = nRT
V = nRT/P
V= .375 moles .0821 L atm 303.0 K
mole K 1.026 Atm
V = 9.09 L
65 g of a CH4 occupies 30.0 L at a temperature of
10.0 C. What pressure must be exerted on this gas?
P=X
V= 30.0 L
n = (65 g CH4)( 1 mole CH4) =
(16 g CH4)
R = .0821 L Atm
Mole K
T = 10.0C + 273 = 283.0 K
4.1 moles CH4
PV = nRT
P = nRT
V
P = (4.1 mole CH4)(.0821 L Atm ) ( 283.0 K)
(30.0 L)
( mole K )
P = 3.1 Atm
What is the molecular weight of a 2.0 g sample of a
gas that occupies 5.0 L at a pressure of 0.303 Atm
at 22 C.
P= 0.303 Atm
V= 5.0 L
n= x
R = 0.0821 L Atm/mole K
T = 22 C + 273 ==295 K
PV= nRT
PV= nRT
PV = n
RT
(0.303 Atm)(5.0 L)(mole K)
(295 K)
(0.0821 L Atm)
n= 0.0625 moles
remember the units for M.W. is g/mole
M.W. = 2.0 g
.0625 moles
= 32 g/mole
e.G How many liters will .639 mole of Cl2 occupy
under standard conditions
PV = nRT
P=
V=
1 Atm
V = nRT
P
x
n = .639 moles Cl2
R=
T=
.0821 L Atm
Mole K
273 K
14.3 L =
V=
(.639 moles Cl2) (.0821 L Atm) (273 K)
(1 Atm)
Mole K
You need your books today
On page 358 – do questions 16-20
Gas Stoichiometry
Mass - Volume problems
What is the volume of CO2 produced when 5.60 g
of glucose are used up in the reaction at STP:
C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
g C6H12O6
mol C6H12O6
mol CO2
V CO2
5.5
1 mole
6 moles
C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
xL
5.60 g
22.4 L/mol
180. g/mole
molar volume
(5.60 gC6H12O6)
(
1 mole C6H12O6
180. g C6H12O6
4.18 L
)(
6 moles CO2
)(
1 mole C6H12O6
22.4 L CO2
1 mole CO2
)
How many liters of Cl2 is required to react
with 50.0 g of K to produce KCl at STP?
2 mole 1 mole
+
Cl 2 -----> 2KCl
2K
50.0 g
XL
39.1 g/mole 22.4 L/mole
When a problem gives you grams - change it to moles
(50.0 g K) ( 1 mole K ) (1 mole Cl2 )
(22.4 L)
(39.1g K) ( 2 moles K ) ( 1 mole Cl )
2
= 14.3 L
Volume - mass problems
During a collision, automobile air bags are inflated by N2
gas formed by the explosive decomposition
of sodium azide at STP
What mass of sodium azide would be needed to inflate
31.0 L
2 NaN3 ====> 2 Na + 3 N2
xg
31.0 L
65.0 g/mole
22.4 L/mole
(65.0 g NaN3)
(2
moles
NaN
)
3
(1
mole
N
)
2
(31.0 L N2)
( 22.4 L N2) ( 3 moles N2) ( 1 mole NaN3)
Try these two
How many liters of Hydrogen are produced from
110.0 g of water at STP?
H2O => H2 +
O2
How many grams of Na are required to produce 200.0 L
of H2 when it reacts with water at STP
Na + H2O => NaOH
+
H2
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
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 = 54.6 g/mol
5.3
Gas Stoichiometry continued
Volume - Volume problems
As the name suggest - you start with a volume and
end with a volume
What volume of O2 is required to completely burn
70.0 L of CH4?
CH4 + 2 O2 ====> CO2 + 2 H2O
As always - make sure you balance all equations
CH4 + 2O2 ====> CO2 + 2 H2O
(70.0 L of CH4) ( 2 liters of O2 )
( 1 liter of CH4 )
= 140. L of O2
You can see that a balanced chemical equation can
be used for a number of things
1
How many liters of O2 are produced when 50.0 L
of H2O2 decomposes
H2O2 ===> H2O + O2
2.
How many liters of O2 are required to burn 700.0
L of propane?
C3H8 + O2 ==> CO2 + H2O
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
5.4
Dalton’s Law of Partial Pressures
V and T
are
constant
P1
P2
Ptotal = P1 + P2
5.6
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 mole fraction (Xi) = ni
nT
nA
XA =
nA + n B
PA = XA PT
nB
XB =
n A + nB
Pi = Xi PT
PB = XB PT
5.6
5.6
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
5.7
GAS DIFFUSION AND
EFFUSION
• diffusion is the
gradual mixing of
molecules of
different gases.
• effusion is the
movement of
molecules through
a small hole into an
empty container.
GAS DIFFUSION AND
EFFUSION
Graham’s law governs
effusion and diffusion
of gas molecules.
KE=1/2 mv2
Rate for A
Rate for B
M of B
M of A
Rate of effusion is
inversely proportional
to its molar mass.
Thomas Graham, 1805-1869.
Professor in Glasgow and London.
GAS DIFFUSION AND
EFFUSION
Molecules effuse thru holes in a
rubber balloon, for example, at
a rate (= moles/time) that is
• proportional to T
• inversely proportional to M.
Therefore, He effuses more
rapidly than O2 at same T.
He
Gas Diffusion
relation of mass to rate of diffusion
• HCl and NH3 diffuse
from opposite ends of
tube.
• Gases meet to form
NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl
forms closer to HCl
end of tube.
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 mols C3H8.
Xpropane = (8.24 mols CH + 0.421mols C H + 0.116 mols C H .)
4
2 6
3 8
Xpropane = 0.0132
Ppropane = 0.0132 x 1.37 atm
= 0.0181 atm
5.6
Bottle full of oxygen
gas and water vapor
2KClO3 (s)
2KCl (s) + 3O2 (g)
PT = PO2 + PH2 O
5.6
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
• Charles’ Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
PaT
5.7
Kinetic Molecular Theory of 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
http://www.ewellcastle.co.uk/science/pages/P_constantM.
5.7
Kinetic theory of gases and …
• Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
Pan
• 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
5.7
What volume of O2 is required to react with 3.00 g of H2
at 22 C and 1.20 atm
2 H2 + O2 ===>2 H2O
(3.00g H2) ( 1 mole H2) ( 1 mole of O2 )
( 2.02 g H2) ( 2 moles H2 )
= .743 moles O2
P=
2.0 Atm
V=
250.0 L
n=
PV = nRT
PV = n
RT
x
.0821 L Atm
R=
Mole K
T = 20.0 C + 273 = 293.0 K
21 moles =
(2.0 Atm)(250.0 L) Mole K
0821 L Atm
293.0 K
1.
How many grams of Na is required to react with water to
produce 250.0 L of H2 at 20.0 C and 2.0 atm?
NaOH is the other product.
2 Na + 2 H2O ------> 2 NaOH
+ H2
Because you are looking for g and start with a volume,
it is a volume - mass problem
Start with the Ideal gas equation
(21 moles of H2)( 2 moles of Na )
(1 mole H2)
= 970 g Na
( 23 g Na)
(1 mole of Na)