Transcript Gas Unit and Pressure

Gas Laws
• The Gas Laws
• Kinetic Molecular Theory
• .Matter is composed of very tiny particles
• .Particles of matter are in continual motion
• .The total KE of colliding particles remains
constant
- When individual particles collide, some lose
energy while others gain energy.
- No overall energy loss = elastic collisons
Properties of a Gas
• Expansion – no definite shape or
volume
• Low density – density of gas is
about 1/1000 of that
same
substance in a liquid or
a solid
phase
• Diffusion – process of spreading
out spontaneously to occupy a
space uniformly
Ideal Gas
• consists of very small independent
particles. These particles move at
random in space and experience
elastic collisions
• Extra –
Approximate speed is 103 meters/sec,
at 0 degrees C and 1 ATM of pressure
• Undergo: 5 x 109 collisions/second
Gases in Real
life
are not so
ideal
At short distance s the nuc leus of one at om is
attracted b y the negat ively charged e lectron
orbitals of another a tom. It is these forces that
cause ga s molecules to come togethe r and
condense to a liquid when the speeds o f the gas
molecules are slowed enou gh to allow the
attract ive forces to ope rate.
When would a gas not act
“ideal”?
• At low temperatures - the molecules may
be moving slow enough to allow attractions
to occur
• At high pressures - the molecules may
once again have attractive forces between
them in effect
Gases
and Pressure
Pressure – the total average force
over a given area is called
pressure. Although the
collisions of any single molecule
of gas within the walls of a container are
intermittent, there are so many molecules in even a
small container, that the tremendous number of
collisions average out to a steady pressure.
Gas and Temperature
• Temperature – the average kinetic energy
or motion energy of its molecules.
• Avogadro’s hypothesis states that at any
given temperature and pressure, equal
volumes of gases, contain the same
number of molecules.
• The following table supports the hypothesis
by showing that one mole (6.02 x 1023
molecules) of 11 different gases occupies
approximately the same volume. These
volumes were derived experimentally at the
same pressure and temperature.
Table 1. Volumes of various gases at 0 C and 760 mm pressure
1 mole:
Oxygen
Sulfur Dioxide
Hydrogen
Helium
Chlorine
Nitrogen
Ammonia
Hydrogen Chloride
Carbon Monoxide
Carbon Dioxide
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
Occupies:
22.393 L
21.888 L
22.430 L
22.426 L
22.063 L
22.403 L
22.094 L
22.248 L
22.402 L
22.262 L
Why should one mole of hydrogen
occupy the same volume as one
mole of oxygen?
Since oxygen molecules are about
16 times as heavy as hydrogen
molecules, it would seem more
reasonable for a mole of oxygen
molecules to occupy much more
volume than a mole of hydrogen
molecules at the same temperature
and pressure.
Remember the Demo?
• The smaller balls make up in speed what
they lack in size in “carving out” an area.
This reasoning can be applied to the
volume occupied by lighter gas molecules
compared to the identical volume occupied
by an equal number of heavier gas
molecules at the same temperature and
pressure.
Van der Waal Forces
• London Dispersion Forces - (Van der Waal’s Forces)
– the attractive forces between gas molecules.
• Even though the total electrical charge of the gas
molecule is neutral, the positively charged nuclei of
its atoms are not completely shielded by negatively
charged electrons.
Van der Waal’s
In general, the
magnitude of
Van Der Waal’s
forces increases with
an increase in the
number of electrons
per atom and with an
increase in size of
the molecule.
Few electrons = lame wave
Many electrons = more
powerful wave!!
STP conditions
• Standard temperature and pressure – useful when
wanting to compare or measure gas volumes
• ST = 0 C = 273K
• SP = 760mm of Hg = 1atm = 760 Torr
• STP = standard temperature and pressure
Boyle’s Law
• Boyle’s Law – the volume of a definite quantity of
dry gas is inversely proportional to the pressure,
provided temperature remains constant.
(V  1/P)
• Formulas or dimensional analysis?
P1V1 = P2V2
Boyle’s Law,
Continues
Examples:
•Heimlich maneuver (increase P in lungs by
decreasing their V)
•Syringe
•Straw (Lungs are acting as the chamber that
increases in V)
•VP
&
VP
Problem 1
Suppose a syringe plunger is pushed all
the way in so that the syringe contains
0.10 mL of air at 1.00 ATM. The plunger
is now pulled back quickly so that the
total volume changes to 3.00 mL.
Calculate the momentary pressure before
any liquid comes into the syringe.
Solve using dimensional analysis
Initial P is 1.00 ATM so you have two possible conversion factors:
A.
0.10 mL
---------3.00 mL
OR
B.
3.00 mL
---------0.10 mL
Which do we choose?
Think: If we are increasing V we expect the gas P to decrease so we
want the conversion factor that is less than one.
1.00 atm
x
0.10 mL = 0.033 atm
---------3.00 mL
The plug-and-chug (no thinking
involved) method
P1V1 = P2V2
1.00 atm x 0.10 mL = P2 x 3.00 mL
P2
=
0.033 atm
Problem 2
A small child carrying an
inflated balloon with a
volume of 2.00 liters gets on
an airplane in N.Y. at a P of
760. Torr. What will the
volume of his balloon be on
arrival at Mexico City,
which is 7347 feet above sea
level and has an average P of
600. Torr?
Answer to problem 2
2.00 L x
760. Torr
----------600. Torr
Solve, p&c method #2 =
=
2.53 Liters
P1V1 = P2V2
760. Torr x 2.00 Liter = V2 x 600. Torr
V2 = 2.53 Liters
Charle’s Law
• The volume of a gas is directly proportional
to the Kelvin Temperature (Pressure
remains a constant).
V T
V1 = V2
T1 T2
Problem 1
 The initial volume of a
piston is 1.5 Liters at 300.K.
What will be the final
volume at 600. K? (P is constant)
– Think: Since volume  temperature, as V
increases then T must increase. So, use the
conversion factor that is greater than 1
Answer
1.5 Liter
Or solve way #2 =
x
600.K = 3.0L
300.K
V1/T1 = V2 / T2
1.5L / 300.K = V2 /600.K
V2 = 3.0L
Problem 2: Suppose a balloon had a
volume of 2 x 105 liter when it was filled
with hot air at a temp. of 500. C. What
volume would it occupy if it were sealed
and cooled at 27 C?
Answer to Problem 2
Think: If T decreases the V would also decrease. So,
choose a factor less than 1.
Solve: 500. C = 773 K
AND 27 C = 300. K
( 2 x 105 L ) x 300.K
773K
= 8 x 104 L
Gay-Lussac’s Law
The pressure inside a
fixed volume is directly
proportional to the
Kelvin temperature.
P T
Example: Spray can warning:
“Do not incinerate!”
P1 = P2
T1 T2
Gay-Lussac’s, continued
The contents of a spray can are all used up and the
can thus contains gas at a pressure of 1.00 atm.
The can is thrown into a fire where it’s T rises to
627 C. Why does the can explode? -- Started at
a temperature of 27 C.
– Think: If T increases, the P must increase if
there is no change in volume. So, solve for
pressure .
Solving for this Law
• The pressure inside the can, right before it blew
up, would have been:
• Remember if
1.00 atm x
T
, P
900. K =
300. K
3.00 atm
Problem 2
Problem: A pressure cooker is closed and is
at 20. C. To what temperature must the gas
be heated in order to create an internal
pressure of 2.00 atm?
Solving problem 2
Solve: P1 = 1.00 atm and it has to increase to
2.00 atm. So, P needs to increase and that
means temperature must also increase.
293 K x 2.00 atm
1.00 atm
= 586 K
Combining the gas laws
• P1V1 = P2V2
T1
T2
Whoops, wrong combining.
Combination problem
• Suppose you have a balloon that has a
volume of 20.0 L at a pressure of 1.00 atm
and a temperature of 32.0 C. If this balloon
rose into the atmosphere where the
temperature was only 12.0 C and the
pressure only 0.82 atm, what would its new
volume be?
First convert the temperatures to Kelvin:
12.0 + 273 = 285 K
32.0 + 273 = 305 K
Second: either plug in values into your combination
formula or do the THINKING method
P
so
V
&
T
so
V
• 20.0 L x 1.00 atm x 285 K = 23 L
0.82 atm
305 K
Dalton’s Law of Partial Pressure
• the total pressure of the mixture of gases is
the sum of their partial pressures
• PTOTAL = PA + PB + PC + ….ETC
Problem 1
A mixture of gases in a 3.00 L flask consists of
Nitrogen at a partial pressure of 100 Torr and
Oxygen at a PP of 300 Torr. What is the volume
of each gas? Calculate the total pressure in the
flask.
Since the presence of another gas has no effect on
the volume available to the first gas, both
Nitrogen and Oxygen have a volume of 3.00L
Ans. PTOTAL = 100 Torr + 300 Torr = 400 Torr
Problem 2
Air has a relative constant proportion of N2
(78.10%), O2 (21.00%), Ar (0.90%), and CO2
(0.03%). The partial pressure of each gas in
normal dry air at sea level is proportional to
the number of molecules. What is the Total
atmospheric pressure?
N2 = 0.7810 atm.
Ar = 0.0090 atm.
+
+
O2 = 0.2100 atm. +
CO2 = 0.0003 atm. =
answer. PTOTAL = 1.0003 atm
Graham’s Law of Diffusion
• Diffusion rate depends on 3 factors:
– Speed of the molecules (the higher the
temperature, the higher the average KE)
– Diameter of the molecule (the larger the
molecule, the slower the diffusion rate)
– Potential Attractive forces (the greater the
potential attractive forces between the
molecules, the slower the diffusion rate)
Graham’s Law
• For an average molecule - A and an average
molecule - B, use the following formula to
determine velocity:
VA
VB
=
MB
MA
Graham’s Law Sample Problem
• At room temperature, an average hydrogen
molecule travels at a speed of
1700. meters/second (about 3000
miles/hr). What is the speed of an average
oxygen molecule under the same
conditions?
Solving for Sample Problem 1
1700. m/sec = 32.0 amu
VB
2.02 amu
So, VB = 425 m/sec
Another awesome formula
• Problem: A steam autoclave is used to sterilize
hospital instruments. Suppose an autoclave with a
volume of 15.0 liters contains pure steam (water) at
a temperature of 121 C and a P of 1550 Torr. How
many grams of water does it contain?
• Solve: PV = nRT
• R = 62.4 L * Torr / Mole * K
Solving using the Ideal Gas Law
• 1550 Torr x 15.0 L = n x 62.4 L * Torr x 394 K
Mol * K
• n = 0.946 moles of H2O
• 0946 moles of H2O x 18.0g H2O = 17.0 g H2O
1 mole H2O
• NOTE: You could solve using your knowledge
of combination gas laws and molar volume (at STP)
Another Sample Problem
• Example 2: A hospital uses an oxygen gas
cylinder containing 3.50 kg of O2 gas in a volume
of 20.0 L and at a T of 24 C. What is the P in atm
in the cylinder?
Problem Solved
PV = nRT
n = 3500 g O2 x 1 mole/32.0 g = 109 moles
P( 20.0 L) = (109 moles)(0.0821 L x atm/mole x K)(297 K)
P = 133 atm
Eudiometer Tubes
(mercury or
water displacement)
• Eudiometer – a tube used to collect a gas.
It is closed at one end and is graduated.
• Reaction = 2HCl (aq) + Ca (s)  H2 (g) +
CaCl2 (aq)
Eudiometer with mercury
Situation #1
• If there is just enough gas to make the
mercury level inside the tube the same as the
level of the mercury in the bowl then the
pressure of the hydrogen gas is the same as
that of the atmosphere!
Eudiometer, Situation #2
• Suppose enough hydrogen gas is added to make the
level inside the tube lower than the level of the
mercury in the bowl. The P of the hydrogen gas is
greater than the atmospheric pressure. (To
determine the hydrogen gas pressure one must add
the level difference to the barometric reading.)
• Problem situation 2: The volume of oxygen in a
eudiometer is 37.0 mL. The mercury level inside the tube
is 25.00mm lower than the outside. The barometric
reading is 742.0 mm Hg.
Solving for Problem 2
Solution 2: 742.0mm + 25.00mm = 767.0mm
Eudiometer with mercury,
situation #3
Suppose there was not enough gas to make the mercury
level the same. Then, the pressure of hydrogen gas
would not be the same as the pressure of the air outside
the tube. (So, we must subtract the level difference
from the barometric reading.)
• Problem situation 3: What is the P of the gas in an
eudiometer, when the mercury level in the tube is 14mm
higher, than that outside? That barometer reads 735mm Hg.
Answer to Problem 3
• Solution: 735mm – 14mm = 721mm
Water in Place of Mercury
• Water is often used in place of mercury.
Calculations are done the same way but the
difference in water levels must first be
divided by 13.6 to convert it to its
equivalent height in terms of a column of
mercury since water is 1/13.6 as dense as
mercury.
New Problem: If you collected some oxygen
gas in the eudiometer by the method known as
water displacement you must:
• Correct for the difference in density between
water and mercury!!!
• Also, water evaporates much more readily than
mercury and so is in with the collected gas. So,
you need to determine the partial pressure of the
dry gas (unmixed with water vapor). The vapor
pressure of water, at the given temperature, must
be subtracted from the total pressure of the gas
within the tube.
• Problem 1: Oxygen is collected using the
water displacement method. The water
level inside the tube is 27.2mm higher than
that outside. The temp. is 25.0 C. The
barometric pressure is 741.0mm. What is
the partial pressure of the dry oxygen in the
eudiometer?
Step 1: 27.2mm/13.6 = 2.00mm
Step 2: 741.0 mm - 2.00mm = 739.0 mm
So: 739.0 mm = P total = Poxygen + Pwater
Step 3: To correct for water vapor pressure, you need to
know the pressure of water vapor at 25.0C and it is
23.8mm (always given - see handout). Subtract this
water partial pressure from the pressure total found in
step 2.
739.0mm - 23.8mm = 715.2 mm = partial pressure
of
the dry oxygen.
• Problem 2: A eudiometer contains 38.4 mL
of air collected by water displacement at a
temperature of 20.0C. The water level
inside the eudiometer is 140. mm higher
than that outside. The barometric reading is
740.0mm. Water vapor pressure at 20.0C is
17.5mm. Calculate the volume of dry air at
STP.
•First Step: 140mm/13.6= 10.3 mm
•Second Step: 740.0mm - 10.3mm = 729.7mm
• Third Step: 729.7mm - 17.5mm = 712.2 mm
 pressure at a Temp. of
293K
• Now it asks you to go a step further and calculate
what the volume would be at standard temperature
and pressure which = 760. mm and 273K. To solve
this problem you need to combine two gas laws –
Charles and Boyles. Think of your possible
conversion factors and solve.
• 38.4mL x ________ x __________ =
Combination finish
• 38.4mL x 273 K x 712.2mm = 33.5 mL
293 K
760. mm
More chemistry problems
involving gases:
1. Gas volume – gas volume
2. Mass-gas volume
Avagadro’s principle: Equal volumes of all
gases, under the same conditions of T and P
contain the same number of molecules.
Further, the number of molecules (of all
gases) are in the same ratio as their
respective gas volumes. (Refer back to
molar volume definition)
Gay Lussac’s law
• Under the same conditions of T and P, the
volumes of reacting gases and their gaseous
products are expressed in ratios of small
whole numbers.
• 2H2 (g) + O2 (g) --> 2H2O(g)
2 volumes of H2 + 1 vol. of O2 
2 vol. of H2O vapor.
Sample problem 1
• What is the volume of oxygen that would
combine with 4.0 liters of hydrogen to form
water vapor?
• Equation =
2H2 (g) + O2(g)  2 H2O
Answer: 4.0 L H2 x 1 vol. O2 = 2.0 L of O2
2 vol. H2
Sample Problem 2
• What is the volume of hydrogen chloride
gas that would be produced, assuming
5.0 liters of chlorine gas fully reacted with
hydrogen gas?
Equation: H2 (g) + Cl2 (g) --> 2HCl(g)
Answer: 5.0 L Cl2 x 2 vol. HCl
1 vol. Cl2
=
10.0 L of HCl
Remember Molar Volume?
• How many grams of CaCO3 must be
decomposed to produce 4.00 liters of CO2 at
STP? (Calcium oxide is also formed)
• CaCO3(s)  CaO (s) + CO2 (g)
• 1 mole--------------------1mole (stoich. ratio)
• 1 mole CaCO3 = 100.1g
• 1 mole CO2 at STP = 22.4 liters (remember?)
Molar volume revisited
(or ideal?)
• 4.00 L x 1 mole/22.4 L = 0.179 moles CO2
• Since CO2 and CaCO3 are on a 1/1 mole ratio,
you will need 0.179 moles of CaCO3
• 0.179 moles CaCO3 x 100.1 g CaCO3
1 mole CaCO3
=
17.9 g