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Unit Two Gases
we use many that were designed with the knowledge of the
A. Properties of Gases
eg)
technologies properties of
gases SCUBA equipment, hot air balloons,jackhammers
Gases and the Kinetic Molecular Theory
gases have several distinct macroscopic (visible)
compressible ie)
gases expandie)
gases have low resistance to flow (viscosity) …allows them to escape quickly through small openings
properties:
gases are
pressure = volume
as temperature increases
temperature = volume (not confined)temperature =pressure (confined)
gases havelow densities
gases mix evenly and completely, they all aremiscible
gases have they of the container they are in
no shape or volume,fill the shape
we need to describe how gases behave on the molecular level
B. Kinetic Molecular Theory
models
the says that all particles are
kinetic molecular theory in motion at all times
an is defined by the following characteristics:
ideal gas (which is hypothetical)
1.the gas molecules are in where they until they
constant random motion move in a straight line collide
2. the gas molecules are “point masses”
3. the only interaction between molecules of the gas and container are collisions where is
with a particle or wall of the container
(they have but no ) mass volume
elastic collisions…kinetic energy
conserved
do not have these perfect characteristics however their behaviour is real gases not that far offof ideal gases
Assignment: p. 101 #1-9
A. Atmospheric Pressure
although gas molecules have , the Earth’s keeps them near the
very little mass gravitational
pull the surface of the planet
force per unit area
pressure is exerted in all directions to the same extent
3.2 Gases and Pressure
pressure =
which creates our atmosphere
atmospheric pressure is the that a column of air exerts on a on the
air is as altitude pressure is exerted
less compressed increases less
forceparticular area
Earth’s surface
B. Measuring Pressure and TemperaturePascal and Perier used to prove that
atmospheric pressure
the work of Pascal, Perier and Torricelli all led to the development of the mercury barometer
Hg()decreases with altitude
there are several different units used to measure pressure: millimetres of mercury (mmHg) or
torr the
the
the
Pascal (Pa)
kilopascal (kPa)
atmosphere (atm)
the bar
you will be using the standard unit of in gas law calculations and therefore you must be able to convert mmHg and atm to kPa
kPa
memorize the following standard atmospheric pressures:
760 mmHg (torr) =1 atm = 101.325 kPa
to convert other units of pressure to kPa, set up aratio
another conversion:1 bar = 100 kPa
Example 1
Convert 650 mmHg to kPa.
101.325 kPa = x
760 mmHg 650 mmHg x = 86.6… kPa
Example 2
Convert 2.5 atm to kPa.
101.325 kPa = x
1 atm 2.5 atmx = 253.3… kPa
Try These:
Convert the following pressures to kPa (unrounded):
1. 4.0 atm
2. 855 mmHg
3. 0.625 atm
4. 150 mmHg
405.3 kPa
113.9…kPa
63.3…kPa
19.9…kPa
TemperatureAll gases have a temperature, usually measured in named after Anders Celsius. When the generalized variable of temperature is discussed, the symbol is used.There is another temperature scale that is very important in gas behavior. It is called the Kelvin scale (K). Note that K does not have a degree sign and Kelvin is capitalized because this was a person's title (Lord Kelvin, his given name was William Thomson).All gas law problems must be done in Kelvin temperatures. Standard Temperature is defined as 0 C or 273.15 K.
degrees Celsius (°C)
T
C. Boyle’s Law
Irish scientist studied the relationship between the and of gases at
Robert Boyle pressure volume
constant temperatures
pressure on the is caused by the
walls of a container collisions of the gas molecules with the walls
as you of a contained gas, there is
reduce the volume less room
more collisions = higher pressure
for the gas particles so theycollide more
http://michele.usc.edu/java/gas/gassim.html
Volume vs. Pressure for a Gas
Volume (L)
Pressure (kPa)
Boyle’s Law states that the volume of a gas varies inversely with the pressure at a constant temperature and mass
eg) lungs – to inhale, we the volume of our chest cavity which the pressure which makes the air move in
increasedecreases
breath-hold diving – all air containing spaces in body as pressure with depth…this doesn’t happen with SCUBA gear
shrink increaseseg)
where:
P1V1 = P2V2
P1, P2 = pressures in kPa V1, V2 = volumes in L
Example 1A balloon is filled with 30.0 L of helium gas at 100 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume constant temperature)
P1 = 100 kPaP2 = 25.0 kPaV1 = 30.0 L
P1V1 = P2V2
(100 kPa)(30.0 L) = (25.0 kPa) V2
V2 = 120 L
Example 2The pressure on 2.50 L of anesthetic gas is 100 kPa. If 6.25 L of gas is the required volume, what pressure must it be under assuming constant temperature? P1 = 100 kPaV1 = 2.50 LV2 = 6.25 L
P1V1 = P2V2
(100 kPa)(2.50 L) = P2 (6.25 L)
P2 = 40.0 kPa
when of a gas are graphed, the plot is
it was also noticed that when these linear plots were down to all the lines at one point
volume vs. temperature linear
A. Volume vs. Temperature3.3 Gases and Pressure
(as long as amount of gas and pressure were constant)
extrapolated zero volume,converged
Volume vs. Temperature for a Gas
Volume (L)
Temperature (C)
-273.15 C
the temperature when the volume of a gas is is
, in 1848, suggested that this is the lowest possible temperature or
he established a new temperature scale which is called the scale in his honour
273.15C
Lord Kelvin
absolute zero
Kelvin
zero
is used for temperature in
is used for temperature in
t C
T K
= = -273.15C absolute zero zero Kelvin
to go from C to K…you add 273.15
eg) 0C = 25C = -30C = -273.15C =
273.15 K298.15 K243.15 K
0 K
B. Charles’ Law (and
Joseph Louis Gay-Lussac) noticed that there was a relationship between the and of a gas
Jacques Charles
temperature volume
as temperature , so does the of the gas molecules
as the molecules move , they exert pressure
the volume of the gas will until it reaches
increases kinetic energy
faster higher
expand under this pressure atmospheric
pressure
where:
V1 = V2 T1 T2
T1, T2 = temperatures in KV1, V2 = volumes in L
Charles’ Law states that the volume of a gas varies directly with the temperature at a constant pressure and mass
Example 1A balloon was inflated at 27C and has a volume of 4.0 L. If it is heated to 57C, what is the new volume? (assume constant pressure)
T1 = 27 C = 300.15 KV1 = 4.0 L T2 = 57 C = 330.15 K
V1 = V2
T1 T2
(4.0 L) = V2
(300.15 K) 330.15 KV2 = 4.39… LV2 = 4.4 L
Example 2A sample of gas occupies 6.8 L at 110C. What will the final temperature be in C when the volume is decreased to 5.6 L?
T1 = 110 C = 383.15 KV1 = 6.8 L V2 = 5.6 L
V1 = V2
T1 T2 (6.8 L) = 5.6 L(383.15 K) T2
T2 = 315.5… K – 273.15T2 = 42C
now we’ll combine Boyle’s Law and Charles’ Law
P1V1 = P2V2
T1 T2
where: V1, V2 = volumes in L T1, T2 = temperatures in KP1, P2 = pressures in kPa
A. Combined Gas Law Calculations4.1 Gases and the Kinetic Molecular Theory
Chapter 4
= standard temperature and pressure
= standard ambient temperature and pressure
= 273.15 K (0 C) and 101.325 kPa
= 298.15 K (25C) and 100.000 kPa
STP
SATP
Example 1A weather balloon is filled with H2(g) at 20C and 100 kPa. It has a volume of 7.50 L. It rises to an altitude where the air temperature is -36C and the pressure is 28 kPa. What is the new volume of the balloon? T1 = 20 C = 293.15 KV1 = 7.50 L P1 = 100 kPaP2 = 28 kPa T2 = -36 C = 237.15 K
P1V1 = P2V2
T1 T2
(100 kPa)(7.50 L) = (28 kPa)V2
(293.15 K) (237.15 K)V2 = 21.6…L V2 = 22 L
Example 2A large syringe was filled with 50.0 mL of ammonia gas at STP. If the gas was compressed to 25.0 mL with a pressure of 210 kPa, what was the final temperature in C?
T1 = 0 C = 273.15 KV1 = 0.0500 L P1 = 101.325 kPaP2 = 210 kPa V2 = 0.0250 L
P1V1 = P2V2
T1 T2
(101.325 kPa)(0.0500 L) = (210 kPa)(0.0250 L) (273.15 K) T2T2 = 283.0…K – 273.15
T2 = 9.91C
Journal #9
How could you explain the difference between covalent and polar covalent bonding?
B. Combining Volumes of Gases Gay-Lussac analyzed
that involved
he studied the of the gaseous reactants and products and concluded that the gases combine in
chemical reactionsgases
volumes
very simple proportions
the states that, when gases react, the of the gaseous reactants and products, measured at constant temperature and pressure, are always in
eg) 1 N2(g) + 3 H2(g) 2 NH3(g)
1 : 3 : 2 is the volume ratio For every one mole of nitrogen gas and three moles of hydrogen gas, 2 moles of ammonia gas is produced
Law of Combining Volumesvolumes
whole number ratios
Formula: v1 = v2
n1 = n2Balancing coefficients
Avogadro’s hypothesis / law
This equation is explain by Avogadro’s law
It States that equal volumes of gases at the same temperature and pressure, contain equal numbers of particles (atoms or molecules).
ExampleWhat volume of nitrogen is used up if 100 mL of ammonia is formed in a composition reaction? (Compare ratios of volume:coefficients for what values you know, and what values you are solving for)
N2(g) + 3H2(g) 2NH3(g)
x mL = 100 mL
1 = 2
x mL = 50.0 mL
x mL 100 mL
Formula: v1 = v2
n1 = n2
this law combines all four variables
where:
PV = nRT
V = volume in L T = temperature in K P = pressure in kPan = number of moles in mol R = universal gas constant = 8.314 kPaL/molK
http://www.youtube.com/watch?v=nO5f50wELjA
(P, V, n and T)
A. Ideal Gas Law Calculations 4.2 Ideal Gas Law
intoone equation:
Example 1What is the volume of 10.8 mol of oxygen gas at 100.00 kPa and 15.5C?
n = 10.8 molT = 15.5 C = 288.65 KP = 100.00 kPaR = 8.314 kPaL/molK
PV = nRT(100.00kPa)V = (10.8mol)(8.314 kPaL/molK)(288.65K)
V = 259.1…L = 259 L
Example 2A rigid steel vessel with a volume of 20.0 L is filled with nitrogen gas to a pressure of 20 000 kPa at 27.0C. What is the number of moles of nitrogen?
T = 27 C = 300.15 KV = 20.0 L P = 20 000 kPaR = 8.314 kPaL/molK
PV = nRT(20 000kPa)(20.0L) = n(8.314 kPaL/molK)(300.15K)
n = 160.2…mol = 160 mol
Example 3What is the pressure exerted by 15.5 g of methane, CH4(g), if it occupies a volume of 10.0 L at 25C?
T = 25 C = 298.15 KV = 10.0 L R = 8.314 kPaL/molKm = 15.5 gM = 16.05 g/mol
PV = nRTP(10.0L) = (0.965…mol)(8.314 kPaL/molK)(298.15K)
P = 239.3…kPa = 2.4 102 kPa
n = m M = 15.5 g 16.05 g/mol
= 0.965…mol
OR PV = mRT M
Example 4What is the mass of hydrogen gas contained in a 4.5 L weather balloon at 25C and 102.0 kPa?
T = 25 C = 248.15 KV = 4.5 L P = 102.0 kPaR = 8.314 kPaL/molKM = 2.02 g/mol PV = mRT
M(102.0 kPa)(4.5L) =m(8.314 kPaL/molK)(248.15K) 2.02 g/mol (102.0)(4.5)= m(1021.346….)
(102.0)(4.5) = m
(1021.346..)= 0.449…g= 0.45 g
C. Ideal Gases and Real Gases for ideal gases we assume that there are no
between the molecules of the gas
in real gases, however, there are between the molecules
we don’t have to worry about considering this in our calculations because at standard P and T conditions, the molecules are and are they much with each other
intermolecular attractions
attractions
very far apart moving very quickly don’t interact
real gases behave like ideal gases at
real gases deviate from ideal gas behaviour at
high temperaturesand low pressures
very low temperatures(moving very slowly)very high pressures
(molecules close together)and
ReviewAn ideal gas is a gas that obeys all the gas laws perfectly:
1. Pressure and volume are inversely proportional to each other (Boyles law)(Boyles law)
2. Volume is directly proportional to absolute temperature (measured in kelvins, K) (Charle’s law)(Charle’s law)
3. Volume is directly proportional to the number of moles of gas (Avogadro’s law)(Avogadro’s law)
Real gases Versus Ideal gases
Real Gases Ideal Gases
Molecules have a distinct size Molecules are point masses (have no size)
Forces of attraction between molecules cause them to condense into a liquid if the temperature gets low enough.
Molecules do not have forces of attraction among them, so there is no cause for them to condense into liquid.
Molecules move about randomly, but because of the attraction between them, they do not travel in straight lines – they are influenced by each other.
Molecules move about randomly; only collisions with container walls and other molecules cause them to change direction.
Collisions between molecules are inelastic – kinetic energy is not conserved
Collisions between molecules are elastic – kinetic energy is conserved
Review assignment p. 180 #1-46