Gases and atmospheric
Gases and atmospheric
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Gases and atmospheric
Chemistry
States of Matter
STATE |
PROPERTIES |
EXAMPLES |
SOLID – particles held tightly together. |
-definite shape & volume -virtually incompressible -does not flow easily |
-crystal of sugar |
LIQUID – particles held more loosely together |
-assumes shape of container but has a definite volume - virtually incompressible -flows readily |
-coloured water |
GAS – appears to be an absence of forces between molecules |
-assumes shape & volume of container -highly compressible -flows readily |
-gas station air pump |
Kinetic Molecular Theory
- all substances contain particles that are in constant, random
motion
- 3 types of motion;
- translational (straight line)
- rotational (spinning)
- vibrational (back & forth motion of atoms in a molecule)
- solids – vibrational-bonds hold molecules tightly
liquids – all 3 types of motion – loosely held
gases – translational (mostly) – very loosely held
9.2 Gas Laws
pressure – Force per unit area
atmospheric pressure – Force per unit area exerted by air on an object
SI UNIT – for atmospheric pressure is the Pascal (Pa) which equals a force 1N/m^{2} (newton/square meter)
* we normally use kilopascals (kPa)
1 kPa=1000Pa=1kN/m^{2}
Standard Pressure=101.325 kPa=1 atm
Standard Ambient Pressure= 100kPa
STP= 0ºC and 101.325 kPa
SATP=25ºC and 100kPa
UNIT NAME |
UNIT SYMBOL |
DEFINITION/CONVERSION |
Pascal |
Pa |
1 Pa = 1N/m^{2} |
Atmosphere |
atm |
1 atm = 101.325kPa |
mm of mercury |
mm Hg |
760mm of Hg=1 atm=101.325kPa |
torr |
torr |
1 torr = 1mm of Hg |
*** Know how to convert these units
e.g. Convert 100 kPa to atm and mm of Hg
a) 100kPa x 1 atm = 0.987 atm
101.325 kPa
b) 100kPa x 760 mm of Hg = 750.mm of Hg
101.325kPa
BOYLE’S LAW
- discovered by Robert Boyle
- “As the pressure on a gas increases, the volume of the gas
decreases proportionally, provided that the temperature and
amount of gas remain constant: The volume and pressure of a
gas is inversely proportional”
- Boyle’s Law mathematically is simply;
P_{1}V_{1} = P_{2}V_{2}
Sample Problem 1
Q: A 2.0L party balloon at 98kPa is taken to the top of a mountain
where the pressure is 75kPa. Assume the temperature is
constant. What is the new volume of the balloon?
A: V_{1}=2.0L
P_{1}=98kPa
P_{2}=75kPa
V_{2}= ?
P_{1}V_{1}= P_{2}V_{2}
V_{2} = P_{1}V_{1}
P_{2}
= 98kPa x 2.0L
75kPa
V_{2} = 2.6L
Therefore the new volume is 2.6L.
KELVIN TEMPERATURE SCALE
- when we work with gases we use a temperature scale called the Kelvin Temperature Scale. K
- When the line is extrapolated downward the line meets the temperature line (X-axis) at a point where the temperature is
- 273ºC.
- this temperature is called Absolute Zero and is the lowest possible temperature
- absolute zero is the basis for the Kelvin temperature scale
- Absolute zero (-273ºC) is Zero Kelvin (0 K)
- To convert ºC to K. add 273
- To convert K to ºC, subtract 273
Do questions 12-14 on page 432.
* For convenience we use 273K for STP
and 298K for SATP
i.e. SATP = 100kPa and 298K
STP = 101.325kPa and 273K
CHARLES’ LAW
“As the temperature of a gas increases, the volume increases proportionally, provided that the pressure & amount of gas remain constant”
V_{1} = k V_{2} =k
T_{1 }T_{2}
_{ }
where k= constant value
therefore V_{1} = V_{2}
T_{1} T_{2}
Sample Problem 3
Q: A gas inside a cylinder with a movable piston is to be heated to
315ºC. The volume of the gas in the cylinder is 0.30L at 25ºC.
What is the final volume when the temperature is 315ºC?
A: V_{1}=0.30L
T_{1}=25ºC=298K
V_{2}=?
T_{2}= 315ºC=588K
V_{1} = V_{2}
T_{1} T_{2}
V_{2}=V_{1}T_{2}
T_{1}
V_{2}=0.30L x 588K
298K
V_{2}=0.59L
Therefore the final volume of the gas at 315ºC is 0.59L.
Pressure and Temperature Law
- sometimes called Gay-Lussac’s Law
- P_{1} = k P_{2} = k
T_{1} T_{2}
- therefore, P_{1} = P_{2}
T_{1} T_{2}
- this represents a direct relationship i.e. heat an aerosol can and
the pressure will increase until the can ruptures.
Sample Problem 4
Q: A sealed storage tank contains argon gas at 18ºC and a pressure
of 875kPa at night. What is the new pressure of the tank and
its contents when it warms up to 32ºC during the day?
A: T_{1}=18ºC=291K
P_{1}=875kPa
P_{2}=?
T_{2}=32ºC=305K
P_{1} = P_{2}
T_{1} T_{2}
P_{2}= P_{1}T_{2}
e 435&436
T_{1}
= 875kPa x 305K
291K
P_{2}= 917kPa
Therefore the new pressure is 917kPa.
COMBINED GAS LAW
P_{1}V_{1} = P_{2}V_{2}
T_{1} T_{2}
Sample Problem 5
Q: A balloon containing hydrogen gas at 20ºC and a pressure of
100kPa has a volume of 7.50L. Calculate the volume of the
balloon after it rises 10km into the upper atmosphere where
the temperature is -36ºC and the outside air pressure is 28kPa.
Assume that no hydrogen gas escapes and that the balloon
expands, etc.
A: T_{1}=20ºC=293K
P_{1}=100kPa
V_{1}=7.50L
T_{2}=-36ºC=237K
P_{2}=28kPa
V_{2}= ?
P_{1}V_{1} = P_{2}V_{2}
T_{1} T_{2}
V_{2}=P_{1}V_{1} x T_{2}
T_{1} P_{2}
=100kPa x 7.50L x 237K
293K x 28kPa
V_{2}=22L
Therefore the new volume of the balloon is 22L.
THE IDEAL GAS LAW
An ideal gas is a hypothetical gas that obeys the gas laws perfectly under all conditions i.e. won’t condense into a liquid when cooled, graphs are all perfect straight lines, etc.
An ideal gas fits the following criteria;
- The volume of a gas is inversely proportional to pressure
v 1/P (Boyle’s Law)
- The volume of a gas is directly proportional to the absolute Kelvin temperature V T (Charles’ Law)
- The pressure of a gas is directly proportional to the absolute Kelvin temperature P T (Pressure-Temperature Law)
- The greater the amount of gas, in moles (temperature & pressure constant), the greater the volume V n
Therefore V 1/P x T x n
V=R[constant] x 1/P x T x n
V= nRT
P
Or PV=nRT
PV=nRT is the Ideal Gas Law.
Where R= the gas constant
To find the value of R, we would substitute known values and solve for R.
Example: At STP 1.00mol of an ideal gas would occupy a volume of
22.414L.
V=22.414L
n=1.00mol
P=101.325kPa
T=0ºC=273.15K
PV=nRT
R=PV
nT
=101.325kPa x 22.414L
1.00mol x 273.15K
R= 8.31kPa*L
mol * K
- If the mass of gas was given, and not moles, we would have to
calculate n first.
Example: If 0.78g of hydrogen at 22ºC and 125kPa is produced, what
volume of hydrogen would be expected?
Step 1: Find moles of hydrogen
n=m/M = 0.78g/2.02g/mol = 0.39mol
Step 2: Use the Ideal Gas Law to find the volume of hydrogen
expected.
PV=nRT
V_{H2}=nRT
P
V_{H2}= 0.39mol x 8.31kPA*L/mol*K x 295K
125kPa
_{}
V_{H2}= 7.6L
Therefore the volume of hydrogen gas expected was 7.6L.
Sample Problem 1
Q: What mass of neon gas should be introduced into an evacuated
0.88L tube to produce a pressure of 90kPa at 30ºC?
A:
V=0.88L
P=90kPa
T=30ºC=303K
m_{Ne}=?
PV=nRT
n_{Ne}=PV
RT
n_{Ne}=90kPa x 0.88L
8.31kPa*L/mol*K x 303K
n_{Ne}=0.031mol
n=m/M , m_{Ne}=n x M
m_{Ne}=0.031mol x 20.18g/mol
=0.63g
Therefore the mass of neon gas needed was 0.63g.
Source : http://www.leafrancis.net/Chem9Notes.doc
Web site link: http://www.leafrancis.net/
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