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PROPERTIES OF GASES &THE GAS LAWS
Chapter 14 Section 1 & 2
MEASURABLE PROPERTIES OF GASES
Pressure (P) Temperature (T)
Volume (V) Moles (n)
Units atm, Pa (SI unit), mmHg, torr
K (SI unit), ºC L, mL, cm3, m3
moles, mol
Instrument barometer, manometer
thermometer graduated cylinder
NA
Meaning force exerted over area by colliding gas particles
amount of average kinetic energy of particles
space occupied by particles, not the volume of particles
1 mol = 6.02×1023 particles
PROPERTIES OF GASES
Compressible Have mass Gas particles always in motion Gas particles exert pressure when they run into a wall Take up any shape and size of container – diffuse (=
to spread out) Are described with four variables: the amount of gas
(n), volume (V), pressure (P), and temperature (T)** variable = something you can change and
represented by a letter
PRESSURE VS. VOLUME(BOYLE’S LAW)
Describe the picture.
BOYLE’S LAW EQUATIONS
P1∙V1 = P2∙V2
“1” and “2” refer to two different sets of condition
Match the unit for each variable The volume is inversely proportional
to the pressure The volume increases (decreases) as the
pressure decreases (increases) The temperature and amount of gas
must remain unchanged for this law to work
BOLYE’S LAW(PRESSURE-VOLUME RELATIONSHIP)
hyperbola
EXAMPLE
A given sample of gas occupies 523 mL at 760 torr. The pressure is increased to 1.97 atm, while the temperature remains the same. What is the new volume of the gas?
PRACTICES ON BOYLE’S LAW1) A flask containing 155 cm3 of hydrogen gas
was collected under a pressure of 22.5 kPa. What pressure would have been required for the volume of the gas to have been 90.0 cm3?
2) A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains the constant?
3) A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume 2.93 L, what will its pressure be, assuming constant temperature?
PRESSURE-VOLUME DATA (BOYLE’S LAW)
Pressure(kPa) Volume (L) PV(kPa×L)
150 0.334 50.1
200 0.250 50.0
250 0.200 50.0
300 0.167 50.1
CHARLES’S LAW(TEMPERATURE – VOLUME
RELATIONSHIP)
Volume (mL) Temperature (K)
V/T (mL/K)
748 373 2.01
567 283 2.00
402 200 2.01
199 100 1.99
CHARLES’S LAW(TEMPERATURE-VOLUME
RELATIONSHIP)
CHARLES’S LAW
1)
2) The volume and temperature are directly proportional
3) The volume of gas increases (decreases) as the temperature increases (decreases)
4) The temperature MUST be in kelvin (K = C + 273)
5) Must keep the pressure and the amount of gas unchanged for this law to work
1 2
1 2
V V
T T
GRAPH OF CHARLES’S LAW(TEMPERATURE-VOLUME
RELATIONSHIP)
ExampleA balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath. What, at −78.5°C, is its volume, assuming the pressure remains constant?
PRACTICES1) A sample of neon gas occupies a volume of
752 mL at 25 ºC. What volume will the gas occupy at 50 ºC if the pressure remains constant?
2) A helium-filled balloon has a volume of 2.75 L at 20 ºC. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature?
3) A gas at 65 ºC occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure?
COMBINED GAS LAW(VOLUME, TEMPERATURE & PRESSURE)
Combine Boyle’s law and Charles’s law:
For this law to work, the amount of gas must remain unchanged
With this law, 2 variables out of 3 can change
1 1 2 2
1 2
1 2
P V P V
V V
T T
1 1 2 2
1 2
P V P V
T T
EXAMPLE
520 mL of hydrogen gas at 750 mmHg and 25 °C is placed in a 1000. mL container and heated to 50 °C. What is the pressure of the gas in the container?
FROM COMBINED GAS LAW TO…. Boyle’s law: at constant temperature Charles’ law: at constant pressure
Gay-Lussac’s law: at constant volume
1 1 2 2
1 2
P V P V
T T
1 1 2 2
1 2
P V P V
T T
1 1 2 2
1 2
P V P V
T T
GAY-LUSSAC’S LAW
GRAPH OF GAY-LUSSAC’S LAW
GAY-LUSSAC’S LAW(TEMPERATURE-PRESSURE
RELATIONSHIP) Temperature and pressure are directly
proportional:
Temperature MUST be in kelvins Works only if the volume and the amount of
gas are kept constant
2
2
1
1
T
P
T
P
EXAMPLE
An aerosol can containing gas at 101 kPa and 22 ºC is heated to 55 ºC. Calculate the pressure in the heated can.
Answer: P2=112kPa
AVOGADRO’S LAW The volume (V) of gas is directly proportional to
the number of moles (n) of gas:
The type of gas doesn’t affect the volume; only the # of moles of gas does 1 mol of ANY gas at 1 atm and 0˚C takes up 22.4 L volume.
For this law to work, pressure and temperature must remain unchanged
1 2
1 2
V V
n n
COMBINE BOYLE, CHARLES, & AVOGADRO’S LAW
P1∙V1 = P2∙V2
1 2
1 2
V V
n n
1 2
1 2
V V
T T
IDEAL GAS LAW
1 mol of ANY gas at 1 atm and 0˚C takes up 22.4 L volume
Combine the two above information and get…
22
22
11
11
nT
VP
nT
VP
EXAMPLE
Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 kPa.
SUMMARY OF GAS LAWS
Combine V, P, n, and T into one law.
From the combined gas law, get(1) Boyle’s law(2) Charles’ law(3) Gay-Lussac’s law(4) Avogadro’s law(5) Ideal gas law
KINETIC MOLECULAR THEORY
1) Gas particles are in constant, rapid, and random motion
2) The distance between gas particles are much larger than the size of atoms*The size of gas particle is almost nothing.
3) Gas particles colliding with surface creates pressure
4) Perfect elastic collisions between gas particles – no loss of energy during collisions but all transferred
5) Average kinetic energy of gas particles is proportional to kelvin temperature
6) Gas particles at the same temperature don’t have the same amount of kinetic energy (See the graph on the next slide)
NON-IDEAL GAS BEHAVIOR
Gas particles attract or repel each other
Gas particles do have a volume Imperfect collision – energy lost Non-ideal behavior becomes more ideal
at high temperature and low pressure
SUMMARY OF GAS LAWS
Boyle’s Charles’s Combined Gay-Lussac’s
Avogadro’s
SUMMARY OF GAS LAWS
Boyle’s Charles’s Combined Gay-Lussac’s
Avogadro’s
volume-pressure
volume-temperature
volume-pressure-temperature
pressure-temperature
volume-mole
V and P inversely proportional
V and T directly proportional
V and P inversely and V and T directly proportional
P and T directly proportional
V and n directly proportional
P1∙V1 = P2∙V2
T and n constant
P and n constant
n constant n and V constant
P and T constant
1 2
1 2
V V
T T 1 1 2 2
1 2
P V P V
T T
1 2
1 2
P P
T T 1 2
1 2
V V
n n