scheffler gas laws l. scheffler lincoln high school ib chemistry 1-2 january 2010 1
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Scheffler
Gas Laws
L. Scheffler
Lincoln High School
IB Chemistry 1-2
January 2010
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Gases Variable volume and shape Expand to occupy volume available Volume, Pressure, Temperature,
and the number of moles present are interrelated
Can be easily compressed Exert pressure on whatever
surrounds them Easily diffuse into one another
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Mercury Barometer
Used to define and measure atmospheric pressure
On the average at sea level the column of mercury rises to a height of about 760 mm.
This quantity is equal to 1 atmosphere
It is also known as standard atmospheric pressure
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Pressure Units & Conversions
The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal.
The US Weather Bureaus commonly report atmospheric pressures in inches of mercury.
Pounds per square inch or PSI is widely used in the United States.
Most other countries use only the metric system. 4
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Boyle’s Law
According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant pressure.
PV = constant. P1V1 = P2V2
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Boyle’s Law
A graph of pressure and volume gives an inverse function
A graph of pressure and the reciprocal of volume gives a straight line
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= 340 kPa
If the pressure of helium gas in a balloon has a volume of 4.00 dm3 at 210 kPa, what will the pressure be at 2.50 dm3?
P1 V1 = P2 V2
(210 kPa) (4.00 dm3) = P2(2.50 dm3)
P2 = (210 kPa) (4.00 dm3) (2.50 dm3)
Sample Problem 1:
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Charles’ Law According to Charles’ Law the volume of a
gas is proportional to the Kelvin temperature as long as the pressure is constant
V = kT
V1
=
T1
V2
T2
Note: The temperature for gas laws must always be expressed in Kelvin where Kelvin = oC +273.15 (or 273 to 3 significant digits)
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Charles’ Law
A graph of temperature and volume yields a straight line. Where this line crosses the x axis (x intercept) is defined
as absolute zero9
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Sample Problem 2 A gas sample at 40 oC occupies a volume of 2.32 dm3. If the temperature is increased to 75 oC, what will be the final volume?
2.58 dm3
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V1 = V2
T1 T2Convert temperatures to Kelvin. 40oC = 313K 75oC = 348K
2.32 dm3 = V2
313 K 349K
(313K)( V2) = (2.32 dm3)(348K)
V2 =
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Gay-Lussac’s Law
Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas.
The pressure and temperature of a gas are directly proportional
P1 = P2
T1T2
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Sample Problem 3:The pressure of a gas in a tank is 3.20 atm at 22 oC. If the temperature rises to 60oC, what will be the pressure in the tank?
3.6 atm12
P1 = P2
T1 T2
Convert temperatures to Kelvin. 22oC = 295K 60oC = 333K
3.20 atm = P2
295 K 333K
(295K)( P2) = (3.20 atm)(333K)
V2 =
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The Combined Gas Law1. If the amount of the gas is constant, then
Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship
2. P1 V1 = P2 V2
T2T1
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Sample Problem 4:
A gas at 110 kPa and 30 oC fills a container at 2.0 dm3. If the temperature rises to 80oC and the pressure increases to 440 kPa, what is the new volume?
V2 = 0.58 dm3
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P1V1 = P2V2
T1 T2
Convert temperatures to Kelvin. 30oC = 303K 80oC = 353K
V2 = V1 P1 T2 P2 T1
= (2.0 dm3) (110 kPa ) (353K) (440 kPa ) (303 K)
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Advogadro’s Law Equal volumes of a gas under the same temperature
and pressure contain the same number of particles. If the temperature and pressure are constant the
volume of a gas is proportional to the number of moles of gas present
V = constant * n
where n is the number of moles of gas
V/n = constant
V1/n1 = constant = V2 /n2
V1/n1 = V2 /n2
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Universal Gas Equation Based on the previous laws there are four factors
that define the quantity of gas: Volume, Pressure, Kevin Temperature, and the number of moles of gas present (n).
Putting these all together:
PVnT
= Constant = R
The proportionality constant R is known as the universal gas constant
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Universal Gas EquationThe Universal gas equation is usually written as
PV = nRTWhere P = pressure
V = volumeT = Kelvin Temperaturen = number of moles
The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include: R = 62.36 dm3 torr mol-1 K-1
= 0.0821 dm3 atm mol-1 K-1
= 8.314 dm3kPa mol-1 K-1
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Standard Temperature and Pressure (STP) The volume of a gas varies with temperature
and pressure. Therefore it is helpful to have a convenient reference point at which to compare gases.
For this purpose standard temperature and pressure are defined as:
Temperature = 0oC 273 K
Pressure = 1 atmosphere = 760 torr
= 101.3 kPa
This point is often called STP18
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Sample Problem 5 Example: What volume will 25.0 g O2 occupy
at 20oC and a pressure of 0.880 atmospheres? :
V = (0.781 mol)(0.08205 dm-3 atm mol-1 K-1)(293K)0.880 atm
V = 21.3 dm3
(25.0 g)n = ----------------- = 0.781 mol (32.0 g mol-1)
V =? P = 0.880 atm; T = (20 + 273)K = 293K R = 0.08205 dm-3 atm mol-1 K-1
PV = nRT so V = nRT/P
Data
Formula
Calculation
Answer
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Density (d) Calculations
d = mV
=PMRT
m is the mass of the gas in gM is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRTP
M = d is the density of the gas in g/L
Universal Gas Equation –Alternate Forms
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A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas?
Sample Problem 6
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A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas?
dRTP
M = d = mV
4.65 g2.10 dm3
= = 2.21 g
dm3
M =2.21
g
dm3
1 atm
x 0.0821 x 300.15 Kdm3•atmmol•K
M = 54.6 g/mol
Sample Problem 6 Solution
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Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is
equal to the sum of the pressures of the individual gases (partial pressures).
PT = P1 + P2 + P3 + P4 + . . . .
where PT = total pressure
P1 = partial pressure of gas 1
P2 = partial pressure of gas 2
P3 = partial pressure of gas 3
P4 = partial pressure of gas 423
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Dalton’s Law of Partial Pressures
Applies to a mixture of gases
Very useful correction when collecting gases over water since they inevitably contain some water vapor.
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Sample Problem 7 Henrietta
Minkelspurg generates Hydrogen gas and collected it over water.
If the volume of the gas is 250 cm3 and the barometric pressure is 765.0 torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP?
(PH2O = 23.8 torr at 25oC)
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Sample Problem 8 -- Solution
Henrietta Minkelspurg generates Hydrogen gas and collected it over water.
If the volume of the gas is 250 cm3 and the barometric pressure is 765.0 torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP?
(PH2O = 23.8 torr at 25oC)
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Sample Problem 9 Henrietta Minkelspurg generated Hydrogen gas and collects it
over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP?
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Sample Problem 9 -- Solution
Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP?
From the previous calculation the adjusted pressure is 742.2 torr
V2 = (250 cm3)(742.2 torr)(273K)
(298K)(760.torr)
V2 = 223.7 cm3
P1= PH2 = 742.2 torr; P2= Std Pressure = 760 torrV1= 250 cm3; T1= 298K; T2= 273K; V2= ?(V1P1/T1) = (V2P2/T2) therefore V2= (V1P1T2)/(T1P2)
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Kinetic Molecular Theory Matter consists of particles (atoms or molecules)
that are in continuous, random, rapid motion The Volume occupied by the particles has a
negligibly small effect on their behavior Collisions between particles are elastic Attractive forces between particles have a
negligible effect on their behavior Gases have no fixed volume or shape, but take
the volume and shape of the container The average kinetic energy of the particles is
proportional to their Kelvin temperature29
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Maxwell-Boltzman Distribution Molecules are in
constant motion Not all particles
have the same energy
The average kinetic energy is related to the temperature
An increase in temperature spreads out the distribution and the mean speed is shifted upward
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The distribution of speedsfor nitrogen gas moleculesat three different temperatures
The distribution of speedsof three different gases
at the same temperature
urms = 3RTM
Velocity of a Gas
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Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH3
17.0 g/mol
HCl36.5 g/mol
NH4Cl
Diffusion
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DIFFUSION AND EFFUSION DiffusionDiffusion is the is the
gradual mixing of gradual mixing of molecules of molecules of different gases.different gases.
EffusionEffusion is the is the movement of movement of molecules through molecules through a small hole into an a small hole into an empty container.empty container.
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Graham’s Law Graham’s law governs Graham’s law governs
effusion and diffusion of effusion and diffusion of gas molecules. gas molecules.
KE=1/2 mv2
Thomas Graham, 1805-1869. Thomas Graham, 1805-1869. Professor in Glasgow and London.Professor in Glasgow and London.
The rate of effusion is The rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.
The rate of effusion is The rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.
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Sample Problem 10 1 mole of oxygen gas and 2 moles of ammonia are placed in a
container and allowed to react at 850oC according to the equation:
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
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Sample Problem 10 Solution 1 mole of oxygen gas and 2 moles of ammonia are placed in a
container and allowed to react at 850oC according to the equation:
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
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Sample Problem 11 What is the rate of effusion for H2 if 15.00 cm3 of CO2
takes 4.55 sec to effuse out of a container?
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Sample Problem 11 Solution What is the rate of effusion for H2 if 15.00 cm3 of CO2
takes 4.55 sec to effuse out of a container?
Rate for CO2 = 15.00 cm3/4.55 s = 3.30 cm3/s
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Sample Problem 12 What is the molar mass of gas X if it effuses
0.876 times as rapidly as N2(g)?
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Sample Problem 12 Solution What is the molar mass of gas X if it effuses
0.876 times as rapidly as N2(g)?
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Ideal Gases v Real Gases
Ideal gases are gases that obey the Kinetic Molecular Theory perfectly.
The gas laws apply to ideal gases, but in reality there is no perfectly ideal gas.
Under normal conditions of temperature and pressure many real gases approximate ideal gases.
Under more extreme conditions more polar gases show deviations from ideal behavior.
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In an Ideal Gas --- The particles (atoms or molecules) in continuous,
random, rapid motion. The particles collide with no loss of momentum The volume occupied by the particles is essentially zero
when compared to the volume of the container The particles are neither attracted to each other nor
repelled The average kinetic energy of the particles is proportional
to their Kelvin temperature
At normal temperatures and pressures gases closely approximate idea behavior
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Real Gases
These deviations occur because Real gases do not actually have zero volume Polar gas particles do attract if compressed
For ideal gases the product of pressure and volume is constant. Real gases deviate somewhat as shown by the graph pressure vs. the ratio of observed volume to ideal volume below.
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van der Waals Equation
(P + n2a/V2)(V - nb) = nRT
The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior
where a => attractive forces between moleculesb => residual volume or molecules
The van der Waals constants for some elements are shown belowSubstance a (dm6atm mol-2) b (dm3 mol-1)
He 0.0341 0.02370
CH4 2.25 0.0428
H2O 5.46 0.0305
CO2 3.59 0.043744
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Sample Problem 13
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
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Sample Problem 13 Solution
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 Kdm3•atmmol•K
1.00 atm= = 4.76 dm3
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