Ideal Gas Law: P V = n R T
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Transcript Ideal Gas Law: P V = n R T
Using PV = nRT
(Honors)
P = Pressure
V = Volume
T = Temperature
N = number of moles
R is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for all the possible unit
combinations, we can just memorize one value and convert the
units to match R.
R = 0.0821
L • atm
Mol • K
Ideal Gas Law: P V = n R T
P = pressure in atm
V = volume measured in Liters
n =# of moles
T = temperature K
R=Universal gas constant
=0.0821 L atm/(mol K)
Ideal
Gas
Law
(Honors)
Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
IDEAL GAS LAW
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
Some Cool Videos (Honors)
Crash Course: Ideal Gas Laws
http://www.youtube.com/watch?v=BxUS1K7xu30&safe=active
Crash Course: Ideal Gas Law Problems
http://www.youtube.com/watch?v=8SRAkXMu3d0
Crash Course: Real Gases
http://www.youtube.com/watch?v=GIPrsWuSkQc&safe=active
Crash Course: Grahams Law
http://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=active
Dalton’s Law of
Partial Pressures
Total pressure of a mixture of gases in a
container equals the sum of the
individual partial pressures of each gas.
Ptotal = P1 + P2 + ...
Patm = PH2 +
PH2O
This is often useful when gases are
collected “over water”
Crash Course: Partial Pressure
and Vapor Pressure
http://www.youtube.com/watch?
v=JbqtqCunYzA&safe=active
You can use Table H
Dalton’s
Law
Hydrogen gas is collected over water at
22.5°C. Find the pressure of the dry gas if the
atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
on for 22.5°C.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Sig Figs: Round to least number
of decimal places.
Dalton’s Law
A gas is collected over water at a temp of 35.0°C when
the barometric pressure is 742.0 torr. What is the partial
pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
DALTON’S LAW
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Look up water-vapor pressure
for 35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
of decimal places.
Using Mole Fraction (Honors)
Graham’s Law
Diffusion
Spreading of gas molecules throughout a
container until evenly distributed.
Effusion
Passing of gas molecules through a tiny
opening in a container
Graham’s Law
Speed of diffusion/effusion
Kinetic energy is determined by the
temperature of the gas.
At the same temp & KE, heavier molecules
move more slowly.
Graham’s Law Formula (Honors)
Graham’s Law
Rate of diffusion of a gas is inversely related
to the square root of its molar mass.
Ratio of gas
A’s speed to
gas B’s speed
vA
vB
mB
mA
Graham’s
Law
Determine the relative rate of diffusion for
krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
vA
vB
v Kr
v Br 2
m Br 2
m Kr
mB
mA
159.80 g/mol
1.381
83.80 g/mol
Kr diffuses 1.381 times faster than Br2.
Graham’s Law
vA
A molecule of oxygen gas has an average speed of 12.3
m/s at a given temp and pressure. What is the average
speed of hydrogen molecules at the same conditions?
vB
mB
vH2
mA
12.3 m/s
32.00 g/mol
2.02 g/mol
vH2
vH2
vO2
m O2
mH2
Put the gas with
the unknown
speed as
“Gas A”.
3.980
12.3 m/s
v H 2 49.0 m/s
Graham’s Law
An unknown gas diffuses 4.0 times faster than O2. Find
its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA
vB
vA
vO2
mB
mA
Square both
sides to get rid
of the square
root sign.
16
m O2
mA
4.0
32.00 g/mol
mA
32.00 g/mol
mA
mA
32.00 g/mol
16
2.0 g/mol
2
Avogadro’s Principle
Equal volumes of gases contain
equal numbers of moles
at constant temp & pressure
true for any gas
Equal volumes of gases at the same T and P have the
same number of molecules.
V
V
n
n
k
Gas Stoichiometry (Honors)
Moles Liters of a Gas
STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP Problems
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv.
Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g of
CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g
CaO
+
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g 1 mol
CaCO3 CaCO3
1 mol
CO2
100.09g 1 mol
CaCO3 CaCO3
CO2
?L
non-STP
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.
Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 1.26 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
Stoichiometry
Problem
How Gas
many grams
of Al2O3 are formed
from 15.0 L of
O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2
2 Al2O3
15.0 L
non-STP
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
NEXT
n = 0.597 mol O2
Gas Stoichiometry Problem
How many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
3 O2
15.0L
Use stoich to convert moles
of O to grams Al O .
non-STP
0.597 2 mol 101.96 g
mol O2 Al2O3
Al2O3
4 Al
2
2
+
2 Al2O3
?g
3
3 mol O2
1 mol
Al2O3
= 40.6 g Al2O3