Transcript Combined and ideal gas laws

Combined and
ideal gas laws
Gas Properties
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Gases have mass
Gases diffuse
Gases expand to fill containers
Gases exert pressure
Gases are compressible
Pressure & temperature are
dependent
Gas Variables
 Volume
(V)
• Units of volume (L)
 Amount (n)
• Units of amount (moles)
 Temperature (T)
• Units of temperature (K)
 Pressure (P)
• Units of pressure (mmHg)
• Units of pressure (KPa)
• Units of pressure (atm)
A Little Review
 Boyle’s
law
• pressure & volume
• as P then V
P 1V 1 =
• at constant T, n
 Charles’ law:
• Temperature & volume
• As T then V
T
V
=
1
2
• At constant P, n
P 2V 2
T2V 1
A Little Review
 Gay-Lussac’s
law:
• Temperature & pressure
• As P then T
P
T
=
1
2
• At constant V, n
P 2T1
Combined gas law
 If
we combine all of the relationships
from the 3 laws covered thus far
(Boyle’s, Charles’s, and GayLussac’s) we can develop a
mathematical equation that can
solve for a situation where 3
variables change :
PV=k1 V/T=k2
P/T=k3
Combined gas law
 Amount
is held constant
 Is used when you have a change in
volume, pressure, or temperature
P 1V 1
T1
=K=
P2V2
T2
Combined gas law
 Amount
is held constant
 Is used when you have a change in
volume, pressure, or temperature
P 1V 1
T1
=
P2V2
T2
P 1 V 1 T2 = P2 V 2 T1
Example problem
A gas with a volume of 4.0L at STP.
What is its volume at 2.0atm
and at 30°C?
- P1  1atm
- V1  4.0 L
- T1  273K
- P2  2.0 atm
- V2  ?
- T2  30°C + 273
= 303K
Example problem
P 1V 1
T1
=
P2V2
T2
2.22L = V2
Avogadro’s Law
 So
far we’ve compared all the
variables except the amount of a
gas (n).
 There is a lesser known law called
Avogadro’s Law which relates V & n.
 It turns out that they are directly
related to each other.
 As # of moles increases then V
increases.
V/n = k
Ideal Gas Law
 Which
leads us to the ideal gas law –
 So far we have always held at least
1 of the variables constant.
 We can set up a much more powerful
eqn, which can be derived by
combining the proportions expressed
by the previous laws.
Ideal Gas Law
 If
we combine all of the laws
together including Avogadro’s Law
mentioned earlier we get:
PV
nT
=R
Normally
written as
Where R is the
universal gas
constant
PV = nRT
Ideal Gas Constant (R)
 R is a constant that connects the
4 variables
 R is dependent on the units of the
variables for P, V, & T
• Temp is always in Kelvin
• Volume is in liters
• Pressure is in either atm or mmHg
or kPa
Ideal Gas Constant

Because of the different pressure
units there are 3 possibilities for
our ideal gas constant
• If pressure is
given in atm
• If pressure is
given in mmHg
R=.0821 L•atm
mol•K
L•mmHg
R=62.4
mol•K
• If pressure is R=8.314 L•kPa
given in kPa
mol•K
Using the Ideal Gas Law
What volume does 9.45g
of C2H2 occupy at STP?
P  1atm
V
n
?
9.45g
26g
L•atm
R  .0821
mol•K
T  273K
= .3635 mol
PV = nRT
(1.0atm)(V) =
L•atm
(.3635mol) (.0821 mol•K
)(273K)
(1.0atm)(V)= (8.147L•atm)
V = 8.15L
A camping stove propane tank holds
3000g of C3H8. How large a
container would be needed to hold
the same amount of propane as a gas
at 25°C and a pressure of 303 kpa?
L•kPa
8.31
P  303kPa R 
V
n
?
3000g
44g
mol•K
T  298K
= 68.2 mol
PV = nRT
(303kPa)(V)=
(68.2 mol) (8.31
L•kPa
)
mol•K
(298K)
(303kPa) (V) = (168,970.4 L•kPa)
V = 557.7L
Ideal Gas Law & Stoichiometry
What volume of hydrogen gas must
be burned to form 1.00 L of
water vapor at 1.00 atm pressure
and 300°C?
PV = nRT
(1.00 atm)(1.00 L)
nH2O=
(.0821L atm/mol K)(573K)
nH2O= .021257 mols
Ideal Gas Law & Stoichiometry
2H2 + O2  2H2O
.021257 mol
2 mol H2
22.4 L H2
2 mol H2O
1mol H2
.476 L H2
=
Loose Ends of Gases
• There are a couple more laws that
we need to address dealing with
gases.
– Dalton’s Law of Partial Pressures
– Graham’s Law of Diffusion and
Effusion.
Dalton’s Law of Partial Pressure
• States that the total pressure of a
mixture of gases is equal to the sum
of the partial pressures of the
component gases.
PT=P1+P2+P3+…
• What that means is that each gas
involved in a mixture exerts an
independent pressure on its
containers walls
Dalton’s Law of Partial Pressure
• Therefore, to find the pressure in
the system you must have the total
pressure of all of the gases
involved.
• This becomes very important for
people who work at high altitudes
like mountain climbers and pilots.
• For example, at an altitude of about
10,000m air pressure is about 1/3
of an atmosphere.
Dalton’s Law of Partial Pressure
• The partial pressure of oxygen at
this altitude is less than 50 mmHg.
• By comparison, the partial pressure
of oxygen in human alveolar blood
needs to be about 100 mmHg.
• Thus, respiration cannot occur
normally at this altitude, and an
outside source of oxygen is needed
in order to survive.
Simple Dalton’s Law Calculation
• Three of the primary components of
air are CO2, N2, and O2. In a
sample containing a mixture of
these gases at exactly 760 mmHg,
the partial pressures of CO2 and N2
are given as PCO2= 0.285mmHg
and PN2 = 593.525mmHg. What is
the partial pressure of O2?
Simple Dalton’s Law Calculation
PT = PCO2 + PN2 + PO2
760mmHg = .285mmHg +
593.525mmHg + PO2
PO2= 167mmHg
Dalton’s Law of Partial Pressure
• Partial Pressures are also
important when a gas is collected
through water.
— Any time a gas is collected
through water the gas is
“contaminated” with water vapor.
— You can determine the pressure of
the dry gas by subtracting out the
water vapor
Atmospheric
Pressure
Ptot = Patmospheric pressure = Pgas + PH2O
— The water’s vapor pressure can be
determined from a list and subtracted from the atmospheric pressure
WATER VAPOR PRESSURES
Temp (°C)
Vapor pressure (kPa)
1
5
10
15
20
25
30
35
40
45
50
0.65176
.87260
1.2281
1.7056
2.3388
3.1691
4.2455
5.6267
7.3814
9.5898
12.344
WATER VAPOR PRESSURES
Temp (°C)
Vapor pressure (kPa)
55
60
65
70
75
80
85
90
95
100
105
15.752
.19.932
25.022
31.176
38.563
47.373
57.815
70.117
84.529
101.32
120.79
Simple Dalton’s Law Calculation
• Determine the partial pressure of
oxygen collected by water displacement if the water temperature is
20.0°C and the total pressure of the
gases in the collection bottle is
730 mmHg.
PH2O at 20.0°C= 2.3388 kPa
We need to convert to mmHg.
Simple Dalton’s Law Calculation
2.3388 kPa
760 mmHg
101.3 kPa
PH2O = 17.5468 mmHg
PT = PH2O + PO2
730mmHg = 17.5468 + PO2
PO2= 712.5 mmHg
Graham’s Law
• Thomas Graham studied the
effusion and diffusion of gases.
– Diffusion is the mixing of gases
through each other.
– Effusion is the process whereby the
molecules of a gas escape from its
container through a tiny hole
Graham’s Law
• Graham’s Law states that the rates
of effusion and diffusion of gases at
the same temperature and pressure
is dependent on the size of the
molecule.
– The bigger the molecule the slower
it moves the slower it mixes and
escapes.
Graham’s Law
• Kinetic energy can be calculated
with the equation ½ mv2
— m is the mass of the object
— v is the velocity.
• If we work with two different at the
same temperature their energies
would be equal and the equation
can be rewritten as:
½ MAvA2 = ½ MBvB2
• “M” represents molar mass
• “v” represents molecular velocity
• “A” is one gas
• “B” is another gas
• If we want to compare both gases
velocities, to determine which gas
moves faster, we could write a ratio
of their velocities.
— Rearranging things and taking the
square root would give the eqn:
vA
vB
=
MB
MA
• This shows that the velocities of two
different gases are inversely proportional to the square roots of their
molar masses.
— This can be expanded to deal with
rates of diffusion or effusion
Rate of effusion of A
Rate of effusion of B
=
MB
MA
Graham’s Law
• The way you can interpret the
equation is that the number of times
faster A moves than B, is the
square root of the ratio of the molar
mass of B divided by the Molar
mass of A
— So if A is half the size of B than it
effuses or diffuses 1.4 times faster.
Graham’s Law Example Calc.
If equal amounts of helium and argon
are placed in a porous container and
allowed to escape, which gas will
escape faster and how much faster?
Rate of effusion of A
Rate of effusion of B
=
MB
MA
Graham’s Law Example Calc.
Rate of effusion of He
Rate of effusion of Ar
=
40 g
4g
Helium is 3.16 times faster than Argon.