Combined and ideal gas laws

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Combined and ideal gas laws
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Gas Properties

• Gases have mass
• Gases diffuse
• Gases expand to fill containers
• Gases exert pressure
• Gases are compressible
• Pressure temperature are dependent

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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)

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A Little Review

• Boyles law
• pressure volume
• as P? then V?
• at constant T, n

P1V1 P2V2

• Charles law
• Temperature volume
• As T? then V?
• At constant P, n

T1V2 T2V1
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A Little Review

• Gay-Lussacs law
• Temperature pressure
• As P? then T?
• At constant V, n

P1T2 P2T1
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Combined gas law

• If we combine all of the relationships from the 3
  laws covered thus far (Boyles, Charless, and
  Gay-Lussacs) we can develop a mathematical
  equation that can solve for a situation where 3
  variables change

PVk1
V/Tk2
P/Tk3
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Combined gas law

• Amount is held constant
• Is used when you have a change in volume,
  pressure, or temperature

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Combined gas law

• Amount is held constant
• Is used when you have a change in volume,
  pressure, or temperature

P1V1T2 P2V2T1
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Example problem
A gas with a volume of 4.0L at STP. What is its
volume at 2.0atm and at 30C?
1atm
2.0 atm
?
4.0 L
273K
30C 273
303K
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Example problem
(1 atm)
(4.0L)
(2 atm)
( V )
2

(273K)
(303K)
2.22L V2
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Avogadros Law

• So far weve compared all the variables except
  the amount of a gas (n).
• There is a lesser known law called Avogadros 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
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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.

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Ideal Gas Law

• If we combine all of the laws together including
  Avogadros Law mentioned earlier we get

Where R is the universal gas constant
Normally written as
PV nRT
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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

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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

• If pressure is given in kPa

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Using the Ideal Gas Law
What volume does 9.45g of C2H2 occupy at STP?
1atm
P ?
R ?
?
V ?
T ?
273K
n ?
.3635 mol
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(1.0atm)
(V)

(.3635mol)
(273K)
V 8.15L
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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 25C and a
pressure of 303 kpa?
303kPa
P ?
R ?
?
V ?
T ?
298K
n ?
68.2 mol
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(303kPa)
(V)

(298K)
(68.2 mol)
V 557.7L
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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 300C?
(1.00 atm)
(1.00 L)
nH2O
(573K)
(.0821L atm/mol K)
nH2O .021257 mols
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Ideal Gas Law Stoichiometry
2H2 O2 ? 2H2O
.021257 mol
.476 L H2
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Loose Ends of Gases

• There are a couple more laws that we need to
  address dealing with gases.
• Daltons Law of Partial Pressures
• Grahams Law of Diffusion and Effusion.

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Daltons 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.

PTP1P2P3

• What that means is that each gas involved in a
  mixture exerts an independent pressure on its
  containers walls

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Daltons 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.

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Daltons 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.

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Simple Daltons 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?

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Simple Daltons Law Calculation
PT PCO2 PN2 PO2
760mmHg .285mmHg 593.525mmHg PO2
PO2 167mmHg
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Daltons 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

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Atmospheric Pressure
Ptot Patmospheric pressure Pgas PH2O

• The waters vapor pressure can be determined from
  a list and subtract-ed from the atmospheric
  pressure

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Simple Daltons Law Calculation

• Determine the partial pressure of oxygen
  collected by water displace-ment if the water
  temperature is 20.0C and the total pressure of
  the gases in the collection bottle is 730 mmHg.

PH2O at 20.0C 2.3388 kPa
We need to convert to mmHg.
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Simple Daltons Law Calculation
2.3388 kPa
PH2O 17.5468 mmHg
PT PH2O PO2
730mmHg 17.5468 PO2
PO2 712.5 mmHg
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Grahams 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

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Diffusion
Effusion
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Grahams Law

• Grahams 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.

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Grahams 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

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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

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vA
MB

vB
MA

• This shows that the velocities of two different
  gases are inversely propor-tional to the square
  roots of their molar masses.
• This can be expanded to deal with rates of
  diffusion or effusion

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Grahams 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.

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Grahams 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?
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Grahams Law Example Calc.
Rate of effusion of He

Rate of effusion of Ar
Helium is 3.16 times faster than Argon.