The gas laws

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The gas laws

• Chapter 5 of AP chemistry

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Gasses

• We are surrounded by gasses.
• We would die if we were released into space where
  there is no gas pressure
• Because much of early chemistry involved
  observations of gas behavior, there are many
  different units of pressure

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Early measurements of pressure

• Pressure was first measured with a meter long
  tube filled with mercury.
• This was the first barometer.
• Ill explain how it works.

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Units of pressure

• The standard unit of pressure here in the States
  is the atmosphere (atm)
• 1 atm 760 mm Hg
• 760 torr
• 101,325 Pa (Pascal)
• 14.7 PSI (pound per square inch)
• 1.01325 bar

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Boyles law

• Boyle found that the pressure and the volume of a
  gas are inversely proportional.
• This makes perfect sense. If you crush a plastic
  bottle with the top on, what happens to the
  pressure in the bottle?
• Lets look at the graph.

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The equation

• PVK
• This law states that the product of pressure and
  volume will be a constant value
• This leads to a more useful equation
• P1V1KP2V2
• OR
• P1V1P2V2

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Charless law

• Jacques Charles determined that the temperature
  and the volume of a gas had a directly
  proportional relationship.
• Again this makes sense. When you heat up a gas
  the particles move faster and they expand their
  container (if they can).
• Well look at a graph.

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Absolute Zero

• Charles did his work in the late 1700s, well
  before modern refrigeration.
• He did note that all his separate gasses shared a
  common x-intercept on the graph
• That x-intercept represented the temperature at
  which all substances would occupy zero volume,
  absolute zero.

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Absolute Zero -273.15 C
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Kelvins

• Whenever temperature is present in gas law
  calculations it should be expressed in Kelvins.
• C 273 K
• IMPORTANT When your temperature increases from
  25 C to 50 C it DID NOT double!!!!!!!!!!!!!!!!!!
  !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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25 C to 50 C

• 25 C 273 25 298 K
• 50 C 273 50 323 K
• 298 K to 323 K is not a doubling of temperature.
• NEVER EVER use C in gas law problems unless
  youre looking up vapor pressure or something
  else on a chart.

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The equation

• The useful equation from Charless law is
• This equation can be combined with Boyles law to
  form the combined gas law

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The combined gas law

• So far we have two equations

• These two can be combined to form the combined
  gas law

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Example problems

1. A helium-filled balloon has a volume of 50.0 L at
   25 C and 1.08 atm. What volume will it have at
   0.885 atm and 10.0 C?
2. A hydrogen-filled zeppelin used on bombing
   missions in WW I had a volume of 4550 m3 at 25 C
   and 1.00 atm. What volume will it have if it
   flies into a low pressure zone with a pressure of
   0.95 atm and 20.0 C?

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Example problems

1. A 500 mL water bottle filled with air at 25 C is
   at a pressure of 0.877 atm. What pressure would
   result if the volume is reduced to 300 mL and the
   temperature is raised to 37 C?
2. A balloon filled with nitrogen gas is placed in a
   freezer at 0 C, and 1 atm. The balloon has a
   volume of 255 mL in the freezer. Once it is
   removed from the freezer it is placed in warm
   water where it warms to the temperature of the
   water and it expands to a volume of 400 mL, and
   as a result, the pressure on the balloon is also
   increased to 2.5 atm? What is the final
   temperature of the balloon? Whats wrong with
   this problem?

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Avogadros law

• Avagadro found that for a gas at constant
  temperature and pressure, the volume is directly
  proportional to the number of moles of gas.
• This is written as

• Where V is volume and n is the number of moles,
  and K is a constant.

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Ramifications

• It may seem strange but all gasses occupy the
  same volume if they are under the same pressure
  and temperature.
• UF6 has a molar mass of over 300 g/mol yet a mole
  of it occupies the same volume as a mole of H2
  which has a molar mass of just over 2 g/mol.
  Well get into how later.

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Example

• Suppose we have a 12.2-L sample containing 0.50
  mol oxygen gas (O2) at a pressure of 1 atm and a
  temperature of 25 C. If all this O2 were
  converted to ozone (O3) at the same temperature
  and pressure, what would be the volume of the
  ozone?

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Solution
Given V1 12.2 L, 0.5 mol O2 Want V2 Need Mol
O3 Will use

• 1) Write the equation

2) Determine moles of O3 produced
3) Solve Avogadros law
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Example problem

• If 1 mole of water vapor occupies 30.0 liters at
  a given pressure and temperature, what volume
  would be occupied by a mole of water if it were
  decomposed into hydrogen and oxygen gas?
• What are you Given? What do you want? What do you
  need? What will you use?

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The ideal gas law

• PVnRT Do you remember me???
• Well look at the laws that weve studied so far

R is the gas constant
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PVnRT

• This is the catch all gas law. This law
  incorporates all the other gas laws into one nice
  neat equation.
• This law should be applies when mass, or number
  of moles of gas is mentioned.
• As always, there are easy problems and there are
  hard problems
• PS Were always going to put pressure in atm
  from now on.

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Easy problems

1. A 5.00 mole sample of argon is put in a steel
   tank with a volume of 100. L. Once the gas fills
   the container the temperature is found to be 25
   ºC. If the container is opened, would air from
   the outside world be sucked in or pushed out?
2. How many moles would be sucked in or pushed out?

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Easy problems

• Givens
• n 5.00 mole
• V 100. L
• T 25 ºC
• R 0.08206
• P ???
• If the P lt 1 atm, are is sucked in, if P gt then
  argon is pushed out.

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Plug and chug

• Givens
• n 5.00 mole
• V 100. L
• T 25 ºC?298K
• R 0.08206
• P ???

• When the container argon will be pushed out of
  the container so as to equalize the pressure.

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A little harder problem

• How much argon will be pushed out?
• Lets think about this. What will happen to the
  pressure in the tank once the tank is opened to
  the outside world?
• It will equalize with the pressure outside the
  tank.
• How will it accomplish this?
• It will release some of the gas (moles)

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Set up

• P2 1.00 atm
• V2 100. L
• T2 298K
• R 0.08206
• n2 ???

• P1 1.22 atm
• V1 100. L
• T1 298K
• R 0.08206
• n1 5.00

• A lot of stuff cancels.

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Plug and chug

• You started with 5.00 mol and ended with 4.10
  mol.
• You she 0.90 mol of argon. You can convert that
  to grams if youre asked to

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Alternatively

• You could also simply apply the ideal gas law to
  the new container
• Givens
• P1.00atm
• V100. L
• T298 K
• R0.08206
• n???

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? ? Gas Stoichiometry ? ?

• What can we now calculate using the ideal gas
  law?
• Now we can do stoichiometry

MOLES!!!
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Sample problem

• Two chambers are mixed. The first occupies a
  volume of 25.0 L and contains ethane gas (C2H6)
  under 2.00 atm of pressure and a temperature of
  20.0 ºC.
• The second chamber is three times as large as the
  first and contains oxygen at STP (standard
  temperature and pressure, 0 ºC, and 1 atm know
  this).
• The gases are mixed, a spark is lit, the whole
  thing explodes in a massive combustion reaction,
  but the container is undamaged. The container is
  cooled to 0 ºC. What is the pressure in the
  container?
• Write and balance the equation, find moles of
  each reactant, determine the limiting reagent and
  calculate the moles of CO2 produced. You have T,
  V, and n. Solve for P

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A note on chemical reactions

• Decomposition of carbonates
• The carbonate ion (CO32-) forms ionic bonds with
  many metals.
• When these metals are heated the carbonate ion
  breaks down into the metal oxide and carbon
  dioxied gas.
• Heres the example of copper (II) carbonate

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Well work a problem

• A carbonate has an unknown metal cation. You
  find that upon thermal decomposition of 57.3705g
  of the carbonate 10.25 L of carbon dioxide can be
  collected at STP.
• What mass of metal oxide would be left behind?

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• First convert L of CO2 to moles
• Next find the mass of CO2 and subtract it from
  the original mass of the carbonate. Done!

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Lets make it harder

• As you will soon see in lab, a metal oxide can be
  liberated from its oxygen by heat in an
  atmosphere of hydrogen or methane gas.
• If upon liberation of the oxygen in a hydrogen
  atmosphere 8.244g of water are produced. If I
  tell you that the oxygen and the metal exist in a
  11 ratio, what metal do you have?

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• First you must find the mass of oxygen lost. For
  that you need moles of oxygen.

• You now know the mass of the oxide and the mass
  of oxygen. So you can solve for the mass of the
  metal.
• 37.24g(mass of MO) -7.32g (mass of oxygen)
• Mass of metal in oxide 29.92g

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The end at last

• If you know the mass of the metal and you know
  that the metal and oxygen exist in a 11 ratio in
  the oxide you have all you need.

• We know that the metal and water will be produced
  in equal molar quantities. Thus we know the mass
  and the mole of metal. So we solve for the molar
  mass.

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Molar mass of a gas

• If you know the density of a gas you can find its
  molar mass.
• Recall Dmass/V
• Also recall PVnRt
• Forget not MMmass/n
• See pg 205 for more

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This is a useful trick

• A gas is found to have a density of 2.68 g/L at a
  temperature of 27 ºC under 1.50 atm of pressure.
  Whats its molar mass?

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Daltons law of partial pressures

• This law states that the total pressure is equal
  to the sum of all its parts. This may seem
  obvious but, it is very useful.
• The equation is below

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More Dalton

• Well look at a former students work.
• He did a good job.

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Those are basic problems

• The problems get harder. Here you go.
• Aluminum will react with hydrochloric acid to
  produce aluminum chloride and hydrogen gas. If
  the reaction takes place at a temperature is 20.0
  ºC and a pressure is 752 torr, what mass of
  aluminum would be required to produce 20.00 L of
  gas if collected by water dispacement.
• Vapor pressure of water at 20.0 ºC is 17.54 torr.

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• Step 1 Stoichiometry

• Step 2 Find partial pressure of H2
• PTPH2OPH2
• Look up PH2O at 20.0 its 17.54 torr
• 750.0 torr-17.54 torr 732.46 torr

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PVnRT

• We now have
• P732.46 torr? 0.9638 atm
• n2.780 mol H2
• R0.08206
• T 20.0 ºC
• Now we solve for v