Gas  Laws  

• Boyle’s  Law  • Charles’s  law  

• Gay-­‐Lussac’s  Law  • Avogadro’s  Law  

• Dalton’s  Law  • Henry’s  Law  

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2

Kinetic-Molecular Theory of Gases  •  KMT  -­‐  the  theory  of  moving  molecules  •  Gases  consist  of  large  numbers  of  conAnuously  moving  parAcles.  •  The  volume  caused  by  the  molecules  is  negligible.  •  The  parAcles  are  in  constant  random  moAon,  colliding  with  the  walls  of  the  container.    These  collisions  with  the  walls  cause  the  pressure  exerted  by  the  gas.  

•  InteracAve  (aHracAve  or  repulsive)  forces  are  negligible.  •  Energy  is  transferred  during  collisions  but  the  average  total  stays  constant.  (Collisions  are  perfectly  elastic.)  

•  The  average  kineAc  energy  of  the  gas  parAcles  is  directly  proporAonal  to  the  Kelvin  temperature  of  the  gas.  

 

Properties of Gases

•  Expand to completely fill their container •  Take the shape of their container •  Low density •  Compressible •  Mixtures of gases are always

homogeneous •  Fluid

Lets  talk  units  for  Gas  Laws  P,  V,  n,  T  

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• We  need  four  variables:  • Volume  (V)  ~  Liters  • Pressure  (P)  ~  next  slide  • Temperature  (T)  ~  Kelvin  (T(K)  =  T(oC)  +  273)      • Moles  (n)  ~  moles    

Gas Pressure

• Pressure = total force applied to a certain area •  Larger force = larger pressure •  Smaller area = larger pressure

• Gas pressure caused by gas molecules colliding with container or surface

• More forceful or more frequent collisions mean higher gas pressure

Units of Gas Pressure

•  Atmosphere (atm) •  Height of a column of mercury (mm Hg, in Hg) •  Torr •  Pascal (Pa) •  Pounds per square inch (psi, lbs/in2)

• 1.000 atm = 760.0 mm Hg = 29.92 in Hg = 760.0 torr = 101,325 Pa = 101.3 kPa = 14.69 psi

Air Pressure

• Constantly present when air present

• Decreases with altitude • Less air = less pressure

• Varies with weather conditions

Barometer •  The mercury in the tube

pushes down with its weight. •  The bottom of the tube is open

to the atmosphere. •  The air pushes on the open

surface of the mercury. •  On an average day, the

pressure of the air equals the pressure exerted by a column of mercury 760 mm high.

•  Above 760 mm, there is a vacuum in the tube.

Weight of mercury

Measuring Air Pressure

Manometer •  A manometer is comprised of a bulb containing a gas and a U-shaped tube.

•  The U-shaped tube is partially filled with mercury. The weight of the mercury puts pressure on the gas.

•  The gas molecules put pressure on the mercury.

PHg

Closed Manometers •  There is a balance

between the weight of the mercury on the left (PHg) and the pressure of the gas on the right (Pgas).

•  The difference between the heights of the mercury on each side of the tube is a measure of the pressure of the gas.

Pgas = Δh

vacuum

PHg

Open Manometers •  When gas pressure is greater

than atmospheric pressure, the mercury is pushed toward the open end.

•  The balance is between the gas on the right, and the air plus mercury on the left.

Pair + PHg = Pgas •  The weight of the mercury is

measured as the height difference:

PHg = Δh

So Pgas = Pair + Δh

Pair

PHg

Open Manometers •  When gas pressure is less

than atmospheric pressure, the mercury is pushed toward the gas reservoir.

•  The balance is between the air on the left and the gas plus mercury on the right:

Pair = Pgas + PHg •  The weight of the mercury is

measured as the height difference:

PHg = Δh So Pgas = Pair- Δh

PHg Pair

Sample Problems

PAIR  =  765  mm

Δh  =  27  mm  Δh  =  13  mm  

closed  

Δh  =    20  mm  

Pair = 790 mm

Find the pressure of the gas in each manometer.

1. 2. 3.

Pgas+ Δh = Pair Pgas= 790 mm - 20 mm =

770 mm Hg

Pgas = Pair + Δh Pgas = 765 + 27 =

792 mm Hg Pgas = Δh

= 13 mm Hg

Units  conversions  1.000 atm = 760.0 mm Hg = 29.92 in Hg = 760.0 torr = 101,325 Pa = 101.3 kPa = 14.69 psi T(K)  =  T(oC)  +  273    

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1.  745 mm Hg to pascals? 2. 5.43 atm to mm Hg? 3. 35oC to Kelvin?

P,  V,  T,  n  (n  =  moles):        Avogadro’s  Law  •  Observe:  blowing  up  a  balloon    

•  When  P  &  T  are  held  constant,    The  relaAonship            between  n  and  V  is  Directly  proporAonal  or  Inversely  proporAonal?  

• Directly  proporAonal! •  Volume directly proportional to the number of gas molecules

As  moles  é,  Volume  é

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V1n1=V2n2

V1n1= constant, so

Avogadro’s Law (cont.)

• Count number of gas molecules by moles • One mole of any ideal gas occupies 22.414 L at

standard conditions = molar volume • Equal volumes of gases contain equal numbers of

molecules. • It doesn’t matter what the gas is!

Practice Problem 4 •  If  2.55  mol  of  helium  gas  occupies  a  volume  of  59.5  L  at  a  parAcular  temperature  and  pressure,  what  volume  does  7.83  mol  of  helium  occupy  under  the  same  condiAons?  

                                                                                 Answer:      183  L  

2

2

1

1

nV

nV

=

Practice Problem 5 •  If 4.35 g of neon gas occupies a volume of 15.0 L at a particular

temperature and pressure, what volume does 2.00 g of neon gas occupy under the same conditions?

Answer: 6.88 L

2

2

1

1

nV

nV

=

•  Observe:  on  on  top  of  the  flask  • When  n  &  P  are  held  constant,    The  relaAonship            between  V  and  T  is  

Directly  proporAonal  or  Inversely  proporAonal?  

Directly  proporAonal!  As  Temperature  é,  Volume  é

   

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P,  V,  T,  n  (n  =  moles):        Charles’  Law  

V1T1=V2T2

V1T1= constant, so

Practice Problem 6 A sample of oxygen gas has a volume of 4.55 L at 25°C.

Calculate the volume of the oxygen gas when the temperature is raised to 45°C. Assume constant pressure

• V1 = V2 T1 T2

Answer: V2 = 4.86 L

P,  V,  T,  n  (n  =  moles):      Boyle’s  Law            

•  Observe:  Squish  syringe  with  marshmallow  in  it  • When  n  &  T  are  held  constant,    The  relaAonship            between  P  and  V  is  

Directly  proporAonal  or  Inversely  proporAonal?  

Inversely  proporAonal!  As  Pressure  é,  Volume  ê

• P x V = constant, so

   

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P1 x V1 = P2 x V2

What is the new volume if a 1.5 L sample of Freon-12 at 56 torr is compressed to 150 torr? P1 x V1 = P2 x V2 V2 = 0.52 L

Practice Problem 7

Practice Problem 8 A sample of helium gas has a pressure of 3.54 atm in a container with a volume of 23.1 L. This sample is transferred to a new container and the pressure is measured to be 1.87 atm. What is the volume of the new container? Assume constant temperature.

Answer: V2 = 43.7 L

Practice Problem 9 A steel tank of argon gas has a pressure of 34.6 atm. If all of the argon is transferred to a new tank with a volume of 456 L, the pressure is measured to be 2.94 atm. What is the volume of the original container? Assume constant temperature.

Answer: V1 = 38.7 L

P,  V,  T,  n  (n  =  moles):    Gay-­‐Lussac’s  Law  •  Observe:  Soda  can  

• When  n  &  V  are  held  constant,    The  relaAonship            between  P  and  T  is  

Directly  proporAonal  or  Inversely  proporAonal?  

Directly  proporAonal!  As  Temperature  ê,  Pressure  ê

   

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P1T1= constant, so P1

T1=P2T2

Ideal Gas Law •  By combining the proportionality constants from the gas

laws we can write a general equation. •  Use the ideal gas law when you have gas at one set of

conditions •  Most gases obey this law when pressure is low (at or below

1 atm) and temperature is high (above 0°C) •  R is called the gas constant. •  The value of R depends on the units of P

and V. •  Generally use R = 0.08206 atm*L/mol*K

PV = nRT

Practice Problem 10 •  A sample of hydrogen gas, H2, has a volume of 8.56 L at a

temperature of 0°C and a pressure of 1.5 atm. Calculate the number of moles of H2 present in this gas sample. (Assume that the gas behaves ideally).

Answer: 0.57 mol

PV = nRT

Practice Problem 11 •  A 2.50 mol sample of nitrogen gas has a volume of 5.50 L at a

temperature of 27°C. Calculate the pressure of the nitrogen gas

Answer: 11.2 atm

PV = nRT

Practice Problem 12 •  A 5.00 mol sample of oxygen gas has a pressure of 1.25 atm at 22°C.

What volume does this gas occupy?

Answer: 96.8 L

PV = nRT

Practice Problem 13 •  A sample of neon gas has a volume of 3.45 L at 25°C and a pressure

of 565 torr. Calculate the number of moles of neon present in this gas sample?

Answer: 0.105 mol

PV = nRT

Practice Problem 14 •  A 0.250 mol sample of argon gas has a volume of 9.00 L at a

pressure of 875 mm Hg. What is the temperature (in °C) of the gas?

Answer: 232 °C

PV = nRT

Standard temperature and pressure (STP) •  STP  commonly  is  used  when  standard  state  condi,ons  are  applied  to  calculaAons.  Standard  state  condiAons,  which  include  standard  temperature  and  pressure,  may  be  recognized  in  calculaAons  

•  standard  temperature  is  273  K  (0°  Celsius)    •  standard  pressure  is  1  atm  pressure.      •  Reminder:  At  STP,  one  mole  of  gas  occupies  22.4  L  of  volume  (molar  volume).  

Combine  the  4  variables  together  mathematically  to  get  The  Combined  Gas  Law  •  Combine  these  four  variables  together  according  to  the  relaAonships  described  on  the  previous  slides  to  get:  

•  Set  the  formula  equal  to  a  constant,  R  and  noAce  how  one  variable  must  change  in  response  to  another  change,  holding  two  others  constant.  

•  Since  a  constant,  R  is  constant,  R  =  R.  This  means  we  can  subsAtute  the  variables  to  get:  

•  The  Combined  Gas  Law   33

P V n T

= R P V n T

= P1 V1 n1 T1

P2 V2 n2 T2

Sample  problems  (‘Before  and  After  Problems’)  The  Combined  Gas  Law  

•  Consider  0.5  L  of  gas  in  syringe    in  a  323  K  water  bath  with  a  pressure    of  200  mmHg.  The  syringe  is  compressed  to  0.1  L,  what  will  be  the    new  pressure?  

• HINT:  Typically  you  would  be  given  all  but  one  piece  of  informaAon  and  then  asked  to  find  the  missing  informaAon.  Be  aware  that  some  of  the  variables  remain  constant  and  you  can  simplify  the  equaAon  by  dropping  those  variables  from  the  formula.  

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P1 V1 n1 T1

P2 V2 n2 T2

=

Sample  problems  The  Combined  Gas  Law  

•  Consider  0.5  L  of  gas  in  syringe  in  a  323  K  water  bath  with  a  pressure  of  200  mmHg.  The  syringe  is  compressed  to  0.1  L,  what  will  be  the  new  pressure?  

 

35

P1 V1 n1 T1

P2 V2 n2 T2

=

Sample  problems  The  Combined  Gas  Law  

•  If  I  iniAally  have  a  gas  at  a  pressure  of  12  atm,  a  volume  of  23  L,  and  a  temperature  of  200  K,  and  then  I  raise  the  pressure  to  14  atm  and  increase  the  temperature  to  300  K,  what  is  the  new  volume  of  the  gas?    

36

P1 V1 n1 T1

P2 V2 n2 T2

=

Scuba divers going deeper than 150ft use a helium and oxygen mixture

•  Divers experience high pressure under several hundred feet of pressure

•  Nitrogen gas dissolves in blood under these conditions

•  Returning to the surface too quickly results in “the bends”

Dalton’s Law of Partial Pressures

•  The total pressure of a mixture of gases equals the sum of the pressures each gas would exert independently.\

•  Partial pressure: the pressure a gas in a mixture would exert if it were alone in the container Ptotal = Pgas A + Pgas B + …

Dalton’s Law (cont.)

•  Particularly useful for determining the pressure a dry gas would have after it is collected over water Pair = Pwet gas = Pdry gas + Pwater vapor Pwater vapor depends on the temperature, look up in table

Partial Pressures

VT x R x n P P P

n n nsame theare mixture in the

everything of volumeand re temperatuTheV

T x R x n P V

T x R x n P

mixture ain B andA gasesFor

totalBAtotal

BAtotal

BB

AA

=+=

+=

==

The partial pressure of each gas in a mixture can be calculated using the Ideal Gas Law:

Practice Problem 13.12 •  Mixtures of helium and oxygen are used in the “air” tanks of

underwater divers for deep dives. For a particular dive, 12 L of O2 at 25°C and 1.0 atm and 46 L of He at 25°C at 1 atm were both pumped into a 5.0 L tank. Calculate the partial pressure of each gas in the tank at 25°C

Answer: 2.4 atm O2; 9.3 atm He; 11.7 atm total

n= RT PV

When two gases are present, the total pressure is the sum of the partial pressures of the gases.

Partial Pressures

The total pressure of a mixture of gases depends on the number of moles of gas particles (atoms or molecules) present, not on the identities of the particles. Note that these three samples show the same total pressure because each contains 1.75 mol of gas. The detailed nature of the mixture is unimportant.

Partial Pressures

Practice Problem 13.12 A •  A 5.00 g sample of helium gas is added to a 5.00 g sample of neon

gas in a 2.50 L container at 27°C. Calculate the partial pressure of each gas and the total pressure

•  Answer: 12.3 atm He; 2.44 atm Ne; 14.7 atm total

PV = nRT

Practice Problem 13.12 B •  Equal  masses  of  oxygen  and  nitrogen  gas  are  present  in  a  container.    Which  gas  exerts  the  larger  parAal  pressure?    By  what  factor?  

 Answer:      N2  exerts  a  par;al  pressure  that  is  1.14  ;mes  as  great  as  the  par;al  pressure  of  O2  

PV = nRT

Water Vapor Pressure A  mixture  of  gases  occurs  whenever  a  gas  is  collected  by  displacement  of  water.  

The  producAon  of  oxygen  by  thermal  decomposiAon  of  KClO3  

2KClO3(s)  à    2KCl(s)    +    3O2(s)    

Practice Problem 13.13 2KClO3(s) à 2KCl(s) + 3O2(s) •  The oxygen produced was collected by displacement of water at 22°C.

The resulting mixture of O2 and H2O vapor had a total pressure of 754 torr and a volume of 0.650L. Calculate the partial pressure of O2 in the gas collected and the number of moles of O2 present. The vapor pressure of water at 22°C is 21 torr.

Practice Problem 13.13 2KClO3(s) à 2KCl(s) + 3O2(s) •  The oxygen produced was collected

by displacement of water at 22°C. The resulting mixture of O2 and H2O vapor had a total pressure of 754 torr and a volume of 0.650L. Calculate the partial pressure of O2 in the gas collected and the number of moles of O2 present. The vapor pressure of water at 22°C is 21 torr.

Ptotal = PO2 + PH2O Answer: PO2 = 0.964 atm

nO2 = 2.59 x 10-2 mol

Practice Problem 13.13 A •  A sample of oxygen gas is saturated with water vapor at 30.0°C.

The total pressure is 753 torr, and the vapor pressure of water at 30.0°C is 31.824 torr. What is the partial pressure of oxygen gas in atm?

Answer: 0.949 atm

Practice Problem 13.13 B •  A sample of oxygen gas is saturated with water vapor at 27°C.

The total pressure is 785.0 torr, and the partial pressure of oxygen is 758.3 torr. What is the vapor pressure of water at 27°C?

Answer: 26.7 torr (0.0351 atm)