Gas Laws

1
Gas Laws

• The Gas Laws
• Kinetic Molecular Theory
• .Matter is composed of very tiny particles
• .Particles of matter are in continual motion
• .The total KE of colliding particles remains
  constant
• When individual particles collide, some lose
  energy while others gain energy.
• No overall energy loss elastic collisons

2
Properties of a Gas

• Expansion no definite shape or volume
• Low density density of gas is about 1/1000 of
  that same substance in a liquid or a solid
  phase
• Diffusion process of spreading out
  spontaneously to occupy a space uniformly

3
Ideal Gas

• consists of very small independent particles.
  These particles move at random in space and
  experience elastic collisions
• Extra Approximate speed is 103 meters/sec,
  at 0 degrees C and 1 ATM of pressure
• Undergo 5 x 109 collisions/second

4
Gases in Real life are not so ideal
5
When would a gas not act ideal?

• At low temperatures - the molecules may be moving
  slow enough to allow attractions to occur
• At high pressures - the molecules may once again
  have attractive forces between them in effect

6
Gases and Pressure
Pressure the total average force over a given
area is called pressure. Although the
collisions of any single molecule of gas
within the walls of a container
are intermittent, there are so many molecules in
even a small container, that the tremendous
number of collisions average out to a steady
pressure.
7
Gas and Temperature

• Temperature the average kinetic energy or
  motion energy of its molecules.
• Avogadros hypothesis states that at any given
  temperature and pressure, equal volumes of gases,
  contain the same number of molecules.

8
Avogadro

• The following table supports the hypothesis by
  showing that one mole (6.02 x 1023 molecules) of
  11 different gases occupies approximately the
  same volume. These volumes were derived
  experimentally at the same pressure and
  temperature.

9
Table 1. Volumes of various gases at 0 C and
760 mm pressure 1 mole Oxygen Occupies
22.393 L Sulfur Dioxide Occupies
21.888 L Hydrogen Occupies
22.430 L Helium Occupies
22.426 L Chlorine Occupies
22.063 L Nitrogen Occupies
22.403 L Ammonia Occupies
22.094 L Hydrogen Chloride Occupies
22.248 L Carbon Monoxide Occupies
22.402 L Carbon Dioxide Occupies
22.262 L
10
Why should one mole of hydrogen occupy the same
volume as one mole of oxygen? Since oxygen
molecules are about 16 times as heavy as hydrogen
molecules, it would seem more reasonable for a
mole of oxygen molecules to occupy much more
volume than a mole of hydrogen molecules at the
same temperature and pressure.
11
Remember the Demo?

• The smaller balls make up in speed what they lack
  in size in carving out an area. This reasoning
  can be applied to the volume occupied by lighter
  gas molecules compared to the identical volume
  occupied by an equal number of heavier gas
  molecules at the same temperature and pressure.

12
Van der Waal Forces

• London Dispersion Forces - (Van der Waals
  Forces) the attractive forces between gas
  molecules.
• Even though the total electrical charge of the
  gas molecule is neutral, the positively charged
  nuclei of its atoms are not completely shielded
  by negatively charged electrons.

13
Van der Waals
In general, the magnitude of Van Der Waals
forces increases with an increase in the number
of electrons per atom and with an increase in
size of the molecule.
Few electrons lame wave
Many electrons more
powerful wave!!
14
STP conditions

• Standard temperature and pressure useful when
  wanting to compare or measure gas volumes
• ST 0 C 273K
• SP 760mm of Hg 1atm 760 Torr
• STP standard temperature and pressure

15
Boyles Law

• Boyles Law the volume of a definite quantity
  of dry gas is inversely proportional to the
  pressure, provided temperature remains constant.
  
• (V ? 1/P)
• Formulas or dimensional analysis?
• P1V1 P2V2

16
Boyles Law, Continues

• Examples
• Heimlich maneuver (increase P in lungs by
  decreasing their V)
• Syringe
• Straw (Lungs are acting as the chamber that
  increases in V)
• V?P? V?P?

17
Problem 1
Suppose a syringe plunger is pushed all the way
in so that the syringe contains 0.10 mL of air at
1.00 ATM. The plunger is now pulled back quickly
so that the total volume changes to 3.00 mL.
Calculate the momentary pressure before any
liquid comes into the syringe.
18
Solve using dimensional analysis
Initial P is 1.00 ATM so you have two possible
conversion factors A. 0.10 mL OR B.
3.00 mL ----------
---------- 3.00 mL 0.10
mL Which do we choose? Think If we are
increasing V we expect the gas P to decrease so
we want the conversion factor that is less than
one.
1.00 atm x 0.10 mL 0.033 atm
----------
3.00 mL
19
The plug-and-chug (no thinking involved) method
P1V1 P2V2 1.00 atm x 0.10 mL
P2 x 3.00 mL P2 0.033 atm
20
Problem 2
A small child carrying an inflated balloon with a
volume of 2.00 liters gets on an airplane in N.Y.
at a P of 760. Torr. What will the volume of his
balloon be on arrival at Mexico City, which is
7347 feet above sea level and has an average P of
600. Torr?
21
Answer to problem 2
2.00 L x 760. Torr
----------- 2.53 Liters
600. Torr
Solve, pc method 2 P1V1 P2V2
760. Torr x 2.00 Liter V2 x 600. Torr
V2 2.53 Liters
22
Charles Law

• The volume of a gas is directly proportional to
  the Kelvin Temperature (Pressure remains a
  constant).
• V ? T
• V1 V2
• T1 T2

23
Problem 1

• The initial volume of a piston is 1.5 Liters
  at 300.K. What will be the final volume at
  600. K? (P is constant)
• Think Since volume ? temperature, as V
  increases then T must increase. So, use the
  conversion factor that is greater than 1

24
Answer
1.5 Liter x 600.K 3.0L 300.K
Or solve way 2 V1/T1 V2 /
T2 1.5L / 300.K V2 /600.K V2
3.0L

• Problem 2 Suppose a balloon had a volume of 2
  x 105 liter when it was filled with hot air at a
  temp. of 500. C. What volume would it occupy if
  it were sealed and cooled at 27 C?

25
Answer to Problem 2
( 2 x 105 L ) x 300.K 8 x 104 L
773K
26
Gay-Lussacs Law

• The pressure inside a fixed volume is directly
  proportional to the Kelvin temperature.
• P ? T
• Example Spray can warning Do not
  incinerate!

P1 P2 T1 T2
27
Gay-Lussacs, continued

• The contents of a spray can are all used up and
  the can thus contains gas at a pressure of 1.00
  atm. The can is thrown into a fire where its T
  rises to 627 C. Why does the can explode? --
  Started at a temperature of 27 C.
• Think If T increases, the P must increase if
  there is no change in volume. So, solve for
  pressure .

28
Solving for this Law

• The pressure inside the can, right before it blew
  up, would have been
• Remember if T , P

1.00 atm x 900. K 3.00 atm
300. K
29
Problem 2

• Problem A pressure cooker is closed and is at
  20. C. To what temperature must the gas be
  heated in order to create an internal pressure
  of 2.00 atm?

30
Solving problem 2

• Solve P1 1.00 atm and it has to increase to
• 2.00 atm. So, P needs to increase
  and that means temperature must also increase.

293 K x 2.00 atm 1.00 atm
586 K
31
Combining the gas laws

• P1V1 P2V2
• T1 T2

Whoops, wrong combining.
32
Combination problem

• Suppose you have a balloon that has a volume of
  20.0 L at a pressure of 1.00 atm and a
  temperature of 32.0 C. If this balloon rose into
  the atmosphere where the temperature was only
  12.0 C and the pressure only 0.82 atm, what would
  its new volume be?

33
First convert the temperatures to Kelvin 12.0
273 285 K 32.0 273 305 K
Second either plug in values into your
combination formula or do the THINKING method
P so V T so V

• 20.0 L x 1.00 atm x 285 K 23 L
• 0.82 atm 305 K

34
Daltons Law of Partial Pressure

• the total pressure of the mixture of gases is the
  sum of their partial pressures
• PTOTAL PA PB PC .ETC

35
Problem 1

• A mixture of gases in a 3.00 L flask consists of
  Nitrogen at a partial pressure of 100 Torr and
  Oxygen at a PP of 300 Torr. What is the volume
  of each gas? Calculate the total pressure in the
  flask.
• Since the presence of another gas has no effect
  on the volume available to the first gas, both
  Nitrogen and Oxygen have a volume of 3.00L

36
Problem 2
Ans. PTOTAL 100 Torr 300 Torr 400 Torr

• Air has a relative constant proportion of N2
  (78.10), O2 (21.00), Ar (0.90), and CO2
  (0.03). The partial pressure of each gas in
  normal dry air at sea level is proportional to
  the number of molecules. What is the Total
  atmospheric pressure?

37
answer. PTOTAL 1.0003 atm
N2 0.7810 atm. O2 0.2100 atm.
Ar 0.0090 atm. CO2 0.0003
atm.
38
Grahams Law of Diffusion

• Diffusion rate depends on 3 factors
• Speed of the molecules (the higher the
  temperature, the higher the average KE)
• Diameter of the molecule (the larger the
  molecule, the slower the diffusion rate)
• Potential Attractive forces (the greater the
  potential attractive forces between the
  molecules, the slower the diffusion rate)

39
Grahams Law

• For an average molecule - A and an average
  molecule - B, use the following formula to
  determine velocity
• VA MB
• VB M A

40
Grahams Law Sample Problem

• At room temperature, an average hydrogen molecule
  travels at a speed of 1700. meters/second
  (about 3000 miles/hr). What is the speed of an
  average oxygen molecule under the same
  conditions?

41
Solving for Sample Problem 1

• 1700. m/sec 32.0 amu
• VB 2.02 amu
• So, VB 425 m/sec

42
Another awesome formula

• Problem A steam autoclave is used to sterilize
  hospital instruments. Suppose an autoclave with
  a volume of 15.0 liters contains pure steam
  (water) at a temperature of 121 C and a P of 1550
  Torr. How many grams of water does it contain?
• Solve PV nRT
• R 62.4 L Torr / Mole K

43
Solving using the Ideal Gas Law

• 1550 Torr x 15.0 L n x 62.4 L Torr x 394
  K Mol K
• n 0.946 moles of H2O
• 0946 moles of H2O x 18.0g H2O 17.0 g H2O
• 1 mole H2O
• NOTE You could solve using your knowledge of
  combination gas laws and molar volume (at STP)

44
Another Sample Problem

• Example 2 A hospital uses an oxygen gas
  cylinder containing 3.50 kg of O2 gas in a volume
  of 20.0 L and at a T of 24 C. What is the P in
  atm in the cylinder?

45
Problem Solved

• PV nRT
• n 3500 g O2 x 1 mole/32.0 g 109 moles
• P( 20.0 L) (109 moles)(0.0821 L x atm/mole x
  K)(297 K)
• P 133 atm

46
Eudiometer Tubes(mercury or water displacement)

• Eudiometer a tube used to collect a gas. It is
  closed at one end and is graduated.
• Reaction 2HCl (aq) Ca (s) ? H2 (g)
  CaCl2 (aq)

47
Eudiometer with mercurySituation 1

• If there is just enough gas to make the mercury
  level inside the tube the same as the level of
  the mercury in the bowl then the pressure of the
  hydrogen gas is the same as that of the
  atmosphere!

48
Eudiometer, Situation 2

• Suppose enough hydrogen gas is added to make the
  level inside the tube lower than the level of the
  mercury in the bowl. The P of the hydrogen gas
  is greater than the atmospheric pressure. (To
  determine the hydrogen gas pressure one must add
  the level difference to the barometric reading.)
• Problem situation 2 The volume of oxygen in a
  eudiometer is 37.0 mL. The mercury level inside
  the tube is 25.00mm lower than the outside. The
  barometric reading is 742.0 mm Hg.

49
Solving for Problem 2
Solution 2 742.0mm 25.00mm 767.0mm
50
Eudiometer with mercury, situation 3

• Suppose there was not enough gas to make the
  mercury level the same. Then, the pressure of
  hydrogen gas would not be the same as the
  pressure of the air outside the tube. (So, we
  must subtract the level difference from the
  barometric reading.)
• Problem situation 3 What is the P of the gas in
  an eudiometer, when the mercury level in the tube
  is 14mm higher, than that outside? That
  barometer reads 735mm Hg.

51
Answer to Problem 3

• Solution 735mm 14mm 721mm

52
Water in Place of Mercury

• Water is often used in place of mercury.
  Calculations are done the same way but the
  difference in water levels must first be divided
  by 13.6 to convert it to its equivalent height in
  terms of a column of mercury since water is
  1/13.6 as dense as mercury.

53
New Problem If you collected some oxygen gas in
the eudiometer by the method known as water
displacement you must

• Correct for the difference in density between
  water and mercury!!!
• Also, water evaporates much more readily than
  mercury and so is in with the collected gas. So,
  you need to determine the partial pressure of the
  dry gas (unmixed with water vapor). The vapor
  pressure of water, at the given temperature, must
  be subtracted from the total pressure of the gas
  within the tube.

54

• Problem 1 Oxygen is collected using the water
  displacement method. The water level inside the
  tube is 27.2mm higher than that outside. The
  temp. is 25.0 C. The barometric pressure is
  741.0mm. What is the partial pressure of the dry
  oxygen in the eudiometer?

55
Step 1 27.2mm/13.6 2.00mm
Step 2 741.0 mm - 2.00mm 739.0 mm
So 739.0 mm P total Poxygen Pwater Step
3 To correct for water vapor pressure, you need
to know the pressure of water vapor at 25.0C and
it is 23.8mm (always given - see handout).
Subtract this water partial pressure from the
pressure total found in step 2. 739.0mm -
23.8mm 715.2 mm partial pressure
of the dry oxygen.
56

• Problem 2 A eudiometer contains 38.4 mL of air
  collected by water displacement at a temperature
  of 20.0C. The water level inside the eudiometer
  is 140. mm higher than that outside. The
  barometric reading is 740.0mm. Water vapor
  pressure at 20.0C is 17.5mm. Calculate the
  volume of dry air at STP.

57

• First Step 140mm/13.6 10.3 mm

• Second Step 740.0mm - 10.3mm 729.7mm

• Third Step 729.7mm - 17.5mm 712.2 mm
• ? pressure at a Temp. of 293K
• Now it asks you to go a step further and
  calculate what the volume would be at standard
  temperature and pressure which 760. mm and
  273K. To solve this problem you need to combine
  two gas laws Charles and Boyles. Think of your
  possible conversion factors and solve.
• 38.4mL x ________ x __________

58
Combination finish

• 38.4mL x 273 K x 712.2mm 33.5 mL
• 293 K 760. mm

59
More chemistry problems involving gases

• 1. Gas volume gas volume
• 2. Mass-gas volume
• Avagadros principle Equal volumes of all gases,
  under the same conditions of T and P contain the
  same number of molecules. Further, the number of
  molecules (of all gases) are in the same ratio as
  their respective gas volumes. (Refer back to
  molar volume definition)

60
Gay Lussacs law

• Under the same conditions of T and P, the volumes
  of reacting gases and their gaseous products are
  expressed in ratios of small whole numbers.
• 2H2 (g) O2 (g) --gt 2H2O(g)
• 2 volumes of H2 1 vol. of O2 ?
• 2 vol. of H2O vapor.

61
Sample problem 1

• What is the volume of oxygen that would combine
  with 4.0 liters of hydrogen to form water vapor?
• Equation

2H2 (g) O2(g) ? 2 H2O
Answer 4.0 L H2 x 1 vol. O2 2.0 L of
O2 2 vol. H2
62
Sample Problem 2

• What is the volume of hydrogen chloride gas that
  would be produced, assuming
• 5.0 liters of chlorine gas fully reacted with
  hydrogen gas?

Equation H2 (g) Cl2 (g) --gt 2HCl(g)
Answer 5.0 L Cl2 x 2 vol. HCl
10.0 L of HCl 1
vol. Cl2
63
Remember Molar Volume?

• How many grams of CaCO3 must be decomposed to
  produce 4.00 liters of CO2 at STP? (Calcium
  oxide is also formed)
• CaCO3(s) ? CaO (s) CO2 (g)
• 1 mole--------------------1mole (stoich. ratio)
• 1 mole CaCO3 100.1g
• 1 mole CO2 at STP 22.4 liters (remember?)

64
Molar volume revisited(or ideal?)

• 4.00 L x 1 mole/22.4 L 0.179 moles CO2
• Since CO2 and CaCO3 are on a 1/1 mole ratio, you
  will need 0.179 moles of CaCO3
• 0.179 moles CaCO3 x 100.1 g CaCO3 17.9 g
• 1 mole CaCO3