1. Gas Laws
Scheffler 1

2. Gases Variable volume and shape Expand to occupy volume available
Volume, Pressure, Temperature, and the number of moles present are interrelated
Can be easily compressed
Exert pressure on whatever surrounds them
Easily diffuse into one another
Scheffler 2

3. Mercury Barometer Used to define and measure atmospheric pressure
On the average at sea level the column of mercury rises to a height of about 760 mm.
This quantity is equal to 1 atmosphere
It is also known as standard atmospheric pressure
Scheffler 3

4. Pressure Units & Conversions
The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal.
The US Weather Bureaus commonly report atmospheric pressures in inches of mercury.
Pounds per square inch or PSI is widely used in the United States.
Most other countries use only the metric system.
Scheffler 4

5. Boyle’s Law
According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant pressure.
PV = constant.
P1V1 = P2V2
Scheffler 5

6. Boyle’s Law A graph of pressure and volume gives an inverse function
A graph of pressure and the reciprocal of volume gives a straight line
Scheffler 6

7. Sample Problem 1:
If the pressure of helium gas in a balloon has a volume of 4.00 dm3 at 210 kPa, what will the pressure be at 2.50 dm3?
P1 V1 = P2 V2
(210 kPa) (4.00 dm3) = P2(2.50 dm3)
P2 = (210 kPa) (4.00 dm3)
(2.50 dm3)
= 340 kPa
Scheffler 7

8. Charles’ Law
According to Charles’ Law the volume of a gas is proportional to the Kelvin temperature as long as the pressure is constant
V = kT
Note: The temperature for gas laws must always be expressed
in Kelvin where Kelvin = oC (or 273 to 3 significant digits) V1 = T1 V2 T2
Scheffler 8

9. Charles’ Law A graph of temperature and volume yields a straight line.
Where this line crosses the x axis (x intercept) is defined as absolute zero
Scheffler 9

10. Sample Problem 2
A gas sample at 40 oC occupies a volume of 2.32 dm3. If the temperature is increased to 75 oC, what will be the final volume?
V1 = V2
T T2
Convert temperatures to Kelvin. 40oC = 313K
75oC = 348K
2.32 dm3 = V2
313 K K
( V2) = (2.32 dm3)(348K)
(313K)
V2 =
2.58 dm3
Scheffler 10

11. Gay-Lussac’s Law P1 = P2 T2 T1
Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas.
The pressure and temperature of a gas are directly proportional
P = P2 T2 T1
Scheffler 11

12. Sample Problem 3:
The pressure of a gas in a tank is 3.20 atm at 22 oC. If the temperature rises to 60oC, what will be the pressure in the tank?
P1 = P2
T T2
Convert temperatures to Kelvin. 22oC = 295K
60oC = 333K
3.20 atm = P2
295 K K
( P2) = (3.20 atm)(333K)
(295K)
P2 =
3.6 atm
Scheffler 12

13. The Combined Gas Law
1. If the amount of the gas is constant, then Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship
P1 V = P2 V2 T1 T2
Scheffler 13

14. Sample Problem 4:
A gas at 110 kPa and 30 oC fills a container at 2.0 dm3. If the temperature rises to 80oC and the pressure increases to 440 kPa, what is the new volume?
P1V1 = P2V2
T T2
Convert temperatures to Kelvin. 30oC = 303K
80oC = 353K
V2 = V1 P1 T2
P2 T1
= (2.0 dm3) (110 kPa ) (353K)
(440 kPa ) (303 K)
V2 = 0.58 dm3
Scheffler 14

15. Advogadro’s Law
Equal volumes of a gas under the same temperature and pressure contain the same number of particles.
If the temperature and pressure are constant the volume of a gas is proportional to the number of moles of gas present
V = constant * n
where n is the number of moles of gas
V/n = constant
V1/n1 = constant = V2 /n2
V1/n1 = V2 /n2
Scheffler 15

16. Universal Gas Equation
Based on the previous laws there are four factors that define the quantity of gas: Volume, Pressure, Kevin Temperature, and the number of moles of gas present (n).
Putting these all together: PV nT
= Constant = R
The proportionality constant R is known as the
universal gas constant
Scheffler 16

17. Universal Gas Equation
The Universal gas equation is usually written as
PV = nRT
Where P = pressure
V = volume
T = Kelvin Temperature
n = number of moles
The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include:
R = dm3 torr mol-1 K-1
= dm3 atm mol-1 K-1
= dm3kPa mol-1 K-1
Scheffler 17

18. Standard Temperature and Pressure (STP)
The volume of a gas varies with temperature and pressure. Therefore it is helpful to have a convenient reference point at which to compare gases.
For this purpose standard temperature and pressure are defined as:
Temperature = 0oC K
Pressure = 1 atmosphere
= torr
= kPa
This point is often called STP
Scheffler 18

19. Sample Problem 5 Example: What volume will 25.0 g O2 occupy
at 20oC and a pressure of atmospheres? :
(25.0 g)
n = = mol
(32.0 g mol-1)
Data
Formula
Calculation
Answer
V =? P = atm; T = ( )K = 293K
R = dm-3 atm mol-1 K-1
PV = nRT so V = nRT/P
V = (0.781 mol)( dm-3 atm mol-1 K-1)(293K)
0.880 atm
V = 21.3 dm3
Scheffler 19

20. d is the density of the gas in g/L
Universal Gas Equation –Alternate Forms
Density (d) Calculations = PM RT
m is the mass of the gas in g m V
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
d is the density of the gas in g/L
M =
Scheffler 20

21. Sample Problem 6
A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas?
Scheffler 21

22. Sample Problem 6 Solution
A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0oC. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 dm3 =
= 2.21 g
dm3
2.21 g
dm3
1 atm
x x K
dm3•atm
mol•K
M =
M =
54.6 g/mol
Scheffler 22

23. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures).
PT = P1 + P2 + P3 + P
where PT = total pressure
P1 = partial pressure of gas 1
P2 = partial pressure of gas 2
P3 = partial pressure of gas 3
P4 = partial pressure of gas 4
Scheffler 23

24. Dalton’s Law of Partial Pressures
Applies to a mixture of gases
Very useful correction when collecting gases over water since they inevitably contain some water vapor.
Scheffler 24

25. Sample Problem 7
Henrietta Minkelspurg generates Hydrogen gas and collected it over water.
If the volume of the gas is 250 cm3 and the barometric pressure is torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP?
(PH2O = 23.8 torr at 25oC)
Scheffler 25

26. Sample Problem 8 -- Solution
Henrietta Minkelspurg generates Hydrogen gas and collected it over water.
If the volume of the gas is 250 cm3 and the barometric pressure is torr at 25oC, what is the pressure of the “dry” hydrogen gas at STP?
(PH2O = 23.8 torr at 25oC)
Scheffler 26

27. Sample Problem 9
Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP?
Scheffler 27

28. Sample Problem 9 -- Solution
Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm3 and the barometric pressure is 765 torr at 25oC, what is the volume of the “dry” oxygen gas at STP?
From the previous calculation the adjusted pressure is torr
P1= PH2 = torr; P2= Std Pressure = 760 torr
V1= 250 cm3; T1= 298K; T2= 273K; V2= ?
(V1P1/T1) = (V2P2/T2) therefore V2= (V1P1T2)/(T1P2)
V2 = (250 cm3)(742.2 torr)(273K)
(298K)(760.torr)
V2 = cm3
Scheffler 28

29. Kinetic Molecular Theory
Matter consists of particles (atoms or molecules) that are in continuous, random, rapid motion
The Volume occupied by the particles has a negligibly small effect on their behavior
Collisions between particles are elastic
Attractive forces between particles have a negligible effect on their behavior
Gases have no fixed volume or shape, but take the volume and shape of the container
The average kinetic energy of the particles is proportional to their Kelvin temperature
Scheffler 29

30. Maxwell-Boltzman Distribution
Molecules are in constant motion
Not all particles have the same energy
The average kinetic energy is related to the temperature
An increase in temperature spreads out the distribution and the mean speed is shifted upward
Scheffler 30

31.  Velocity of a Gas 3RT urms = M The distribution of speeds
of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
urms =
3RT M 
Scheffler 31

32. Diffusion
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17.0 g/mol
HCl
36.5 g/mol
Scheffler 32

33. DIFFUSION AND EFFUSION
Diffusion is the gradual mixing of molecules of different gases.
Effusion is the movement of molecules through a small hole into an empty container.
Scheffler 33

34. Graham’s Law
Graham’s law governs effusion and diffusion of gas molecules.
KE=1/2 mv2
The rate of effusion is inversely proportional to its molar mass.
Thomas Graham, Professor in Glasgow and London.
Scheffler 34

35. 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Sample Problem 10
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850oC according to the equation:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
Scheffler 35

36. Sample Problem 10 Solution
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850oC according to the equation:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
Scheffler 36

37. Sample Problem 11
What is the rate of effusion for H2 if cm3 of CO2 takes 4.55 sec to effuse out of a container?
Scheffler 37

38. Sample Problem 11 Solution
What is the rate of effusion for H2 if cm3 of CO2 takes 4.55 sec to effuse out of a container?
Rate for CO2 = cm3/4.55 s = cm3/s
Scheffler 38

39. Sample Problem 12
What is the molar mass of gas X if it effuses times as rapidly as N2(g)?
Scheffler 39

40. Sample Problem 12 Solution
What is the molar mass of gas X if it effuses times as rapidly as N2(g)?
Scheffler 40

41. Ideal Gases v Real Gases
Ideal gases are gases that obey the Kinetic Molecular Theory perfectly.
The gas laws apply to ideal gases, but in reality there is no perfectly ideal gas.
Under normal conditions of temperature and pressure many real gases approximate ideal gases.
Under more extreme conditions more polar gases show deviations from ideal behavior.
Scheffler 41

42. In an Ideal Gas ---
The particles (atoms or molecules) in continuous, random, rapid motion.
The particles collide with no loss of momentum
The volume occupied by the particles is essentially zero when compared to the volume of the container
The particles are neither attracted to each other nor repelled
The average kinetic energy of the particles is proportional to their Kelvin temperature
At normal temperatures and pressures gases closely approximate idea behavior
Scheffler 42

43. Real Gases
For ideal gases the product of pressure and volume is constant. Real gases deviate somewhat as shown by the graph pressure vs. the ratio of observed volume to ideal volume below.
These deviations occur because
Real gases do not actually have zero volume
Polar gas particles do attract if compressed
Scheffler 43

44. van der Waals Equation (P + n2a/V2)(V - nb) = nRT
The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior
(P + n2a/V2)(V - nb) = nRT
where a => attractive forces between molecules
b => residual volume or molecules
The van der Waals constants for some elements are shown below
Substance
a (dm6atm mol-2)
b (dm3 mol-1) He
0.0341
CH4
2.25
0.0428
H2O
5.46
0.0305
CO2
3.59
0.0437
Scheffler 44

45. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Sample Problem 13
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
Scheffler 45

46. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Sample Problem 13 Solution
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
dm3•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 dm3
Scheffler 46