1. The Gas Laws

2. An Overview of Gases
Like solids and liquids, gases have a temperature, volume, and an amount that can be measured in moles.
Unlike solids and liquids, gases have a pressure that will affect the nature of the gas.
Gases have many units: atmospheres, mm Hg (torr), kPa, Pa, lbs/in2, bar, mbar, and more.

3. Pressure Pressure = Force per unit area.
Gas molecules will always fill any container they are in.
Molecules move around and hit the insides of the container. These collisions cause the pressure.
Measured with a barometer or a manometer.

4. Barometer
Vacuum
The pressure of the atmosphere at sea level will raise a column of mercury 760 mm Hg.
1 atm = 760 mm Hg
760 mm Hg
1 atm of Air Pressure

5. Manometer Column of mercury to measure pressure.
‘h’ is how much lower the pressure is than outside. h
Gas

6. Manometer
h is how much higher the gas pressure is than the atmosphere.
The Hg is pushed by the gas with the greater pressure. h
Gas

7. The difference in the height of mercury on both sides is the difference in pressure between the two gases.
The gas with the higher pressure will be able to move the mercury more than the gas with the lower pressure.
Remember our “Law of Phyics” The stronger force always wins !!
Pressure Readings

8. Units of pressure
These values are Standard Pressure and need to be known.
1 atmosphere = 760 mm Hg = 760 torr
1 atm = 101,325 Pascals = kPa
They also need to be able to be converted from one unit to another.

9. Converting Pressure Units
Convert 3.5 atm to ___ kPa
Like any conversion, set up a proportion.
The known values are on the left.
1 atm / kPa = 3.5 atm / x kPa
x = kPa

10. The Gas Laws
Our first look at Gas Laws will be a group of laws that focus on simple relationships between two variables.
When these two variables are studied, it implies that the other variables are constant. So Boyle’s Law of P1V1 = P2V2 means that while Pressure and Volume are being examined, moles and temperature remain constant.

11. The Gas Laws Boyle’s Law
Pressure and volume are inversely related at constant temperature. OR
As one goes up, the other goes down with the same factor.
PV= k was Boyle’s original work and it turned into P1V1 = P2 V2

12. Boyle’s Law
The following two graphs show the data when Pressure v. Volume are plotted.
In the first graph, it shows Pressure graphed v. Volume.
The second graph shows the straight line plot of Volume v. 1 / Pressure

13. Pressure v. Volume V
P (at constant T)

14. Pressure v. 1 / Volume
Slope = k V
1/P (at constant T)

15. Looking at those graphs
The second graph uses the reciprocal and makes a straight line.
Straight lines allow us to calculate a slope easily and form relationships and enables us to calculate much more easily.

16. Examples
20.5 L of nitrogen at 742 torr are compressed to 9.8 atm. What is the new volume ?
Pressure is changing and a new volume is needed = Boyle’s Law
Make sure the units match up as torr does not equal atm.
(20.5 L) (742 torr) = (7448 torr) V2
V2 = L

17. More examples
30.6 mL of carbon dioxide at 740 torr is expanded to 750 mL What is the final pressure in kPa ?
(30.6 mL) (740 torr) = (750 mL) P2
P2 = torr (needs to be in kPa)
30.19 torr = kPa

18. Charles’ Law
Before we explore Charles’ Law, lets work a simple problem, given the Charles’ Law formula of
V1 = V T T2
If V1 = 10 L and T1 = -10 what is the new volume when the temperature is raised to 30 ?

19. Charles’ Law
We see we have a problem as we can’t have a negative volume.
For quite some time, scientists knew that there was a positive relationship between the temperature of a gas and the volume it would occupy.
The problem was that they couldn’t determine how to mathematically and accurately express this relationship.

20. Charles’ Law
A scientist named Lord Kelvin was working on the ideas of temperature and determined that the lowest possible temperature that an object can reach is ˚C.
Charles took this idea and did the rest.

21. Charles’ Law
Volume of a gas varies directly with the absolute temperature at constant pressure.
V = kT (if T is in Kelvin)
V1 = V T T2
Graphically it looks like this.

22. Charles’ Law Graph He
H2O
CH4
V (L) H2 ºC
T (ºC)

23. Charles’ Law
Given the following formula
V1 = V2
T T2
If V1 = 10 L and T1 = -10 what is the new volume when the temperature is raised to 30 ?

24. Examples
What would the final volume be if 247 mL of gas at 22ºC is heated to 98 ºC ?
247 / 295 = V2 / V2 = mL
Remember that all Gas Law temperatures must be in Kelvin

25. Avogadro's Law
At constant temperature and pressure, the volume of gas is directly related to the number of moles. When air is added to a ball to pump it up, it is putting more moles of air in it !!
V = k n (n is the number of moles)
V1 = V n1 = n2

26. Gay- Lussac Law
At constant volume, pressure and absolute temperature are directly related.
P = k T
P1 = P T T2

27. Combined Gas Law
If the moles of gas remains constant, use this formula and cancel out the other things that don’t change.
P1 V1 = P2 V2
. T T2

28. Examples
A cylinder has a volume of 175 mL and a pressure of 3.8 atm at 22 ºC. What would the pressure be if the cylinder’s volume was increased to 250 mL as it was heated to 100 ºC ?
(175)(3.8) / 295 = V2 (250) / 373
V2 = 3.36 atm.

29. Molar Volume Standard Pressure = 1 atmosphere
Standard Temperature = 273 K
Together they from STP conditions
1 mole of any gas at STP occupies 22.4 Liters of Volume

30. What have we seen so far ?
We have looked at several gas law formulas so far. Have you noticed something peculiar about them so far ?
Is there anything about them that might make it difficult to work with from a practical point of view.

31. What have we seen so far ?
What has to be happening in our situations to be able to solve ?
So far, we are only working on problems where a change in conditions are taking place.

32. Ideal Gas Law PV = nRT R is the ideal gas law constant.
R = ? It depends on the pressure unit.
L atm / mol K OR
L torr / mol K OR
J / mol K
The other laws tell you about a gas when it changes.

33. Examples
1. A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temp. ?
2. Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8 ºC. What is the mass of Kr ?

34. Solved Answers
1. PV = nRT (1.85)(47.3) = 1.62 R T T = K or °C 2. PV = nRT (8.61) (18.5) = n (.08205)(297.8) n = 6.52 moles = grams

35. Gas Density and Molar Mass
Sometimes we have to combine our gas laws with formulas that we learned earlier, such as :
Density = m / V OR moles = g / M
PV = nRT substitute for n.
M = g RT / PV Another algebra manipulation is
P x M = gRT/ V = P x M = DRT

36. Gas Density and Molar Volume
It is strongly recommended that you don’t memorize the two derived formulas from the last slide.
You will be much better off by learning the Ideal Gas Law and the formulas for Density and n = g / M which you should already know very well.
These are also given to you on tests.

37. Density Example
What is the density of ammonia (NH3) at 23 ºC and 735 torr ?
For this we need to remember that Density is an intensive property.
This means that it doesn’t matter what the amount of ammonia we use.
So we might as well use a convenient amount.

38. Solved Answer Assume 1 mole (17 grams) and PV = nRT to get volume.
What is the density of ammonia (NH3) at 23 ºC and 735 torr ?
Assume 1 mole (17 grams) and PV = nRT to get volume.
PV = nRT  (735)V = 1 R (295)
Volume = Liters
Density = 17 / = .677 g / L

39. Gases and Stoichiometry
Reactions happen in moles.
At Standard Temperature and Pressure (STP, 0ºC and 1 atm.) 1 mole of gas occupies L.
Recall Molar Volume of a gas.
If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

40. Examples
Mercury can be achieved by the following reaction:
2 HgO  2 Hg + O2
What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP ?
What volume at 400 ºC and 740 torr ?

41. Answers First we need to find moles of HgO. n = g/M .0189 moles
Use a mole ratio from the balanced eq.
2 HgO / 1 O2 = / x x = .0095
Proportion: 1 n / 22.4 L = / x
Use molar volume (1 n  22.4 Liters)
.212 Liters of Oxygen Gas
For the second problem: L of gas

42. Examples Using the following reaction:
NaHCO3 + HCl  NaCl + CO2 (g) + H2O (l)
Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25 ºC and 2.00 atm.
If 27 L of gas are produced at 26 ºC and 745 torr when 2.6 L of HCl are added what is the concentration of HCl ?

43. Answers PV = nRT 2 (2.87) = n R (298) n = .235 n 1:1 ratio SO
.235 moles of NaHCO3
Converting to grams – grams
PV = nRT (.98) (27) = n R (299)
n = n 1:1 ratio SO
M = n / L  / M = .415 M

44. Examples Consider the following reaction:
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
What volume of H2O at 1.0 atm and 1000 º C can be produced from 10.0 L of NH3 and excess O2 at the same temperature and pressure ?
What volume of O2 measured at STP will be consumed when 10.0 kg NH3 is reacted ?

45. Answers PV = nRT 1 (10) = n R (1273) n = .0957 n of NH3
4 NH3 / 6 H2O = / x x = .144 n H2O
Use PV = nRT Liters
10 kg = moles of NH3
Use molar volume since it is at STP
1 / 22.4 = / x x = n O2
Convert to Liters n x 22.4 L  L of O2

46. The Same Reaction 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
What mass of H2O will be produced from 65.0 L of O2 and 75.0 L of NH3 both measured at STP ?
What volume of NO would be produced ?
What mass of NO is produced from 500. L of NH3 at ºC and 3.00 atm ?

47. Dalton’s Law
The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container.
The total pressure is the sum of the partial pressures.
PTotal = P1 + P2 + P3 + P4 + P5 ...
Make sure that the units match up.

48. Dalton’s Law Problem
In a flask filled with a combination of Nitrogen, Oxygen, and Hydrogen, it was found to have a total pressure of kPa. If the oxygen is determined to have a partial pressure of 2.1 atm, and the hydrogen is found to have a partial pressure of 138 torr, what is the partial pressure of nitrogen measured in kPa.

49. Solved Answer We need to get all the gases in the same units.
In a flask filled with a combination of Nitrogen, Oxygen, and Hydrogen, it was found to have a total pressure of kPa. If the oxygen is determined to have a partial pressure of 2.1 atm, and the hydrogen is found to have a partial pressure of 138 torr, what is the partial pressure of nitrogen measured in kPa.
We need to get all the gases in the same units.
2.1 atm = kPa
138 torr = kPa
Now we have: = N2
Solving for N2 = kPa

50. Mole Fraction
Some calculations are done with the concentration term of Mole Fraction.
Similar but not identical to % composition.
DO NOT multiply by 100 %
Ratio of moles of the substance to the total moles.

51. Mole Fraction Symbol is Greek letter chi c
It is the same as % except you don’t multiply by 100 %
c1 = n1 = P nTotal PTotal

52. Examples
The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen ?
What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm ?

53. Solved Answers 592 / 752 = .787 (notice there are no units)
The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen ?
592 / 752 = (notice there are no units)
What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm ?
.787 x 5.25 = atm.

54. Dalton’s Law of Partial Pressure
CH4
1.50 L N2
3.50 L O2
0.752 atm
2.70 atm
4.58 atm
When these valves are opened, what is each partial pressure and the total pressure ?

55. Dalton’s Law of Partial Pressure
When these valves are opened, what is each partial pressure and the total pressure ?
The total volume of the flasks is 9 L.
To solve for the partial pressure of each gas, we can use Boyle’s Law.
P1V1 = P2V2

56. Dalton’s Law of Partial Pressure
For CH4 : atm ( 4 L ) = x atm ( 9 L )
Pressure is 1.2 atm.
Doing similar work for N2 and O2 gives
PNitrogen = .763 atm and POxygen = atm.
According to Dalton’s Law:
PTotal = 1.2 atm atm atm.
= atm.

57. Vapor Pressure Water evaporates at any temperature !
Even ice will evaporate
Hot water evaporates more than cold water (of course)
Even though the material started as a liquid, once it reaches the gas phase, it acts just like a gas. In this instance, that means it has a pressure.

58. Application of Dalton’s Law
When that water evaporates, the vapor has a pressure.
Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure.
Check out the visual on the next slide to see how gases are often isolated in a lab setting.

59. Collecting a Gas Sample

60. Example N2O can be produced by the following reaction :
NH4NO3 (s)  N2O (g) + 2 H2O (l)
What volume of N2O collected over water at a total pressure of 94 kPa and 22 ºC can be produced from 2.6 g of NH4NO3? ( the vapor pressure of water at 22 ºC is 21 torr )

61. Dalton’s Law in a Lab: Solved
N2O can be produced by the following reaction :
NH4NO3 (s)  N2O (g) + 2 H2O (l)
What volume of N2O collected over water at a total pressure of 94 kPa and 22 ºC can be produced from 2.6 g of NH4NO3? ( the Pwater at 22 ºC is 21 torr )
We must first determine the pressure of just the N2O alone. We use Dalton’s Law
94 kPa = P kPa (21 torr  2.8 kPa)
P1 = 91.2 kPa
Using PV = nRT  Liters

62. Kinetic Molecular Theory (KMT)
Explains why ideal gases behave the way they do.
Assumptions that simplify the theory, but don’t work in real gases, especially under specific conditions.
These conditions will be looked at later in the chapter.

63. Kinetic Molecular Theory
The particles are so small we can ignore their volume.
- This implies that a molecule of gas has the opportunity to go anywhere in the container.
- However, it is limited to places where a molecule is not found.
The particles are in constant motion and their collisions with the inside of the container causes pressure.

64. Kinetic Molecular Theory
The particles do not affect each other, neither attracting or repelling.
- This is called an elastic collision.
- No energy is lost.
- Imagine throwing a tennis ball at a
wall – it bounces back.
- No imagine throwing a lump of clay against a wall. This is called an inelastic collision.

65. Kinetic Molecular Theory
The average kinetic energy is proportional to the Kelvin temperature.
- So we have a new concept of temperature and it shows that temperature is determined by the movement of the molecules in the sample
We need the formula KE = 1/2 mv2

66. What it tells us (KE)avg = 3/2 RT R is in Joules !
This means use J / n K
This is the meaning of temperature.
u is the particle velocity.
u is the average particle velocity.
u 2 is the average particle velocity squared.
the root mean square velocity is
Ö u = urms

67. Combine these two equations
(KE)avg = NA(1/2 mu 2 )
(KE)avg = 3/2 RT
M is the formula weight in kg / mole, and R has the units J / K mol.
The velocity (u) will be in m/s

68. Example
Calculate the root mean square velocity of carbon dioxide at 25 ºC.
Calculate the root mean square velocity of hydrogen at 25 ºC.
Calculate the root mean square velocity of chlorine; Cl2 at 25 ºC.

69. Answers For CO2 the root mean square speed is 411 m / s
For H2 the root mean square speed is 1,927.8 m / s
For Cl2 the root mean square speed is m / s

70. Range of velocities
The average distance a molecule travels before colliding with another particle is called the mean free path and is small (near 10-7)
Temperature is an average. There are molecules of many speeds in the average.
Shown on a graph called a Boltzmann distribution

71. 273 K
number of particles
Molecular Velocity

72. 273 K
320 K
number of particles
Molecular Velocity

73. 273 K
320 K
number of particles
350 K
Molecular Velocity

74. Velocity Average increases as temperature increases.
The difference in temperature between the lowest and the highest temperature increases as the temperature increases.

75. Diffusion Diffusion is the passage of gas through a small hole.
The Diffusion rate measures how fast this happens.
Graham’s Law of Diffusion is inversely proportional to the square root of the mass of its particles.

76. Graham’s Law of Diffusion
Where the rate is measured in m / s
M1 and M2 is the formula weight in kg.
Sometimes the rate of change between the two rates is being asked for.

77. Graham’s Law The rate of Diffusion should be proportional to urms
Effusion Rate 1 = urms 1 Effusion Rate 2 = urms 2

78. Diffusion The spreading of a gas through a room.
Slow considering molecules move at 100’s of meters per second.
Collisions with other molecules slow down diffusions.
Best estimate is Graham’s Law.

79. Examples
A compound effuses through a porous cylinder 3.20 times slower than helium. What is the formula weight of the compound ?
If mol of NH3 effuse through a hole in 2.47 min, how much HCl would effuse in the same time ?
A sample of N2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions ?

80. Answer to Problem # 1
A compound effuses through a porous cylinder 3.20 times slower than helium. What is the formula weight of the compound ?
Let’s examine this problem: The rate of the individual gases is not given, so we can write the equation like this:
3.2 = M21/2 / 41/2  = M21/2 / 2
M21/2 = 6.4 M2 = g / n

81. Answer to Problem # 2
If mol of NH3 effuse through a hole in 2.47 min, how much HCl would effuse in the same time ?
This problem is describing a situation where two gases are ‘racing’ past a defined point.
The moles are a direct reflection of the rate of the velocity of the gases.
/ x = (36.45 / 17)1/2
/ x = x = n

82. Answer to Problem # 3
A sample of N2 effuses through a hole in 38 seconds. what must be the molecular weight of a gas that effuses in 55 seconds under identical conditions ?
The situation with this one is that while time is a reflection of the rate, it is not a direct reflection: The faster gas has a lower time !
So we need to take that into account.
While time is a reflection of rate, if we were simply using the time, we would have the faster rate with the slower value. Time traveled and rate are inversely proportional.

83. Answer to Problem # 3
Since they are inverse, our formula would look like this:
( 1 / 38 ) / ( 1 / 55 ) = ( x / 28 ) ½
55 / 38 = ( x / 28 ) ½
x = g / n
This agrees with the idea that the heavier gas moves slower.

84. Real Gases
Real molecules do take up space and they do interact with each other (especially polar molecules).
Recall the KMT
Need to add correction factors to the ideal gas law to account for these.
Also called the van der Waals Equation.

85. Volume Correction
The actual volume free to move in is less because of particle size.
More molecules will have more effect.
Corrected volume V’ = V - nb
b is a constant that differs for each gas.
P’ = nRT ( V-nb )

86. ( ) Pressure correction n Pobserved = P - a V 2
Because the molecules are attracted to each other, the pressure on the container will be less than ideal.
Depends on the number of molecules per liter.
Since two molecules interact, the effect must be squared. ( ) 2 V n
Pobserved
= P - a

87. ( ) Summary Pobs= nRT - a n 2 V-nb V
Called the Van der Waal’s equation if rearranged
Corrected Corrected Pressure Volume

88. Where does it come from
‘a’ and ‘b’ are determined by experiment and are different for each gas.
Bigger molecules have larger ‘b’.
‘a’ depends on both size and polarity.
The van der Waals equation is preferred when the temperature is low and/or the pressure is high.
The College Board will not have you solve these, but you should be familiar with the theory.

89. Example
Calculate the pressure exerted by mol Cl2 in a L container at 25.0 ºC
Using the ideal gas law.
Van der Waal’s equation
a = 6.49 atm L2 /mol2
b = L/mol