1. Chapter 5: the Gaseous state
Vanessa Prasad-Permaul
Valencia Community College
CHM 1045

2. GAS LAWS: PRESSURE AND MEASUREMENT
Elements that exist as gases at 250C and 1 atmosphere

3. GAS LAWS: PRESSURE AND MEASUREMENT
Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.

4. GAS LAWS: PRESSURE AND MEASUREMENT
a) Gas is a large collection of particles moving at random throughout a volume
b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure

5. GAS LAWS: PRESSURE AND MEASUREMENT
HOW IS PRESSURE DEFINED?
The force the gas exerts on a given area of the container in which it is contained.
The SI unit for pressure is the Pascal, Pa.
Pressure =
Force
Area
If you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).

6. GAS LAWS: PRESSURE AND MEASUREMENT
Barometer
760mm
mercury
Height is proportional
to the barometric pressure
Units of Pressure
1 pascal (Pa) = 1 kg/m·s2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
1 bar = 105 Pa
1 atm = lb/in2
Hg is used instead of H2O :
more dense
better visibility
accuracy

7. GAS LAWS: PRESSURE AND MEASUREMENT
EXERCISE 5.1
A GAS CONTAINER HAD A MEASURED PRESSURE OF
57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM
AND mmHg.
First, convert to atm (57 kPa = 57 x 103 Pa).
57 x 103 Pa x atm = = atm
x 105 Pa
Next, convert to mmHg.
57 x 103 Pa x mmHg = = 4.3 x 102 mmHg

8. GAS LAWS: EMPIRICAL GAS LAWS
You can predict the behavior of gases based on the following properties:
Pressure
Volume
Amount (moles)
Temperature
* If two of these physical properties are held constant,
it is possible to show a simple relationship between the other two…

9. GAS LAWS: EMPIRICAL GAS LAWS
Boyle’s experiment:
A manometer to study the relationship between
pressure (P) & Volume (V) of a gas Hg Hg
As P (h) increases
V decreases

10. GAS LAWS: EMPIRICAL GAS LAWS
BOYLE’S LAW
the volume of a sample of gas at a given temperature
varies inversely with the applied pressure
So….
For a given amount of gas (n)
&
@ constant temperature (T) :
If pressure (P) increases, the volume (V) of the gas decreases
P1 * V1 = P2 * V2

11. Pressure–Volume Law (Boyle’s Law):

12. GAS LAWS: EMPIRICAL GAS LAWS
Boyle’s Law
P a 1/V
P * V = constant
P1 * V1 = P2 * V2
Constant temperature
Any given amount of gas

13. Application of Boyle’s law gives
GAS LAWS: EMPIRICAL GAS LAWS
EXERCISE 5.2
A volume of carbon dioxide gas equivalent to 20.0 L was 23oC and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm?
P1 * V1 = P2 * V2
Application of Boyle’s law gives
V2 = V1 x P1 = L x 1.00atm = = L
P atm

14. Charles’ Law: relating volume and temperature
GAS LAWS: EMPIRICAL GAS LAWS
Charles’ Law: relating volume and temperature
As T increases
V increases

15. Temperature–Volume Law (Charles’ Law):

16. GAS LAWS: EMPIRICAL GAS LAWS
Charles’ Law
Variation of gas volume with temperature
V a T
For a given amount of gas
at constant pressure
Temperature must
be in Kelvin
V / T = constant
T (K) = t (0C)

17. GAS LAWS: EMPIRICAL GAS LAWS
CHARLES’ LAW
the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature
So….
For a given amount of gas (n)
&
@ constant pressure (P) :
If temperature (T) increases, the volume (V) of the gas increases
V V2
T1 = T2

18. GAS LAWS: EMPIRICAL GAS LAWS
EXERCISE 5.3
A chemical reaction is expected to produce 4.3dm3 of oxygen at 19oC and 101kPa. What will the volume be at constant pressure and 25oC?
First, convert the temperatures to the Kelvin scale.
Ti = ( ) = 292 K
Tf = ( ) = 298 K
Following is the data table.
Vi = dm3 Pi = 101 kPa Ti = 292 K
Vf = ? Pf = 101 kPa Tf = 298 K
Apply Charles’s law to obtain
Vf = Vi x Tf = dm3 x 298K = = dm3
Ti K

19. COMBINED GAS LAW Relating Volume, Temperature and Pressure
GAS LAWS: EMPIRICAL GAS LAWS
COMBINED GAS LAW
Relating Volume, Temperature and Pressure
Taking Boyle’s Law and Charles’ Law:
The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure:
V = constant x T or PV = constant (for a given amount of gas)
P T
P1V1 = P2V2
T T2

20. Use Kelvins for temp, any pressure, any volume
Combined Gas Law
We can combine Boyle’s and charles’ law to come up with the combined gas law
Use Kelvins for temp, any pressure, any volume

21. EXERCISE 5.4 A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa.
GAS LAWS: EMPIRICAL GAS LAWS
EXERCISE 5.4
A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa.
Suppose the gas in the balloon is heated to 35oC and the
pressure is now 102.8kPa, what is the volume of the gas?
First, convert the temperatures to kelvins.
Ti = ( ) = 297 K
Tf = ( ) = 308 K
Following is the data table.
Vi = dm3 Pi = kPa Ti = 297 K
Vf = ? Pf = kPa Tf = 308 K
Apply both Boyle’s law and Charles’s law combined to get
Vf = Vi x Pi x Tf =5.41 dm3 x 101.5kPa x 308K = = 5.54 dm3
Pf Ti kPa K

22. AVOGADRO’S LAW relating volume and amount
GAS LAWS: EMPIRICAL GAS LAWS
AVOGADRO’S LAW relating volume and amount
equal volumes of any two gases at the same
temperature & pressure contain the same number of molecules
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
Constant temperature
Constant pressure

23. The Volume–Amount Law (Avogadro’s Law):

24. V1 = V2 n1 n2 The Volume–Amount Law (Avogadro’s Law):
At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present.
Use any volume and moles
V1 = V2
n n2

25. GAS LAWS: EMPIRICAL GAS LAWS
Avogadro’s Number = one mole of any gas contains the same number of molecules x 1023
Must occupy the same volume at a given temperature and pressure
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.

26. IDEAL GAS LAW GAS LAWS: THE IDEAL GAS LAW
Boyle’s law: V a (at constant n and T) 1 P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the molar
gas constant
PV = nRT

27. IDEAL GAS EQUATION GAS LAWS: THE IDEAL GAS LAW PV = nRT PVnT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)
R = J / (mol • K)
R = cal / (mol • K)
*The units of pressure times volume are the units of energy
joules (J) or calories (cal)

28. 1 mole of an ideal gas occupies 22.414 L at STP
The Ideal Gas Law
Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.
1 mole of an ideal gas occupies L at STP
STP conditions are K and 1 atm pressure
The gas constant R = L·atm·K–1·mol–1
P has to be in atm
V has to be in L
T has to be in K

29. Or, expressing this as a proportion,
GAS LAWS: THE IDEAL GAS LAW
EXERCISE 5.5
Show that the moles of gas are proportional to the pressure for
constant volume and temperature
Use the ideal gas law, PV = nRT, and solve for n:
n = PV RT
n = V x P
Note that everything in parentheses is constant. Therefore, you can write
n = constant x P
Or, expressing this as a proportion,
n  P

30. GAS LAWS: THE IDEAL GAS LAW
EXERCISE 5.6
What is the pressure in a 50.0L gas cylinder that contains
3.03kg of oxygen at 23oC?
T = 23oC + 273K = 296K
V = 50.0L
R = L . atm/K . mol
n = 3.03kg x 1000g x 1mol = mol O2
1 kg g
P = ?
PV= nRT
P = nRT = mol x L . atm x 296K
V K . mol = 46.0 atm
50.0L

31. Density and Molar Mass Calculations:
The Ideal Gas Law
Density and Molar Mass Calculations:
You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.

32. GAS LAWS: THE IDEAL GAS LAW
EXERCISE 5.7
Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium?
Variable
Value P
752 mmHg x 1 atm = atm
760mmHg V
1 L (exact number) T
( ) = 294 K n ? 32

33. GAS LAWS: THE IDEAL GAS LAW
Using the ideal gas law, solve for n, the moles of helium.
n = PV = atm x 1L =
RT L . atm/ K . mol x 294K
mol
Now convert mol He to grams.
mol He x g He = g He = g/L
1.00 mol He
1 L
Therefore, the density of He at 21°C and 752 mmHg is g/L.
The difference in mass between one liter of air and one liter of He:
mass air − mass He = g − g = = g difference

34. GAS LAWS: THE IDEAL GAS LAW
EXERCISE 5.8
A sample of a gaseous substance at 25oC and atm has a density of 2.26 g/L. What is the molecular mass of the substance?
Variable Value
P atm
V L (exact number)
T ( ) = 298 K
n ?
From the ideal gas law, PV = nRT, you obtain
n = PV = atm x 1 L = mol
RT L . atm/K . Mol x 298K 34 34

35. GAS LAWS: THE IDEAL GAS LAW
Dividing the mass of the gas by moles gives you the mass per mole (the molar mass).
Molar mass = grams of gas = g = g/mol
moles of gas mol
Therefore, the molecular mass is 64.1 amu.

36. Dalton’s Law of Partial Pressures
GAS LAWS: GAS MIXTURES
Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2

37. Dalton’s Law of Partial Pressures
For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB).
Mole fraction is related to the total pressure by:

38. Each gas obeys the ideal gas law.
GAS LAWS: GAS MIXTURES
EXERCISE 5.10
A 10.0L flask contains 1.031g O2 and 0.572g CO2 at 18oC. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture?
Each gas obeys the ideal gas law.
1.031 g O2 x 1 mol = mol O2
32.00g
P = nRT = mol x L.atm/K.mol x 291K = atm
V L
0.572 g CO2 x 1 mol = mol CO2
44.01g
P = nRT = mol x L.atm/K.mol x 291K = atm

39. The total pressure is equal to the sum of the partial pressures:
GAS LAWS: GAS MIXTURES
The total pressure is equal to the sum of the partial pressures:
PT = PO2 + PCO2 = atm atm =
= atm
The mole fraction of oxygen in the mixture is
Mole fraction O2 = PO2 = atm = =
PT atm
0.712 x 100% = mole % of O2 in this gas mixture

40. Kinetic Molecular Theory
This theory presents physical properties of gases in terms of the motion of individual molecules.
Average Kinetic Energy  Kelvin Temperature
Gas molecules are points separated by a great distance
Particle volume is negligible compared to gas volume
Gas molecules are in constant random motion
Gas collisions are perfectly elastic
Gas molecules experience no attraction or repulsion

41. Kinetic Molecular Theory

42. Average Kinetic Energy (KE) is given by:
U = average speed of a gas particle
R = J/K mol
m = mass in kg
MM = molar mass, in kg/mol
NA = x 1023

43. The Root–Mean–Square Speed: is a measure of the average molecular speed.
Taking square root of both sides gives the equation

44. Example 17:
Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C.

45. Maxwell speed distribution curves.
Kinetic Molecular Theory
Maxwell speed distribution curves.

46. Graham’s Law
Diffusion is the mixing of different gases by random molecular motion and collision.

47. Graham’s Law
Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum.

48. Graham’s Law: Rate of effusion is proportional to its rms speed, urms.
For two gases at same temperature and pressure:

49. Behavior of Real Gases
At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.

50. Behavior of Real Gases
The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

51. IntermolecularAttractions
Behavior of Real Gases
Corrections for non-ideality require van der Waals equation.
Excluded
Volume
IntermolecularAttractions

52. Example 1: Boyle’s Law
A sample of argon gas has a volume of 14.5 L at 1.56 atm of pressure. What would the pressure be if the gas was compressed to 10.5 L? (at constant temperature and moles of gas)

53. Example 2: Charles’ Law
A sample of CO2(g) at 35C has a volume of 8.56 x10-4 L. What would the resulting volume be if we increased the temperature to 85C? (at constant moles and pressure)

54. Example 3: Avogadro’s Law
6.53 moles of O2(g) has a volume of 146 L. If we decreased the number of moles of oxygen to 3.94 moles what would be the resulting volume? (constant pressure and temperature)

55. Example 4:Combined Gas Law
Oxygen gas is normally sold in 49.0 L steel containers at a pressure of atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?

56. Example 5: Gas Laws
An inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?

57. Example 6: Ideal Gas Law
Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C.

58. What is the volume (in liters) occupied by 7.40 g of CO2 at STP?
Example 7: Ideal Gas Law
What is the volume (in liters) occupied by 7.40 g of CO2 at STP?

59. What is the molar mass of a gas with a density of 1.342 g/L at STP?
Example 8: Density & MM
What is the molar mass of a gas with a density of g/L at STP?

60. Example 9: Density & MM
What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP?

61. Example 10: Density & MM
The density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass?

62. Example 11: Dalton
Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25oC. What are the partial pressures of each gas and the total pressure?

63. Example 12: Dalton
A sample of natural gas contains 6.25 moles of methane (CH4), moles of ethane (C2H6), and moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases?

64. Example 13: Mole Fraction
What is the mole fraction of each component in a mixture of g of H2, g of N2, and 2.38 g of NH3?

65. Example 14: Partial Pressure
On a humid day in summer, the mole fraction of gaseous H2O (water vapor) in the air at 25°C can be as high as Assuming a total pressure of atm, what is the partial pressure (in atm) of H2O in the air?

66. Gas Stoichiometry & Example
In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles.
Example: Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)

67. Example 15:
All of the mole fractions of elements in a given compound must add up to?
100 1 50 2

68. Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)
Example 16:
Hydrogen gas, H2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react?
Zn(s) HCl(aq)  H2(g) + ZnCl2(aq)

69. Example 18:
Under the same conditions, an unknown gas diffuses times as fast as sulfur hexafluoride, SF6 (MM = 146 g/mol). What is the identity of the unknown gas if it is also a hexafluoride?

70. Example 19: Diffusion
What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne?

71. Deviations result from assumptions about ideal gases.
Behavior of Real Gases
Deviations result from assumptions about ideal gases.
1. Molecules in gaseous state do not exert
any force, either attractive or repulsive, on
one another.
2. Volume of the molecules is negligibly small
compared with that of the container.

72. Example 20: Ideal Vs. Van Der Waals
Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using
the ideal gas equation
(b) the van der Waals equation. (a = 4.17, b = )

73. Example 21: Ideal Vs. Van Der Waals
Assume that you have mol of N2 in a volume of L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation.
For N2, a = 1.35 L2·atm mol–2, and b = L/mol.