1. Chapter 5 The Gaseous State

2. Contents and Concepts Gas Laws
We will investigate the quantitative relationships that describe the behavior of gases.
Gas Pressure and Its Measurement
Empirical Gas Laws
The Ideal Gas Law
Stoichiometry Problems Involving Gas Volumes
Gas Mixtures; Law of Partial Pressures
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3. Kinetic-Molecular Theory
This section will develop a model of gases as molecules in constant random motion.
Kinetic Theory of Gases
Molecular Speeds; Diffusion and Effusion
Real Gases
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4. Gases are compressible.
Gas volume depends on temperature and pressure.
Pressure, volume, temperature, and amount of a gas are related by the ideal gas law:
PV = nRT
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5. The force exerted per unit area It can be given by two equations:
Pressure, P
The force exerted per unit area
It can be given by two equations:
The SI unit for pressure is the pascal, Pa.
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6. Pressure exerted by a column of fluid
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8. A barometer is a device for measuring the pressure of the atmosphere.
A manometer is a device for measuring the pressure of a gas or liquid in a vessel.
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9. The water column would be higher because its density is less by a factor equal to the density of mercury to the density of water.
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10. Empirical Gas Laws
All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.
The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century.
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11. The mathematical relationship: In equation form:
Boyle’s Law
The volume of a sample of gas at constant temperature varies inversely with the applied pressure.
The mathematical relationship:
In equation form:
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12. Figure A shows the plot of V versus P for 1. 000 g O2 at 0°C
Figure A shows the plot of V versus P for g O2 at 0°C. This plot is nonlinear.
Figure B shows the plot of (1/V) versus P for g O2 at 0°C. This plot is linear, illustrating the inverse relationship.
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13. At one atmosphere the volume of the gas is 100 mL
At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to one-third, 33 mL.
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14. When a 1.00-g sample of O2 gas at 0C is placed in a containerat a pressure of 0.50 atm, it occupies a volume of 1.40 L.
When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.
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15. A volume of oxygen gas occupies 38. 7 mL at 751 mmHg and 21°C
A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
Vf = ?
Pf = 359 mmHg
Tf = 21°C
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16. (3 significant figures)
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
Vf = ?
Pf = 359 mmHg
Tf = 21°C
= 81.0 mL
(3 significant figures)
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17. A graph of V versus T is linear
A graph of V versus T is linear. Note that all lines cross zero volume at the same temperature, °C.
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18. The temperature -273. 15°C is called absolute zero
The temperature °C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.
This is the basis of the absolute temperature scale, the Kelvin scale (K).
When working with gas laws (and almost anything related to thermochemistry), the first thing you should do is convert the temperature to Kelvin
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19. The mathematical relationship: In equation form:
Charles’s Law
The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).
The mathematical relationship:
In equation form:
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20. As the air inside warms, the balloon expands to its orginial size.
A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.
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21. A 1. 0-g sample of O2 at a temperature of 100 K and a pressure of 1
A 1.0-g sample of O2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L.
When the absolute temperature of the sample is raised to 200 K, the volume of the O2 is doubled to 0.52 L.
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22. You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3
You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
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23. (3 significant figures)
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
= 87.3 mL
(3 significant figures)
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24. The mathematical relationship: In equation form:
Combined Gas Law
The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature.
The mathematical relationship:
In equation form:
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25. Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?
Vi = 5.0 L
Pi = 5.0 × 101 atm
Ti = 4°C = 277 K
Vf = ?
Pf = 1.0 atm
Tf = 11°C = 284 K
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26. (2 significant figures)
Vi = 5.0 L
Pi = 5.0 × 101 atm
Ti = 4°C = 277 K
Vf = ?
Pf = 1.0 atm
Tf = 11°C = 284. K
= 2.6 x 102 L
(2 significant figures)
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27. Avogadro’s Law
Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.
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28. Standard Temperature and Pressure (STP)
The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.
The molar volume, Vm, of a gas at STP is 22.4 L/mol.
The volume of the yellow box is 22.4 L. To its left is a basketball.
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29. Ideal Gas Law
P1V1=P2V2 or PV=constant (at constant n and T) V1/T1=V2/T2 or V/T=constant (at constant n and P) Avogadro’s Law V1/V2=n1/n2 or V/n=constant (ant constant P and T) PV/nT=constant=R PV=nRT
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30. The ideal gas law is given by the equation PV=nRT
The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P.
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31. (3 significant figures)
A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?
V = 50.0 L
P = 17.1 atm
T = 23°C = 296 K
mass = 986 g
(3 significant figures)
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32. Gas Density and Molar Mass
Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to density (grams per liter).
To find molar mass, find the moles of gas, and then find the ratio of mass to moles.
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33. What is the density of methane gas (natural gas), CH4, at 125°C and 3
What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?
Mm = g/mol
P = 3.50 atm
T = 125°C = 398 K
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34. A mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane?
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35. d = 1.57 g/ L
= g/L
P = 634 mmHg
= atm
T = 95.0°C = K
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36. Empirical formula molar mass = 57 g/mol
Empirical formula: C4H9
Empirical formula molar mass = 57 g/mol
Molecular formula: C8H18
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37. Stoichiometry and Gas Volumes
Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.
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38. When a 2. 0-L bottle of concentrated HCl was spilled, 1
When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C?
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39. First, write the balanced chemical equation: CaCO3(s) + 2HCl(aq) 
CaCl2(aq) + H2O(l) + CO2(g)
Second, calculate the moles of CO2 produced:
Molar mass of CaCO3 = g/mol
Moles of CO2 produced = mol
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40. (2 significant figures)
n = 12.0 mol
P = 735 mmHg
= atm
T = 20°C = 293 K
= 3.0 × 102 L
(2 significant figures)
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41. Gas Mixtures
Dalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.
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42. Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg
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43. The pressure exerted by a particular gas in a mixture
Partial Pressure
The pressure exerted by a particular gas in a mixture
Dalton’s Law of Partial Pressures
The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture:
P = PA + PB + PC
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44. A mL sample of air exhaled from the lungs is analyzed and found to contain g N2, g O2, g CO2, and g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?
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46. P = 1.00 atm
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47. The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, mmHg; oxygen, mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?
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48. 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg
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49. Mole fraction of N2 Mole fraction of O2 Mole fraction of CO2
Mole fraction of H2O
Mole fraction N2 =
Mole fraction O2 =
Mole fraction CO2 =
Mole fraction O2 =
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50. Collecting Gas Over Water
Gases are often collected over water. The result is a mixture of the gas and water vapor.
The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.
The partial pressure of water depends only on temperature and is known (Table 5.6).
The pressure of the gas can then be found using Dalton’s law of partial pressures
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51. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.
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53. The variable Pshould be italic (but not the subscripts)
The variable Pshould be italic (but not the subscripts). In lines 5 and 6, there should be equal space around the minuses.
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54. You prepare nitrogen gas by heating ammonium nitrite:
NH4NO2(s)  N2(g) + 2H2O(l)
If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
Molar mass NH4NO2
= g/mol
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55. P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K
Molar mass NH4NO2
= g/mol
= mol CO2 gas
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56. (3 significant figures)
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
n = mol
= 2.32 L of CO2
(3 significant figures)
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57. Kinetic-Molecular Theory (Kinetic Theory)
A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.
Kinetic energy is related to the mass and velocity:
m = mass
v = velocity
Program demonstrating kinetic-molecular theory:
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58. Postulates of the Kinetic Theory
1. Gases are composed of molecules whose sizes are negligible.
2. Molecules move randomly in straight lines in all directions and at various speeds.
3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.
4. When molecules collide with each other, the collisions are elastic.
5. The average kinetic energy of a molecule is proportional to the absolute temperature.
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59. An elastic collision is one in which no kinetic energy is lost
An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy.
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60. frequency of collision x average force
Each of the gas laws can be derived from the postulates.
For the ideal gas law:
frequency of collision x average force
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61. The average force depends on the mass of the molecules, m, and its average speed, u; it depends on momentum, mu.
The frequency of collision is proportional to the average speed, u, and the number of molecules, N, and inversely proportional to the volume, V.
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62. Rearranging this relationship gives
The average kinetic energy of a molecule of mass m and average speed u is 1/2mu2.
Thus PV is proportional to the average kinetic energy of the molecule.
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63. Inserting the proportionality constant, R, gives
However, the average kinetic energy is also proportional to the absolute temperature and the number of molecules, N, is proportional to moles of molecules. We now have
Inserting the proportionality constant, R, gives
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64. Root-Mean Square (rms) Molecular Speed, u
Molecular Speeds
According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases.
Root-Mean Square (rms) Molecular Speed, u
A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy
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65. When using the equation R = 8.3145 J/(mol  K) T must be in kelvins
Mm must be in kg/mol
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66. What is the rms speed of carbon dioxide molecules in a container at 23°C?
T = 23°C = 296 K
CO2 molar mass = kg/mol
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68. Both hydrogen and helium have been used as the buoyant gas in blimps
Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?
Hydrogen will diffuse more quickly by a factor of 1.4.
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70. a. He will reach the end first because it has a smaller molar mass.
b. Open the valves at two different times, allowing Ar more time by a factor equal to the square root of the ratio of molar masses of Ar to He, or approximately 3.16 times longer.
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71. Maxwell predicted the distributions of molecular speeds at various temperatures. The graph shows 0°C and 500°C.
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72. Diffusion
The process whereby a gas spreads out through another gas to occupy the space uniformly
Below NH3 diffuses through air. The indicator paper tracks its progress.
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73. Effusion
The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion.
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74. Graham’s Law of Effusion
At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas.
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75. Real Gases
At high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below.
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76. At high pressure, some of the assumptions of the kinetic theory no longer hold true:
1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.
2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.
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77. The term V becomes (V – nb). The term P becomes (P + n2a/V2).
Van der Waals Equation
An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior
The term V becomes (V – nb).
The term P becomes (P + n2a/V2).
Values for a and b are found in Table 5.7
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78. Other Resources
Visit the student website at
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