1. John Bookstaver St. Charles Community College Cottleville, MO
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 10 Gases
John Bookstaver
St. Charles Community College
Cottleville, MO
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2. Characteristics of Gases
Unlike liquids and solids, gases
expand to fill their containers;
are highly compressible;
have extremely low densities.
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3. Gas Pressure gas molecules are constantly in motion
as they move and strike a surface, they push on that surface
push = force
if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting
pressure = force per unit area

4. Pressure: force per unit area
newton (N): Unit of force
pascal (Pa): Unit of pressure

5. Units of Pressure mm Hg or torr
These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.
Atmosphere
1.00 atm = 760 mm Hg
Barometer
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6. Standard Pressure
Normal atmospheric pressure at sea level is referred to as standard pressure.
It is equal to
1.00 atm
760 torr (760 mm Hg)
kPa
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7. Manometer
This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.
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8. The Gas Laws Gas laws – empirical relationships among gas parameters
Volume (V)
Pressure (P)
Temperature (T)
Amount of a gas (n)
Copyright McGraw-Hill 2009

9. Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.
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10. As P and V are inversely proportional
A plot of V versus P results in a curve.
Since
V = k (1/P)
This means a plot of V versus 1/P will be a straight line.
PV = k
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11. Example
When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change)

12. Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.
i.e., V
= kT
A plot of V versus T will be a straight line.
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13. Avogadro’s Law V a number of moles (n) V = constant x n
Constant temperature
- Constant pressure
V a number of moles (n)
V = constant x n

14. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
Mathematically, this means
V = kn
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15. Ideal-Gas Equation V  nT P So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
V  n (Avogadro’s law)
Combining these, we get
V  nT P
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16. Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship
then becomes or
PV = nRT
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17. PV = nRT Ideal Gas Law P = pressure in atm V = volume in liters
n = moles
R = Gas constant = L. atm. Mol-1. K-1
T = temperature in Kelvin

18. Ideal-Gas Equation
The constant of proportionality is known as R, the gas constant.
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19. For an ideal gas, calculate the pressure of the gas if
0.215 mol occupies 338 mL at 32.0ºC.
n = mol
V = 338 mL = L
T = = K
P = ?
PV = nRT 
= atm

20. Example
Air occupies 24.8 cm3 at 1.12 atm. The pressure has been increased to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?
P2 = 2.64 atm
P1 = 1.12 atm
V1 = 24.8 cm3
V2 = unknown
PV = nRT
→ PV = Constant
P1V1 = P2V2
P1V1 P2
V2 =
1.12 atm
2.46 atm = L = L

21. Example
2 NO2(g)
N2O4(g)
Consider the above chemical reaction. If 25 ml of NO2 gas completely converted to N2O4 gas under the same conditions, what volume will N2O4 will occupy? V RT
T = Constant
P = Constant
PV = nRT =
= constant n P
VNO2
VN2O4
VN2O4
0.025 =
= constant =
n NO2
n N2O4
n N2O4
2 x n N2O4
2 x VN2O4 = 0.025
VN2O4 = L = 12.5 ml

22. A steel cylinder with a volume of 68.0 L contains O2 at
a pressure of 15,900 kPa at 23ºC. What is the volume
of this gas at Pressure = 1 atm and Temperature = 0 ºC ? P1
P2 = 1 atm
T1 = = 296 K
T2 = 273 K
V1 = L
V2 = ?
PV = nRT 

23. Standard Condition STP
The condition 0 °C and 1 atm called standard condition or standard temperature and pressure (STP)
STP
- Standard pressure = 1 atm
- Standard temperature = 273 K = 0 °C

24. Molar Volume Molar Volume is volume of 1 mole of gas at STP
(1 atm, K)
Calculate the volume of 1 mole of gas at STP condition?
n =1 mole
R = L. atm. Mol-1. K-1
at STP
T= K
P = 1 atm
nRT
PV = nRT V= P
1 mole x L.atm. Mol-1.K-1 x K
V =
= 22.4 L
1 atm

25. Molar Volume

26. Example How many grams of O2 are present in 8.0 L of gas at STP? n =PV RT
PV = nRT
n =
1 atm x 8.0 L
0.082 L.atm. Mol-1.K-1 x K
= 0.36 mole
Mass of O2 = 0.36 mole x 32 g/mole
= 11.4 g of O2

27. Example
30.2 mL of 1.00 M HCl are reacted with excess FeS. What is the volume of H2S gas generated at STP?
2 HCl + FeS  FeCl2 + H2S(g)
Moles of H2S produced = ½ moles of HCl
= ½ x 1.00 mole/L x L
= mole
V =
nRT P
PV = nRT =
mole x L.atm. Mol-1.K-1 x K
1 atm
= L

28. Densities of Gases n P V = RT
If we divide both sides of the ideal-gas equation by V and by RT, we get n V P RT =
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29. Densities of Gases n   = m P RT m V = We know that
moles  molecular mass = mass
n   = m
So multiplying both sides by the molecular mass ( ) gives P RT m V =
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30. Densities of Gases P RT m V = d =
Mass  volume = density
So, P RT m V =
d =
Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.
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31. Molecular Mass P d = RT dRT P  =
We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT
d =
Becomes
dRT P
 =
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32. Example Find the density of CO2 (in g/L) at STP (00C and 1 atm)
at room conditions (20.0C and 1.00 atm). d = RT
M x P
44.01 g/mol
x 1atm
atm*L
mol*K
0.0821
x K
(a)
= 1.96 g/L
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33. Continue d = 44.01 g/mol x 1 atm x 293K atm*L mol*K 0.0821 (b)
= 1.83 g/L
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34. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.
In other words,
Ptotal = P1 + P2 + P3 + …
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35. Schematic of Dalton’s Law
PN2
Ptotal = PN2 + PO2
Add O2

36. Partial Pressures
When one collects a gas over water, there is water vapor mixed in with the gas.
To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.
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38. Oxygen was produced and collected over water at 22ºC
and a pressure of 754 torr.
2 KClO3(s)  2 KCl(s) O2(g)
325 mL of gas were collected and the vapor pressure of
water at 22ºC is 21 torr. Calculate the number of moles
of O2 and the mass of KClO3 decomposed.

39. Ptotal = PO2 + PH2O = PO2 + 21 torr = 754 torr
PO2 = 754 torr – 21 torr = 733 torr = 733 / 760 atm
V = 325 mL = L
T = 22ºC = 295 K
2 KClO3(s)  2 KCl(s) O2(g)

40. Effusion
Effusion is the escape of gas molecules through a tiny hole into an evacuated space.
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41. Effusion
The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.
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42. Diffusion
Diffusion is the spread of one substance throughout a space or throughout a second substance.
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43. Graham's Law KE1 KE2 = 1/2 m1v12 1/2 m2v22 = = m1 m2 v22 v12 m1 m2
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