1. Chapter 10 Gases

2. Characteristics of Gases
Unlike liquids and solids, gases
Expand to fill their containers.
Are highly compressible.
Have extremely low densities.

3. Pressure
Pressure is the amount of force applied to an area:
P = F A

4. Common Units of Pressure
Atmospheric Pressure
Scientific Field
pascal(Pa); kilopascal(kPa)
x105Pa; kPa
SI unit; physics, chemistry
chemistry
atmosphere(atm)
1 atm*
millimeters of mercury(Hg)
760 mm Hg*
chemistry, medicine, biology
torr
760 torr*
chemistry
pounds per square inch (psi or lb/in2)
14.7lb/in2
engineering
bar
bar
meteorology, chemistry, physics
*This is an exact quantity; in calculations, we use as many significant figures as necessary.

5. Sample Problem 5.1
Converting Units of Pressure
PROBLEM:
A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals.
PLAN:
Construct conversion factors to find the other units of pressure.
SOLUTION:
291.4 mmHg
1torr
1 mmHg
= torr
1 atm
760 torr
291.4 torr
= atm
kPa
1 atm
atm
= kPa

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 mmHg)
kPa

7. The Gas Laws
Gas experiments revealed that four variables affect the state of a gas:
Temperature, T
Volume, V
Pressure, P
Quantity of gas present, n (moles)
These variables are related through equations know as the gas laws.

8. Boyle’s Law 1662 Boyle’s Law:
The volume occupied by a gas is inversely related to its pressure
P1 x V1 = P2 x V2

9. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg
5.3

10. Charles’ Law 1787 Charles’s Law:
At constant pressure, the volume occupied by a fixed amount of gas directly proportional to its absolute temperature
First conceived by Guillaume Amontons in 1702 and published by Joseph Gay-Lussac 1802.
V1 /T1 = V2 /T2

11. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K
5.3

12. Avogadro’s Law Avogadro’s Law:
At fixed temperature and pressure, equal volumes of any ideal gas contain equal number of particles (or moles)
V1 /n1 = V2 /n2

13. Figure 5.7
Standard molar volume.

14. 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

15. Ideal-Gas Equation
The constant of proportionality is known as R, the gas constant.

16. Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship
then becomes or
PV = nRT

17. Becomes  molar mass of a gaseous substance
Molecular Mass
We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT
d =
Becomes  molar mass of a gaseous substance
dRT P
 =

18. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.6 g/mol
5.4

19. 1)What is the volume (in liters) occupied by 49.8 g of HCl at STP?
2) Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
5.4

20. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.6 L
5.4

21. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm
5.4

22. Dalton’s Law of Partial Pressures
Formulated in 1801
The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it was alone.

23. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

24. PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
mole fraction (Xi) = ni nT
5.6

25. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
5.6

26. In Groups

27. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C 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
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L
5.5