1. Gases
Chapter 5
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Acknowledgement
Thanks to The McGraw-Hill Companies, Inc. for allowing usage in the classroom.

2. Substances that exist as gases
Elements that exist as gases at 250C and 1 atmosphere
5.1

3. 5.1

4. Sea level
1 atm
4 miles
0.5 atm
10 miles
0.2 atm
Our Atmosphere:
• exerts pressure on earth
• more at sea level
• less on mountain top
• The air we breathe:
• 80% N2
• 20% O2

5. Pressure of Gas 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa
Atmospheric pressure
1 atm = 760 mmHg = 760 torr
(1 torr = 1mm Hg)
1 atm = 101,325 Pa
1 atm = 101 kPa
5.2

6. Gas Laws 1. Boyle’s law: A fixed amount of gas at constant temperature
P1V1 = P2V2
2. Charles’s law:
A fixed amount of gas at constant pressure
T1/V1 = T2/V2
3. Gay-Lussac’s law:
A fixed amount of gas at constant volume
P1/T1 = P2/T2
4. Avogadro law:
At constant pressure and temperature
n1/V1 = n2/V2
5. Combined gas law:
A fixed amount of gas
P1V1/T1= P2V2/T2

7. From the Gas Laws: 4 variables are involved: • P = pressure
• V = volume
• n = # of moles
• T = temperature (in Kelvin)
Pressure and Temperature
• Pressure = force/area
(Commonly used units in chemistry: torr, mm Hg, cm Hg & atm)
• Always use Kelvin temperature (K)
K = ° C + 273
*Absolute temperature Kelvin temperature; absolute zero is 0 K.

8. Derive the ideal gas equation from Boyle’s, Charles’s and Avogadro’s laws:
Above is to derive the Ideal gas equation. Due to software comparability issue, the ones can’t be saved and shown here. Will be shown in lecture.
Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.
Ideal gas assumingly has mass but does not have volume and no attractive or repulsive forces between each gas molecule.
5.4

9. Example:
What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of 20.0 L at 27 oC?
n= 5.0mole, V=20.0L, T= 27 oC=( )K=300.15K
PV=nRT
P=nRT/V
= 5.0mole*0.082 L• atm / (mol • K)*300.15K/20.0L
=6.15 atm

10. In calculation, the units of R must match those for P,V,T and n.
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 L.
The conditions 25 0C and 1 atm are called normal temperature and pressure (NTP). Experiments show that at NTP, 1 mole of an ideal gas occupies 24.5 L.
Molar volume of gas
• 1 mole of gas at STP = 22.4 Liters
• Example:
2 moles of gas at STP = 44.8 L
R = L • atm / (mol • K)=8.314J/(K·mol)
= L·kPa/(K·mol)
In calculation, the units of R must match those for P,V,T and n.
5.4

11. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L
Example: 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

12. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Ex.: 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

13. d is the density of the gas in g/L
Density (d) Calculations:
from PV = nRT PV=(m/M)RT PM=(m/V)RT  PM=dRT m V = PM RT
m is the mass of the gas in gram.
d =
M is the molar mass of the gas.
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L
5.4

14. Ex. What is the density of HCl gas in grams per liter at 700 mmHg and 25 oC?V = PM RT
d =
P = 700 mmHg = (700/760) atm = 0.92 atm
T = 25 oC = ( ) K = K
0.92 atm x 36.45g/mol
x K
0.0821
L•atm
mol•K d=
=1.37g/L

15. T = 313.15K P = 741torr = (741/760) atm = 0.975 atm
Ex. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 oC?
dRT P
M =
d = m V
7.10 g
5.40 L =
= 1.31 g L
T = K P = 741torr = (741/760) atm = atm
1.31 g L
0.975 atm
x x K
L•atm
mol•K
M =
M =
34.6 g/mol
5.4

16. 15 mole CH4 ---------------- 15 mole CO2
Gas Stoichiometry
The combustion process for methane is  
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) 
If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 oC and atm?
x 1CO2/1CH4
15 mole CH  15 mole CO2
15mol x x K
L•atm
mol•K
0.985 atm =
nRT P
V =
= L
5.5

17. V and T are constant P1 P2 Ptotal = P1 + P2
Dalton’s Law of Partial Pressures
Partial pressure is the pressure of the individual gas in the mixture.
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

18. 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

19. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
Ex. 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

20. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
u2 = (u12 + u22 + …+ uN2)/N
KE a T
KE = ½ mu2
Mean square speed
5.7

21. Graham’s Law Nickel forms a gaseous compound of the formula Ni(CO)x
Graham’s Law Nickel forms a gaseous compound of the formula Ni(CO)x. What is the value of x given the fact that under the same conditions of temperature and pressure, methane (CH4) effuses 3.3 times faster than the compound? Ans. 4 Hint: 10th ed. p.p Example This question is taken from p. 219 (5.84); 9th ed. p.p Example This question is taken from p. 213 (5.84). Due to the software comparability issue, the details can’t be saved or shown here. Please see lecture for details.