1. CHE101
Gases
Instructor: HbR

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

3. 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.
NO2 gas

4. Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the earth’s atmosphere.
Depends on location, temperature and weather conditions.
The denser the air is, the greater the pressure it exerts.
By increasing height, density decreases.
Barometer: measure the atmospheric pressure.
Standard atmospheric pressure (1atm): pressure that supports a column of mercury exactly 760mm (or 76 cm) high at 0⁰C at sea level.
Barometer

5. (force = mass x acceleration)
Pressure of a Gases
Pressure =
Force
Area
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

6. Atmospheric Pressure
10 miles
0.2 atm
4 miles
0.5 atm
Sea level
1 atm

7. Manometers Used to Measure Gas Pressures

8. Calculate the pressure of the gas..
Pgas= =779mmHg
Pgas= =728mmHg

9. The Gas laws Boyel’s Law: the Pressure- Volume relationship
Charls’s and Gay-Lussac’s Law: Temperature-Volume relationship
Avogadro’s Law: Volume- Amount relationship

10. Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P increases
V decreases

11. Boyle’s Law P a 1/V Constant temperature Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2

12. 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

13. Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases

14. Variation of Gas Volume with Temperature at Constant Pressure
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C)

15. 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

16. Another form of Charle’s Law
For constant amount of gas and volume, the pressure is proportional to the temperature.

17. Avogadro’s Law V a number of moles (n)
Constant temperature Constant pressure
V = constant x n
V1 / n1 = V2 / n2

18. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO

19. Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) 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 gas constant
PV = nRT

20. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 00C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

21. Can you calculate the density from this equation??
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 = 30.7 L
Can you calculate the density from this equation??

22. d is the density of the gas in g/L
Density (d) Calculations n V = P RT n = m M
PV = nRT, 
Again,
m is the mass of the gas in g, M is the molar mass of the gas m VM = P RT
d = m V = PM RT
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L

23. 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 =
54.5 g/mol

24. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 37oC 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

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

26. 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
where n, the total no of moles; n = nA + nB, and PA, PB are partial pressures of gases A and B.
Total pressure of a mixture of a gas depends on the total number of moles of gas, not on the nature of the gas.

27. In general, total pressure of a mixture of a gases,
where P1, P2, P3, … are the partial pressure of the compo-nents1, 2, 3, …
where XA is mole fraction of A.
To see how each partial pressure is related to the total pressure, consider a mixture of gases A and B.
The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. The mole fraction of component i in a mixture,

28. PA = XA PT PB = XB PT Pi = Xi PT So the partial pressures of A and B,
The sum of the mole fractions for a mixture of gases must be unity. If two components are present,
If a system contains more than two gases,
mole fraction (Xi ) = ni nT
Pi = Xi PT
From mole fractions and total pressure, we can calculate the partial pressures of individual components.

29. 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

30. You are expected to practice relevant maths from the Book
By: Raymond Chang
You might get problems not discussed in the class,
Presuming we can’t cover all problems in our class.
So going through the book is of major important
Your equations will NOT provided in questions this time
As derivations were shown in our lesson!
You are always more than welcome to discuss
problems at my office…

31. Thank You!