1. Chapter 5- Gases
By
Dr. Mohammad Tariq Saeed
5.1: Existence of gases
5.2: Pressure of a gas
5.3: The Gas Laws
5.4: The Ideal Gas
Equation
5.5: Gas Stoichiometry
5.6: Dalton’s Law of
Partial pressure
2. 2
5.1 Elements that exist as gases at 250C
and 1 atmosphere, Ionic compounds are not, strong
attraction
3. 3`
• Molecular compounds like CO, CO2, HCl, NH3, and CH4 are
gases but majority of molecular compounds are liquids or
solids at room temp.
• 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.
5.2: Physical Characteristics of Gases
NO2 gas
Units of Pressure
Pressure =
Force
Area
(force = mass x acceleration)
1 pascal (Pa) = 1 N/m2 (SI unit of pressure)
1 atm = 760 mmHg = 760 torr ( Std. Atm pressure
1 atm = 101,325 Pa, = 1.01325 x 102 kPa
1kPa = 1000 Pa
1N = 1kg / m2
4. Example 5.1
The pressure outside a jet plane flying at high altitude falls
considerably below standard atmospheric pressure.
Therefore, the air inside the cabin must be pressurized to
protect the passengers.
What is the pressure in atmospheres in the cabin if the
barometer reading is 688 mmHg?
Sea level 1 atm
4 miles 0.5 atm
10 miles 0.2 atm
The atmospheric pressure in San Francisco on a certain day was 732 mmHg.
What was the pressure in kPa?, 1 atm = 1.0135 x 102 kPa, = 760 mmHg
Example 5.2
5. 5
Manometers Used to Measure Gas Pressures other than atmosphere
closed-tube open-tube
Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
V decreasesAs P (h) increases
Use to measure
pressure equal or
greater than
atmospheric press
Use to measure
pressure below
atmospheric press
Patm = Ph
Patm = Ph + Patm
6. Gas Pressure (P gas) less
than atmospheric
pressure
P gas = P atm - h
Gas Pressure (Pgas)
greater than
atmospheric pressure
P gas = P atm + h
Gas pressure and Atmospheric pressure
7. 7
5.3: Gas Laws:
Pressure and volume are inversely related at constant
temperature and moles
PV= k
As one goes up, the other goes down
Boyle noted that , if pressure (P) increased at const. temperature,
the volume(V) decrease ,
1. Boyle’s Law: Pressure – volume relationship
aP
V
1
PinitialVinitial=k= PfinalVfinal
P = (nRT) x
𝟏
𝑽
nRT = constant
P = k x
𝟏
𝑽
10. 10
As T increases V increases
Variation in Gas Volume with Temperature at Constant Pressure
V ∝ T
Charles’ Law: Temperature – Volume relationship
Volume is directly proportional to the temperature,
If the volume increase then temperature increase and vice versa.
11. 11
Variation of Gas Volume with Temperature
at Constant Pressure, “Charles’s & Gay-Lussac’s
Law”
Volume of a gas expand
when the gas heated and
contract when cooled “V a T”
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C) + 273.15
Temperature must be
in Kelvin
Example: Study of temperature-
Volume relationship at various
pressure. At any given pressure,
plot of the volume versus
temperature yields a straight line.
By extending the line to zero
volume, , we find the intercept on
the temperature exist to be – 273.15
oC.
13. 13
Avogadro’s Law: [Volume – Amount relationship]
Avogadro’s complimented the studies of Boyle, Charles, and Gay-
Lussac.: He published a hypothesis stating that at the same temperature
and pressure , equal volume of all gases contain the same number of
molecules (or atoms).
V a number of moles (n)
V = k x n V1 / n1 = V2 / n2
Constant temperature &Constant pressure
n represent the no. of moles and k is the
proportionality constant.
3H2 (g) + N2 (g) 2NH3 (g).
3mol 1 mol 2 mol
3H2 (g) + N2 (g) 2NH3 (g).
3 volume 1volume 2 volume
15. 15
5.4: Ideal Gas Equation
Charles’s law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Boyle’s law: P a (at constant n and T)1
V
V a
nT
P
V = constant x = R
nT
P
nT
P
R is the gas constant
PV = nRT
The molecules of an ideal gas do not attract or repel
one another, and their volume is negligible compared
with the volume of the container.
Ideal gas equation describe the relationship among
the four variables P, V,T, and n. An ideal gas is a
hypothetical gas whose pressure–volume –
temperature behavior can be completely accounted
for by an ideal gas equation.
16. 16
The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
PV = nRT
R =
PV
nT
=
(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.
17. 5.3
Sulfur hexafluoride (SF6) is a
colorless and odorless gas.
Calculate the pressure (in atm)
exerted by 1.82 moles of the
gas in a steel vessel of volume
5.43 L at 69.5°C.(change to K)
R = 0.0821L.atm/K.mol
P = nRT / V
18. 5.4
Calculate the volume (in L) occupied by 7.40 g of NH3 at STP.
g mole L Std V= 22.41L
An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to
rise to a height of 6.5 km, where the pressure is about 0.40 atm.
Assuming that the temperature remains constant, what is the final volume of the
balloon? P1V1 = P2 V2
5.5
19. 5.6
Argon is an inert gas used in light bulbs to retard the vaporization
of the tungsten filament.
A certain light bulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. P1/T1 = P2 / T2
Calculate its final pressure (in atm).
Electric light bulbs
are usually filled
with argon.
A small bubble rises from the bottom of a lake, where the temperature and pressure
are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the
pressure is 1.0 atm. P1V1 / T1 = P2V2 / T2
Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.
5.7
20. 20
Density (d) Calculations [n /V = P / RT], n = m/MM
m
MV
P
RT=
m is the mass of the gas in g
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
d = m
V
=
PM
RT
21. 5.8
Calculate the density of carbon dioxide (CO2) in grams per liter
(g/L) at 0.990 atm and 55°C. d = PM / RT
5.5: Gas Stoichiometry: Use the relationship between
amounts( in moles) and masses (in grams) of reactant and products to solve the
stoichiometric problems, the reactants and products are gases, then we use amounts
(in moles) and volume(V) to solve the such problems.
22. 5.11
Calculate the volume of O2 (in liters) required for the complete
combustion of 7.64 L of acetylene (C2H2) measured at the same
temperature and pressure.
The reaction of
calcium carbide
(CaC2) with
water produces
acetylene
(C2H2), a
flammable gas.
23. 5.12
Sodium azide (NaN3) is used in some automobile air bags. The impact
of a collision triggers the decomposition of NaN3 as follows:
The nitrogen gas produced quickly inflates the bag between the driver
and the windshield and dashboard.
Calculate the volume of N2 generated at 80°C and 823 mmHg by the
decomposition of 60.0 g of NaN3., mmHg atm, oC oK
g NaN3 mole NaN3 mole N2
An air bag
can protect
the driver in
an
automobile
collision.
24. 5.6: Dalton’s Law of Partial Pressures-1801
V and T are constant
P1
Ptotal = P1+ P2
Partial Pressure: that is the pressure of individual gas components in the
mixture exerted on the wall of container.
Mole fraction: mole of one gas divided by moles of all gases present in a
container
Total gas Pressure: Which states that the total pressure of a mixture of
gases is just the sum of the pressure of each gas would exert in a
container. .
P1 P2
Ptotal = P1+ P2
25. 25
Consider a case in which two gases, A and B, are in a
container of volume V.
PA =
nART
V
PB =
nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB
XA =
nA
nA + nB
XB =
nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT mole fraction (Xi ) =
ni
nT
Partial pressure of A = mole fraction of A x Pt
26. 5.14
A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole
of argon (Ar), and 2.15 moles of xenon (Xe).
Calculate the partial pressures of the gases if the total pressure
is 2.00 atm at a certain temperature. PNe/Ar/Xe = XNe/Ar/Xe x PT
Volume of oxygen collected at 24 oC and atm pressure of 762
mmHg is 128 mL. Calculate the mass (in gram) of oxygen gas
obtained . The pressure of water vapor at 24 oC is 22.4 mmHg
PO2 = PT – PH2O PV = nRT = m/M x RT
m = PVM/RT
5.15