Chapter 2
Gases
Characteristics of gases
 Expand to fill and assume the shape of their container
 Compressible
 Readily forms homogeneous mixtures with other gases
These behaviors are due to large distances between the gas
molecules.
Physical behavior of gases
For gas sample there are three measurable variables
1. Volume, V, is the capacity of the container
enclosing it. Unit of volume: is the cubic meter (m3),
a cubic decimeter (dm3), or liter. For smaller volumes
we will use the cubic centimeter (cm3), which is the same
as the milliliter (ml).
2. Temperature: Three temperature scales
Fahrenheit (ºF), Celsius (ºC), Kelvin
K (no
degree symbol)
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Always use K when performing calculations with the gas law
equations.
K = (ºC) + 273.15
Example: ºC = 20º
K = 293.15
Absolute or Kelvin scale: Temperature scale that has 273.15 ºC as its zero. Temperature interval of one Kelvin
equals one degree Celsius. 1 K = 1 ºC
3. Pressure, P, is defined as force per unit area
𝑝 = 𝐹/𝐴
(F = force A = area)
The molecules in a gas are in
constant motion. The gaseous
atoms are colliding with each other
and the walls of the container.
"Pressure" is a measure of the collisions of the atoms
with the container.
Barometer
 Barometer is a device used to measure the pressure
exerted by the atmosphere.
 Height of mercury varies with atmospheric conditions
and with altitude
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Mercury Barometer
 Standard atmospheric pressure is the pressure
required to support 760 mm of Hg in a column.
 There are several units used for pressure:
 Pascal (Pa), N/m2
 Millimeters of Mercury (mmHg) = torr
 Atmospheres (atm) = 760 mmHg = 100 KPa
Measurement of Gas Pressure
Manometers
Closely related to the barometer is the manometer, a device
used to measure the pressure of a gas.
Types of Manometers
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1. Closed-end manometer
The gas pressure is equal to the difference in height (Δh) of
the mercury column in the two arms of the manometer
open-end manometer
2. Open-end Manometer
The difference in mercury levels (Δh) between the two arms
of the manometer gives the difference between barometric
pressure and the gas pressure
Three Possible Relationships
1. Heights of mercury in both columns are equal if gas
pressure and atmospheric pressure are equal. Pgas
= Pbar
2. Gas pressure is greater than the barometric
pressure.
∆P > 0
Pgas = Pbar + ∆P
3. Gas pressure is less than the barometric pressure.
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∆P < 0
(a) Gas pressure
equal to barometric
pressure
Pgas = Pbar + ∆P
(b)
Gas pressure
greater than
barometric
pressure
(c) Gas pressure
less than
barometric
pressure
• The gas laws
There are four variables required to describe a gas:
 Amount of substance: moles
 Volume of substance: volume
 Pressures of substance: pressure
 Temperature of substance: temperature
The gas laws will hold two of the variables constant and see
how the other two vary
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1. Boyles Law (The Volume - pressure relationship):
“The Volume of fixed quantity of gas is inversely
proportional to its pressure.”
Vα
Processes that occur
1
p
at constant
constant
P
PV  constant
V
temperature are said
to be isothermal.
These pressure-
P1V1 = P2V2
volume data are often
displayed on a plot of P versus V.
Example 2.1
An ideal gas is enclosed in a Boyle's-law apparatus. Its
volume is 247 ml at a pressure of 625 mmHg. If the
pressure is increased to 825 mmHg, what will be the new
volume occupied by the gas if the temperature is held
constant?
Solution
Method 1: According to Boyle's law, P1V1 = P2V2
P V
V  1 1
2
p
2
Gases
V 
2
625 mmHgX247 ml
 187 ml
825 mmHg
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Method 2: The pressure of the gas increases from 625 to
825 mmHg, or in other words, by a factor 825/625 because
volume and pressure are inversely proportional, the
volume must decrease by a factor of 625/825 Defining V1 and V2
as in Method 1 above, we can write
V2= V1 X (ratio of pressures)
625 mmHg
V  247 x
 187 ml
2
825 mmHg
Example 2.2
Suppose 4.63 liters of an ideal gas at 1.23 atm is expanded at
constant temperature until the pressure is 4.14 x 10-2 atm.
What is the final volume of the gas?
Solution
P1V1= P2V2
V2 
V2 
P1 V1
p2
1.23 atm X 4.63 L
 138 liters
4.14 x 10 -2 atm
Example 2.3
Suppose 10.9 ml of an ideal gas at 765 mmHg is expanded at
constant temperature until its volume is 38.1 ml. What is the
final pressure?
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Solution
P1V1 = P2V2
P2 
or
P2 
P1 V1
V2
765 mmHgX 10.9 ml
 219 mmHg
38.1 ml
Method 2: The volume increases by a factor of
38.1
/10.9.
Because we know that the pressure and volume are
inversely proportional, we predict that the pressure will
decrease, and by a factor of
P2  765 mmHg
10.9
/38. 1.
10.9 ml
 219 mmHg
38.1 ml
(Note that
10.9
/38. 1 is
a fraction less than 1.)
2. Charles Law (The Temperature-Volume Relationship):
“The volume of a fixed quantity of gas at constant
pressure increase as the temperature increases”
V T
V1 V2

T1 T2
or V1T2 = V2T1
V = a (t + 273) (at constant P, n)
Where V is the volume of the gas, t is the; Celsius
temperature, and a is the slope of the straight line. Since the
temperature on the Celsius scale is related to that on the
Kelvin scale by
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T (Kelvin) = T (degrees by Celsius +273) we can write
V = a T (at constant P, n)
Example 2.4.
A 4.50-L sample of gas is warmed at constant pressure from
300 K to 350 K. What will its final volume be?
Given: V1 = 4.50 L, T1 = 300. K, T2 = 350. K
Equation:
V1 V2

T1 T2
(4.50 L)(350. K) = V2 (300. K)
or V1T2 = V2T1
V2 = 5.25 L
Example 2.5
An ideal gas occupies a volume of 1.28 liters at 25 °C. If the
temperature is raised to 50 °C,
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what is the new volume of the gas if the pressure remains
constant?
Solution
V1 V2

T1 T2
Solving for V2
V2 
T2 V1
T1
The absolute temperatures are T1= 25 + 273 = 298 K and T2 =
50 + 273 = 323 K. Thus, we have
V2 
(1.28 liters)(323K)
298 K
 1.39 L
3. Gay-Lussac’s Law:
“The pressure of a sample of gas is directly
proportional to the absolute temperature when
volume remains constant”
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P1 P2

T1 T2
Or
P1T2 = P2T1
The amount of gas and its volume are the same in either
case, but if the gas in the ice bath (0 ºC) exerts a pressure of
1 atm, the gas in the boiling-water bath (100 ºC) exerts a
pressure of 1.37 atm. The frequency and the force of the
molecular collisions with the container walls are greater at
the higher temperature.
4. Combined gas law
Pressure and volume are inversely proportional to each
other and directly proportional to temperature.
P1V1 P2 V2

T1
T2
or
P1V1T2 = P2V2T1
Example 2.6
Suppose 2.65 liters of an ideal gas at 25 °C and 1 .00 atm is
simultaneously warmed and compressed until the final
temperature is 75 °C and the final pressure is 2.00 atm. What
is the final volume?
Solution: From combined gas law
P1V1 P2 V2

T1
T2
Solving the above equation for V2 we have
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V2 
T2P1V1 1.00 atom x 2.65 liters x 348 K

 1.55 liters
P2 T1
2.00 atm x 298 K
Example 2.7
A sample of gas is pumped from a 12.0 L vessel at 27ºC
and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is
the final pressure?
Given: P1 = 760 Torr , V1 = 12.0 L, V2 = 3.5 L, T1 = 300 K
and T2 = 325 K
Equation:
P1V1 P2 V2

T1
T2
or
P1V1T2 = P2V2T1
(760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K)
P2 = 2.8 x 10³ Torr
5. Avogadro’s Law
(The Quantity-Volume Relationship):
“At constant temperature and pressure the volume of a
sample of gas is proportional to the number of molecules
(or the number of moles) in the sample “
V = c. n
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The ideal gas equation
Combine the gas laws (Boyle, Charles, Avogadro) yields a
new law or equation.
Ideal gas equation:
PV = nRT
R = gas constant = 0.08206 L.atm/mol-K
P = pressure (atm)
n = moles
V = volume (L)
T = temperature (K)
 Gas that obeys this equation is said to be an ideal gas (or
perfect gas).
 No gas exactly follows the ideal gas law, although many
gases come very close at low pressure and/or high
temperatures.
 Ideal gas constant, R, is
R
R
PV
nT
PV
1 atm x 22.4 L

= 0.082058L·atm/mol·K
nT 1mol x 273.15 K
R = 0.0821 liter atm K -1mol-1
Example 2.8
Suppose 0.176 mol of an ideal gas occupies 8.64 liters at a
pressure of 0.432 atm. What is the temperature of the gas in
degrees Celsius?
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Solution
PV = nRT

(0.432 atm)(8.64 liters)
(0.176 mol)(0.082 1 liter atm K -1mol -1 )
 258 K
To degrees Celsius we need only subtract 273 from the
above result:
=258 - 273 = -15OC
Example 2.9
Suppose 5.00 g of oxygen gas, O2, at 35 °C is enclosed
in a container having a capacity of 6.00 liters. Assuming
ideal-gas behavior, calculate the pressure of the oxygen
in millimeters of mercury. (Atomic weight: 0 = 16.0)
Solution
One mole of O2 weighs 2(16.0) = 32.0 g. 5.00 g of O2 is,
therefore, 5.00 g/32.0 g mol-1, or 0.156 mol. 35 °C is 35
+ 273 = 308 K
PV = nRT
(0.156 mol)(0.082 1 liter atm K - I mol - 1)( 308 K)
 0.659 atm
6.00 liters
760 mmHg
 0.659 atm
 500 mm Hg
1 atm
p
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Molar volume of an ideal gas at STP
Using the ideal-gas law we can calculate what volume would
be occupied by 1 mol of an ideal gas at any temperature and
pressure. A reference condition which is commonly used for
describing gas properties is 0°C (273.15 K) and 1.0000 atm
(760.00 mmHg), called standard temperature and pressure,
or more briefly, STP.
The volume occupied by one mole, or molar volume, of an
ideal gas at STP is
V
nRT
P
(1.0000 mol)(0.082 057 liter atm K -I mol -1 )( 273.15 K)

1.0000 atm
 22.414 liters
Daltons Law (Gas Mixtures and Partial
Pressures):
“In a gas mixture the total pressure is given by the sum of
partial pressure of each component
Pt= P1 + P2 + P3…….”
- The pressure due to an individual gas is called a partial
pressure.
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Gas 1 P1 = n1RT
P1 =
n1RT
V
(1)
Gas 2 P1 = n2RT
P2 =
n2RT
V
(2)
Gas 3 P1 = n3RT
P3 =
n3RT
V
(3)
Pt =
n1RT n2RT n3RT


V
V
V
= (n1 + n2 + n3)
RT
RT
= nt
V
V
(4)
By dividing equations (1), (2), and (3) by (4) we get,
n1 P1

nt Pt
n2 P2

nt Pt
and
and
P1  Pt
n1
nt
(5)
P2  Pt
n2
nt
(6)
n1 P1
 = X1 = mole fraction of the first gas
Where nt Pt
n 2 P2

= X2 = mole fraction of the second gas
n t Pt
Note that the sum of all mole fractions in a system will be
equal to unity
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X 1+ X 2+ X 3 = 1
Therefore, the final equation of Dalton's law of partial
pressures becomes:
P1 = P t X1
and P2 = Pt X2
Example 2.10
A gaseous mixture made from 6.00 g O2 and 9.00 g CH4
is placed in a 15.0 L vessel at 0ºC. What is the partial
pressure of each gas, and what is the total pressure in
the vessel?
Step 1: nO2 = 6.00 g O2 x
nCH4 = 9.00 g CH4 x
1mol O2
= 0.188 mol O2
32 g O2
1 mol CH4
= 0.563 mol CH4
16 g CH4
Step 2: Calculate pressure exerted by each
PO2 =
=
nRT
V
(0.188 mol O2)(0.0821 L - atm/mol - K)(273 K)
= 0.281
15.0 L
atm
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PCH4=
(0.563 mol)(0.082 1L - atm/mol - K)(273 K)
=0.841 atm
15.0 L
Step 3: Add pressures
Ptotal = PO2 + PCH4 = 0.281 atm + 0.841 atm
Ptotal = 1.122 atm
Try ?
1. A mixture of 40.0g of oxygen and 40.0g of helium has
a total pressure of 0.9 atm. What is the partial pressure
of oxygen? (Answer= 0.101 atm)
2. A sample of air tias the following partial pressures of
components; 593.4 mm Hg N2/ 159.2 mm Hg 02, 7.1
mm Hg Ar and 0.3 mm Hg CO2. What is the
composition of the air in mole fraction?. (Answer=
0.7808 + 0.2095 + 0.0093 + 0.0004 = 1.0000)
3. Suppose 1 g of H2, N2, O2 are placed together in 10
liter container at 125oC. Assume ideal gas , calculate
the total pressure (At.wt, H=1.01 , O=16, N=14)
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Graham’s Law (Molecular Effusion and
Diffusion):
“The rate of effusion of a gas is inversely
proportional to the squre root of its molecular mass”
Effusion – The escape of gas
through a small opening.
Diffusion – The spreading of
one
substance
through
another.
𝒓𝟏
𝑴𝟐
=√
𝒓𝟐
𝑴𝟏
d
m
V
𝒂𝒏𝒅
so V 
𝒓𝟏
𝒅𝟐
=√
𝒓𝟐
𝒅𝟏
m
d
Ideal Gas Equation: PV = nRT and n 
Substituting
p
“m” cancels out
Gases
m
M
m mRT

d
M
d
PM
RT
Page 43
Example 2.11
The rate of effusion of an unknown gas (X) through a
pinhole is found to be only 0.279 times the rate of effusion of
hydrogen (H2) gas through the same pinhole, if both
gases are at STP. What is the molecular weight of the
unknown gas? (Atomic weight, H = 1.01.)
Solution
𝒓𝟏
𝑴𝟐
=√
𝒓𝟐
𝑴𝟏
√𝑴𝒙 =
𝒓𝑯𝟐
𝟏
𝒙 √𝑴𝑯𝟐 =
𝒙 √𝟐(𝟏. 𝟎𝟏) = 𝟓. 𝟎𝟗
𝒓𝒙
𝟎. 𝟐𝟕𝟗
Mx= 26.0
Try ?
1. What is the relative rate of diffusion of H2 and CO2
under the same condition?
2. What is the density of gas which it’s diffusion is 1.414
times of the rate of diffusion of CO2 at STP?.
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Deviations from Ideal Behavior
To describe the behavior of a gas, we must first
describe what a gas is:
– Gases consist of a large number of molecules in
constant random motion.
– Volume of individual molecules negligible compared
to volume of container.
– Intermolecular
forces
(forces
between
gas
molecules) negligible.
– Energy can be transferred between molecules, but
total
kinetic
energy
is
constant
at
constant
temperature.
– Average kinetic energy of molecules is proportional
to temperature.
Ideal gas on obeys the gas law PV =
constant (at constant T)
Gases may be
Real gas Deviate (not obey to gas law
PV ≠ Constant)
• Real gas deviate from ideal gases when (T is very
low and P is very high).
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• The dependence of real gases from ideal behavior
can be account by two factors which were ignored
by the kinetic theory.
1. Real gases posse’s attractive forces between
molecules.
2. Every molecule in a real has a real volume.
There are 2 correction factors should be taken into
consideration:
1. Volume of gas molecules is not negligible:
True volume (Real) =
V container – non copressiable volume (b)
= V- b
Actual volume
Gases
for 1 mole
= V – nb ( for n mole)
Page 46
1. There is a force of attraction between the molecules
of gases: (pressure)
Actual pressure = measured pressure + pressure due
to attraction force.
p
a
V2
1 mole
an2
 p  2 n moles
V
N.B bulk surface force α
1 1 1
a
. α 2  2
V V V
V
Van der Walls Equation
For ideal gases PV = n RT but in case of real gases PV≠
nRT
When introduce actual volume & actual pressure
P ideal . V effective = n RT
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The Van der Waal's equation is represented as:
p
a
(V  b )  RT
V2
for 1 mole
an2
p  2 (V  nb )  nRT
V
for n mole
The last equation is known as Van der Waal's equation
where a, b are Van Waals constants depend on the
nature of the gas.
Note that:
According to Van der Waal's, the persevere of a real gas
will be lower than that of an ideal gas because attraction
to neighboring molecules tends to decrease the impact of
a real molecule that it makes with the wall of the
container, this can be expressed.
an 2
P real = P Ideal V2
Therefore, P real  P ideal
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Van der Waals constants for some gases
Gas
CO2
Ethane C2H6
Methane CH4
Helium He
Hydrogen H2
Oxygen O2
Sulfur dioxide SO2
a
(L2.atm/mol2)
3.658
5.570
2.25
0.0346
0.2453
1.382
6.865
b
(L/mol)
0.04286
0.06499
0.0428
0.0238
0.02651
0.03186
0.05679
Example 2.12
Calculate the pressure exerted by 10.0 g of methane, CH4,
when enclosed in a 1.00-liter container at 25 °C by using (a)
the ideal-gas law and (b) the van der Waals equation. (Using
the above table)
Solution
The molecular weight of CH4 is 16.0; so n, the number of
moles of methane, is 10.0 g/16.0 g mol-1, or 0.625 mol.
(a) Considering the gas to be ideal and solving for P, we
obtain
nRT (0.625 mol)(0.082 1 liter atm K -1mol 1 )(298K)
P

V
1.00 liter
 15.3 atm
(b) Treating the gas as a Van der Waals gas and
solving for P, we have
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nRT n 2 a
P

V  nb V 2
(0.625 mol)(0.082 1 liter atm K -1mol 1 )(298K)
P
1.00 liter - (0.625 mol) ( 0.248 liter mol -1 )
(0.625 mol) 2 (2.25 liters 2 atm mol 2 )

(1.00 liter) 2
 14.8atm
Try ?
1. The measured pressure of one mole of a gas of
volume 1.0 L at 503 K is 30 atm: a) Show if the gas
under such conditions behaves as an ideal gas. and b)
given that a = 17.0 L2 atm/mol2 and b= 0.136 L/mol.
Calculate the pressure of the gas using Van der
Waal's equation. Comment on the results.
2. Compare the pressure predicted for 1 mole of n-
Octane confined to 20.0 liter at 200°C the combined
gas law and by Van der Waal's equation. Where a:
37.32 liter 2 atm. mole2, b= 0.2368 liter mole-1.
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