Gases
Dr. K. Shahzad Baig
MUN-Canada
Properties of Gases
Gases have five properties
1. Low Density…
2. Indefinite Shape or Volume. ...
3. Compressibility and Expandability. ...
4. Diffusivity. ...
5. Pressure.
The physical behavior of a gas can be determined by four properties:
• pressure of the gas
• the amount of the gas (in moles)
• the volume
• temperature, and
Pressure
𝐹
𝐴
Pressure is defined as a force per unit area. 𝑃 =
that is, a force divided by the area over which the force is distributed.
In SI, the unit of force is a newton (N),
A Newton is a force, F, required to produce an acceleration of one meter per second per
second (1 ms-2) in a one kilogram mass (1 kg)., i.e 1N = 1 kg ms-2.
The corresponding force per unit area, pressure, is expressed in the unit N/m2. A
pressure of one newton per square meter is defined as one pascal (Pa)
𝐹 (𝑁)
𝑃 𝑃𝑎 =
… 6.1
2
𝐴 (𝑚 )
it is difficult to measure the total force exerted by gas molecules
Pressure of a gas
The pressure of a gas is usually measured indirectly, by comparing it with a liquid
pressure.
To confirm this statement, consider a liquid with density d, contained in a cylinder
with cross-sectional area A, filled to a height h.
𝐹
𝑊
𝑔𝑥𝑚
𝑔𝑥𝑉𝑥𝑑
𝑔𝑥𝐴𝑥ℎ𝑥𝑑
𝑃 =
=
=
=
=
= 𝑔𝑥ℎ 𝑥𝑑
𝐴
𝐴
𝐴
𝐴
𝐴
𝑷 = 𝒈𝒙𝒉 𝒙𝒅
because g is a constant,
liquid pressure is directly proportional to the liquid density and
the height of the liquid column.
Barometric Pressure
To measure the pressure exerted by the atmosphere the device is called a barometer
1 atm = 760 mmHg
The pressure exerted by a column of mercury that is exactly 760 mm in height when the
density of mercury is equal to 13.595 g/cm3(0 °C) and the acceleration due to gravity, g,
is equal to 9.8065 m/s2 .
Notice that this unit of pressure assumes specific values for the density of mercury and
the acceleration due to gravity.
This is because the density of Hg(l) depends on temperature and
g depends on the specific location.
The density of mercury is, d = 13.5951 g/cm3 = 1.35951 104 kg/m3 and
g = 9.80655 m/s2
𝑃 = 𝑔𝑥ℎ 𝑥𝑑
𝑃 = 9.80665 𝑚 𝑠 −2
0.760000 𝑚 ( 1.35951 104 𝑘𝑔 𝑚−3 )
𝑃 = 1.01325 105 𝑘𝑔 𝑚−1 𝑠 −2 )
A pressure of exactly 101,325 Pa or 101.325 kPa, has special significance because in
the SI system of units, one standard atmosphere (atm)
Measurement
of atmospheric
pressure with
a mercury
barometer
(a) The liquid mercury levels are equal inside and outside the open-end tube.
(b) A column of mercury 760 mm high is maintained in the closed-end tube, regardless of
the overall height of the tube
(c) as long as it exceeds 760 mm.
(d) A column of mercury fills a closed-end tube that is shorter than 760 mm.
Manometers
It is difficult to place a barometer inside the container of gas whose pressure is to be
measured. However, the pressure of the gas to be measured is compared with barometric
pressure by using a manometer
In manometer
𝑷𝒈𝒂𝒔 = 𝑷𝒃𝒂𝒓 + ∆𝑷
A, the pressure inside the flask is equal to atmospheric pressure
B, the pressure inside the flask balances the atmospheric pressure plus an additional pressure
C is less than atmospheric pressure, so the excess mercury is in the inside arm of manometer
The lower the liquid density, the greater the height of the liquid column.
Mercury is 13.6 times as dense as water (13.6 g/cm3 versus 1.00 g/cm3)
𝑔 𝑥 ℎ𝐻2𝑂
𝑔
𝑔
𝑥 1.00
= 𝑔 𝑥 76. 𝑐𝑚 𝑥 13.6
3
𝑐𝑚
𝑐𝑚3
ℎ𝐻2 𝑂 = 1.03 𝑥 103 𝑐𝑚 = 10.3 𝑚
Using a Manometer to Measure Gas Pressure
Example 6.2
When the manometer is filled with liquid mercury ( d= 13.6 g/cm3), it was observed that
the barometric pressure is 748.2 mmHg, and the difference in mercury levels is 8.6
mmHg. What is the gas pressure Pgas ?
𝑷𝒈𝒂𝒔 = 𝑷𝒃𝒂𝒓 + ∆𝑷
𝑃𝑔𝑎𝑠 = 748.2 𝑚𝑚𝐻𝑔 − 8.6 𝑚𝑚 𝐻𝑔 = 739.6 𝑚𝑚𝐻𝑔
The gas pressure is less than the barometric pressure. Therefore, we subtract 8.6 mmHg
from the barometric pressure to obtain the gas pressure.
Boyle s law
For a fixed amount of gas at a constant temperature,
the gas volume is inversely proportional to the gas pressure
1
𝑃 ∝
𝑉
𝑃𝑉 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
… 6.5
When the proportionality sign is replaced with an equal sign and a proportionality constant,
If write equation (6.5) for the initial state (i) and for the final state (f), we get
and Because both PV products are equal to the same value of a, we obtain the result
𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓
( 𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝑇 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
The equation above is often used to relate pressure and volume changes.
Charles s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to the
Kelvin (absolute) temperature.
𝑇 𝐾 = 𝑡 (𝑜𝐶) + 273.15
The temperature at which the volume of a hypothetical* gas becomes zero is the absolute
zero of temperature: -273.15 𝑜𝐶 on the Celsius scale or 0 K on the absolute, or Kelvin, scale
𝑉 ∝𝑇
𝑉 = 𝑏𝑇
(𝑤ℎ𝑒𝑟𝑒 𝑏 𝑖𝑠 𝑎𝑜𝑐𝑛𝑠𝑡𝑎𝑛𝑡)
for the initial state (i) and once for the final state (f), we get
𝑉𝑓
𝑉𝑖
=
𝑇𝑖
𝑇𝑓
𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝑃 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The equation above is often used to relate volume and temperature changes
Standard conditions of temperature and pressure
For the gases,
standard temperature is taken to be 0 oC = 273.15 K and
standard pressure, 1 bar = 1K Pa = 105 Pa.
Standard conditions of temperature and pressure are usually abbreviated as STP
Avogadro’s Law
At a fixed temperature and pressure,
the volume of a gas is directly proportional to the amount of gas.
𝑉 ∝𝑛
𝑉= 𝑐𝑥𝑛
If the number of moles of gas (n) is doubled, the volume doubles, and so on. A
mathematical statement of this fact is
The constant c, which is equal to V/n
Which is the volume per mole of gas, a quantity we call the molar volume.
For given values of T and P, the molar volumes of all gases are approximately equal.
The Ideal Gas Equation
In accord with the three simple gas laws, the volume of a gas is directly proportional to
the amount of gas, directly proportional to the Kelvin temperature, and inversely
proportional to pressure. That is,
𝑛 𝑇
𝑉 ∝
𝑃
𝑛 𝑇
𝑛𝑅𝑇
𝑉 =𝑅
=
𝑃
𝑃
𝑃𝑉 = 𝑛𝑅𝑇
6.11
A gas whose behavior conforms to the ideal gas equation is called an ideal, or perfect, gas.
R, the gas constant
to obtain ‘R’ is to substitute into equation (6.11) the molar volume of an ideal gas at and 1 atm
𝑃𝑉
1 𝑎𝑡𝑚 𝑥 22.4140 𝐿
𝑅=
=
= 0.082057 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙−1 𝐾 −1
𝑛𝑇
1 𝑚𝑜𝑙 𝑥 273.15 𝐾
Using the SI units of for volume and Pa for pressure gives
𝑃𝑉
101325 𝑃𝑎 𝑥 22.4140 𝑥 10−2 𝑚3
𝑅=
=
= 8.3145 𝑃𝑎 𝑚3 𝑚𝑜𝑙−1 𝐾 −1
𝑛𝑇
1 𝑚𝑜𝑙 𝑥 273.15 𝐾
𝑅=
8.3145 𝐽𝑚𝑜𝑙−1 𝐾 −1
Calculating a Gas Pressure with the Ideal Gas Equation
What is the pressure, in kilopascals, exerted by 1.00 x 1020 molecules of N2 in a 305 mL
flask at 175 oC
The first step is to convert from molecules to moles of a gas, n.
1 𝑚𝑜𝑙 𝑁2
𝑛 = 1.00 𝑥 1020 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁2 𝑥
= 0.000166 𝑚𝑜𝑙 𝑁2
23
6.022 𝑥 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁2
Convert from milliliters to liters and then to cubic meters
1𝐿
1 𝑚3
𝑉 = 305 𝑚𝐿 𝑥
𝑥
= 305 𝑥 10−4 𝑚3
1000 𝑚𝐿 1000 𝐿
Rearrange the ideal gas equation to the form and substitute the above data.
𝑛𝑅𝑇
𝑃=
=
𝑉
0.000166 𝑚𝑜𝑙 𝑥 8.3145 𝑃𝑎 𝑚3 𝑚𝑜𝑙 −1 𝐾−1 𝑥 448 𝐾
3.05 𝑥 10−4 𝑚3
= 2.03 𝑥 103 Pa
Finally, convert the pressure to the unit kilopascal
1 𝑘 𝑃𝑎
3
𝑃 = 2.03 𝑥 10 𝑃𝑎 𝑥
= 2.03 𝑘𝑃𝑎
1000 𝑃𝑎
Molar Mass Determination
If we know the volume of a gas at a fixed temperature and pressure, we can
solve the ideal gas equation for the amount of the gas in moles.
𝑚
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠
𝑛 =
=
𝑀
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑃𝑉 = n RT
𝑚𝑅𝑇
𝑃𝑉 =
𝑀
Problem statement
Propylene is an important commercial chemical (about 9th in the amount produced among
manufactured chemicals) used in the synthesis of other organic chemicals and in
production of plastics (polypropylene). A glass vessel weighs 40.1305 g when clean, dry,
and evacuated; it weighs 138.2410 g when filled with water at 25.0 oC (density of water =
0.9970 g/ml) and 40.2959 g when filled with propylene gas at 740.3 mmHg and 24.0 oC.
What is the molar mass of propylene?
Solution
Determine the mass of water required to fill the vessel
= 138.241 𝑔 − 40.1305 𝑔 = 98.1105 𝑔
Use the density of water in a conversion factor to obtain the volume of water (and hence,
the volume of the glass vessel).
1 𝑚 𝐿 𝐻2 𝑂
= 98.1105 𝑥
= 98.41 𝑚𝐿 = 0.09841 𝐿
0.9970 𝑔 𝑜𝑓 𝐻2 𝑂
The mass of the gas is the difference between the weight of the vessel filled with
propylene gas and the weight of the empty vessel.
= 𝟒𝟎. 𝟐𝟗𝟓𝟗 𝒈 − 𝟒𝟎. 𝟏𝟑𝟎𝟓 𝒈 = 𝟎. 𝟏𝟔𝟓𝟒 𝒈
The values of temperature = 24.0 oC + 273.15 = 297.2 K
𝑃 = 740.3 𝑚𝑚𝐻𝑔 𝑥
1 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
= 0.9741 𝑎𝑡𝑚
molar mass of propylene
𝑚𝑅𝑇
𝑀=
𝑃𝑉
0.1654 𝑔 𝑥 0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙−1 𝐾 −1 𝑥 297.2 𝐾
𝑀=
0.9741 𝑎𝑡𝑚 𝑥 0.09841 𝐿
𝑀= 42.08 g mol -1
Gas Densities
To determine the density of a gas, we can start with the density equation,
𝑚
𝑛𝑥𝑀
𝑑 =
=
=
𝑉
𝑉
𝑛
𝑥𝑀
𝑉
with the ideal gas equation, we can replace n/V by its equivalent, P/RT
𝑚
𝑀𝑃
𝑑 =
=
𝑉
𝑅𝑇
The density of gases differs from that of solids and liquids in two important ways.
1. Gas densities depend strongly on P and T.
Gas density increases as the gas P increases and decreases as the T increases.
Densities of liquids and solids also depend somewhat on temperature, but
they depend far less on pressure
2. The density of a gas is directly proportional to its molar mass. No simple relationship
exists between density and molar mass for liquids and solids
Using the Ideal Gas Equation in Reaction Stoichiometry
What volume of N2 measured at 735 mmHg and 26 oC is produced when 75.0 g NaN3
is decomposed?
2 NAN3(s) → 2 NA(s) + 3 N2(g)
1 𝑚𝑜𝑙 𝑁𝑎𝑁3
3 𝑚𝑜𝑙 𝑁2
𝑚𝑜𝑙 𝑁2 = 75.0 𝑁𝑎 𝑁3 𝑥
𝑥
= 1.73 𝑚𝑜𝑙 𝑁2
65.01 𝑔 𝑁𝑎𝑁3 2 𝑚𝑜𝑙 𝑁𝑎𝑁3
1 𝑎𝑡𝑚
𝑃 = 735 𝑚𝑚𝐻𝑔 𝑥
= 0.967 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔
𝑇 = 26𝑜𝐶 + 273 = 299 𝐾
𝑉 =?
n = 1.73 mol
R = 0.08206 atm L mol-1 K-1
𝑛𝑅𝑇
1.73 𝑚𝑜𝑙 𝑥 0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙 −1 𝐾 −1 𝑥 299 𝐾
𝑉=
=
= 43.9 𝐿
𝑃
0.967 𝑎𝑡𝑚
Mixtures of Gases
Dalton s law of partial pressures states that the total pressure of a mixture of gases is
the sum of the partial pressures of the components of the mixture.
For a mixture of gases, A, B, and so on
𝑃𝑡𝑜𝑡 = 𝑃𝐴 + 𝑃𝐵 + …
… 6.16
In a gaseous mixture of nA moles of A, nB moles of B, and so on, the volume each gas
would individually occupy at a pressure equal to Ptot is
𝑉𝐴 =
𝑛𝐴 𝑅𝑇
𝑛𝐵 𝑅𝑇
; 𝑉𝐵 =
𝑃𝑡𝑜𝑡
𝑃𝑡𝑜𝑡
𝑉𝑡𝑜𝑡 = 𝑉𝐴 + 𝑉𝐵 + …
We can derive a particularly useful expression
𝑃𝐴
𝑛𝐴 𝑅𝑇 𝑉𝑡𝑜𝑡
=
𝑃𝑡𝑜𝑡
𝑛𝑡𝑜𝑡 𝑅𝑇 𝑉𝑡𝑜𝑡
which means that
𝑛𝐴
=
𝑛𝑡𝑜𝑡
𝑎𝑛𝑑
𝑛𝐴
𝑃𝐴
𝑉𝐴
=
=
= 𝑥𝐴
𝑛𝑡𝑜𝑡
𝑃𝑡𝑜𝑡
𝑉𝑡𝑜𝑡
𝑉𝐴
𝑛𝐴 𝑅𝑇 𝑃𝑡𝑜𝑡
=
𝑉𝑡𝑜𝑡
𝑛𝑡𝑜𝑡 𝑅𝑇 𝑃𝑡𝑜𝑡
𝑛𝐴
=
𝑛𝑡𝑜𝑡
Applying the Ideal Gas Equation to a Mixture of Gases
Example 6.11
What is the pressure, in bar, exerted by a mixture of 1.0 g H2 and 5.00 g He when the
mixture is confined to a volume of 5.0 L at 20 oC
𝑛𝑡𝑜𝑡 =
1 𝑚𝑜𝑙 𝐻2
1.0 𝑔𝐻2 𝑔 𝑥
2.02 𝑔 𝐻2
+
1 𝑚𝑜𝑙 𝐻𝑒
5.00 𝑔 𝐻𝑒 𝑥
4.003 𝑔 𝐻𝑒
= 0.50 𝑚𝑜𝑙 𝐻2 + 1.25 𝑚𝑜𝑙 𝐻𝑒 = 1.75 𝑚𝑜𝑙 𝑔𝑎𝑠
1.75 𝑚𝑜𝑙 𝑥 0.0831 𝑏𝑎𝑟 𝐿 𝑚𝑜𝑙−1 𝐾 −1 𝑥 293 𝐾
𝑃 =
= 8.5 𝑏𝑎𝑟
5.0 𝐿
It is also possible to solve this problem by starting from equation (6.12).
𝑃𝑓 𝑉𝑓
𝑃𝑖 𝑉𝑖
=
𝑛𝑖 𝑇𝑖
𝑛𝑓 𝑇𝑓
Collecting a Gas over a Liquid (Water)
In the following reaction, 81.2 mL of O2 (g) is collected over water at and barometric
pressure 751 mmHg. What mass of Ag2O (s) decomposed? (The vapor pressure of water at
23 oC is 21.1 mmHg.)
2𝐴𝑔2 𝑂 𝑠
→ 4 𝐴𝑔 𝑠 + 𝑂2 𝑔
The key concept is that the gas collected is wet, that is, a mixture of and water vapor. Use
𝑃𝑏𝑎𝑟 = 𝑃𝑂2 + 𝑃𝐻2𝑂 to calculate and then use the ideal gas equation to calculate the
number of moles of
𝑃𝑂2 = 𝑃𝑏𝑎𝑟 − 𝑃𝐻2𝑂
= 751 𝑚𝑚𝐻𝑔 − 21.1 𝑚𝑚𝐻𝑔 = 730 𝑚𝑚𝐻𝑔
𝑃𝑂2
1 𝑎𝑡𝑚
= 730 𝑚𝑚 𝐻𝑔 𝑥
= 0.961 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
𝑉 = 81.2 𝑚𝐿 = 0.0812 𝐿
n= ?
R = 0.08206 atm L mol -1 K-1
𝑇 = 23 𝑜𝐶 + 273 = 296 𝐾
𝑃𝑉
𝑛=
=
𝑅𝑇
=
0.961 𝑎𝑡𝑚 𝑥 0.0812 𝐿
0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙 −1 𝐾 −1 𝑥 296 𝐾
=
= 0.00321 𝑚𝑜𝑙
From the chemical equation we obtain a factor to convert from moles of O2 to moles of
Ag2O The molar mass of Ag2O provides the final factor.
2 𝑚𝑜𝑙 𝐴𝑔2𝑂
231.7 𝑔 𝐴𝑔2𝑂
𝐴𝑔2𝑂 𝑔 = 0.00321 𝑚𝑜𝑙 𝑂2 𝑥
𝑥
= 1.49 𝑔 𝐴𝑔2𝑂
1 𝑚𝑜𝑙 𝑂2
1 𝑚𝑜𝑙 𝐴𝑔2𝑂
Kinetic-Molecular Theory of Gases
It is based on the model outlined as follows:
A gas is composed of a very large number of extremely small particles (molecules or, in
some cases, atoms) in constant, random, straight-line motion.
Molecules of a gas are separated by great distances. The gas is mostly empty space. (The
molecules are treated as so-called point masses, as though they have mass but no volume.)
Molecules collide only fleetingly with one another and with the walls of their container,
and most of the time molecules are not colliding.
They are assumed to be no forces between molecules except very briefly during collisions.
That is, each molecule acts independently of all the others and is unaffected by their
presence, except during collisions.
Individual molecules may gain or lose energy as a result of collisions. In a collection of
molecules at constant temperature, however, the total energy remains constant.
Derivation of Boyle s Law
Boyle s law was stated mathematically in equation
𝑃𝑉 = 𝑎
The value of ‘a’ depends on the number, N, of molecules in the sample and the temperature, T.
Let’s focus on a molecule moving in the x- direction toward a wall perpendicular to its path.
The speed of the molecule is denoted by ux.
The force exerted on the wall by the molecule depends on the following 2 factors.
1. The frequency of molecular collisions
𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ∝ (𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑) 𝑥 (𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒)
𝑁
𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ∝ 𝑢𝑥 𝑥
𝑉
2. The momentum transfer, or impulse.
𝑖𝑚𝑝𝑢𝑙𝑠𝑒 𝑜𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 ∝ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑥 (𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑)
𝑖𝑚𝑝𝑢𝑙𝑠𝑒 ∝ 𝑚𝑢𝑥
The pressure of a gas (P) is the product of impulse and collision frequency. Thus, the
complete proportionality expression for factors that affect pressure is
𝑃 ∝
𝑚𝑢𝑥 𝑥 𝑢𝑥
𝑁
𝑥
𝑉
𝑁
∝
𝑚 𝑢𝑥 2
𝑉
At any instant, however, the molecules in a gas sample are traveling at different speeds.
Therefore, we must replace 𝑢𝑥 2 with the average value of which is denoted by (𝑈𝑥 2 )
𝑁 2
𝑃 ∝ 𝑚 𝑢𝑥
𝑉
we should expect that 𝑈𝑥
value of U2. Thus
2
= 𝑈𝑦
2
= 𝑈𝑧
2
=
1 𝑁𝑚 2
𝑃 =
𝑢𝑥
3 𝑉
1
𝑃𝑉 = 𝑁 𝑚 𝑢𝑥 2
3
1 2
𝑈 ,
3
where the quantity is the average
PV = nRT
We also know that
𝑃𝑉 =
1
3
𝑁𝐴 𝑚 𝑢2
3 𝑃𝑉 = 𝑁𝐴 𝑚 𝑢2
3 𝑅𝑇 = 𝑁𝐴 𝑚 𝑢2
the product NAm represents the mass of 1 mol of molecules, the molar mass, M.
3 𝑅𝑇 = 𝑀𝑢2
𝑢𝑟𝑚𝑠 =
𝑢2
=
3𝑅𝑇
𝑀
6.20
Distribution of Molecular Speeds
The Maxwell equation for the distribution of
speeds gives us fraction of molecules that
have speed ‘u’ is
𝐹(𝑢) = 4𝜋
𝑀
2𝜋𝑅𝑇
3/2
𝑀𝑢2
𝑢3 𝑒 −( 2𝑅𝑇 )
The distribution depends on the product of two opposing factors: a factor that is
proportional to U and an exponential factor, 𝑒
𝑀𝑢2
−( 2𝑅𝑇 )
.
As u increases, the factor increases from a value of zero, while the exponential factor
decreases from a value of one. The factor favors the presence of molecules with high
speeds and is responsible for there being few molecules with speeds near zero
The distributions for O2 (g) at 273 K and at
1000 K and H2 (g) at 273 K reveal that the
lighter the gas, the broader the range of speeds.
The root-mean-square speed, Urms, is obtained
from the average of u2
𝑢𝑟𝑚𝑠 =
𝑢2 .
The root-mean-square, speed is of particular
interest because we can use the value of urms to
calculate the average kinetic energy, 𝑒𝑘 of a
collection of molecules.
𝑒𝑘
1
=
2
𝑚 𝑢2
=
1
2
𝑚 𝑢𝑟𝑚𝑠 2
the lighter the gas,
the broader the range of speeds
Equation (6.20) shows that urms of a gas is directly proportional to the square root of its
Kelvin temperature and inversely proportional to the square root of its molar mass.
From this we infer that, on average,
1. molecular speeds increase as the temperature increases, and
2. lighter gas molecules have greater speeds than do heavier ones
The Meaning of Temperature
The basic equation of the kinetic-molecular theory
𝑃𝑉 =
modify it slightly
𝑃𝑉 =
2
1
𝑥
3
2
2
𝑃𝑉 =
𝑁𝐴
3
1
3
𝑁𝐴 𝑚 𝑢2
𝑁𝐴 𝑚 𝑢2
1
𝑚 𝑢2
2
=
2
3
𝑁𝐴 𝑒𝑘
1
The quantity 𝑚 𝑢2 represents the average translational kinetic energy, 𝑒𝑘 of a
2
collection of molecules.
𝑅𝑇
=
2
3
𝑁𝐴 𝑒𝑘
3 𝑅𝑇
𝑒𝑘 =
2 𝑁𝐴
Because R and are constants, equation (6.21) simply states that
𝑒𝑘 = constant * T
… 6.22
The Kelvin temperature (T)of a gas is directly proportional to the average translational
kinetic energy 𝑒𝑘 of its molecules
Molecules in the hotter object have, on average, higher kinetic energies than do the
molecules in the colder object. When the two objects are placed into contact with each
other, molecules in the hotter object give up some of their kinetic energy through
collisions with molecules in the colder object until the temperatures become equalized
the absolute zero of temperature, is the temperature at which translational molecular
motion should cease.
Gas Properties Relating to the Kinetic-Molecular Theory
Diffusion is the migration of molecules as a result of random molecular motion
The diffusion of two or more gases results in an intermingling of the molecules and, in a
closed container, soon produces a homogeneous mixture.
Effusion, is the escape of gas molecules from their container through a tiny orifice or pinhole
The rate at which effusion occurs is directly proportional to molecular speeds. That is,
molecules with high speeds effuse faster than molecules with low speeds.
Let us consider the effusion of two different gases at the same temperature and pressure
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐴
=
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐵
𝑈𝑟𝑚𝑠
𝑈𝑟𝑚𝑠
𝐴
𝐴
=
3𝑅𝑇/𝑀𝐴
3𝑅𝑇/𝑀𝐵
=
𝑀𝐴
𝑀𝐵
The rate of effusion of a gas is inversely proportional to the square root of
its molar mass … Graham’s law
Graham s law has serious limitations :
It can be used to describe effusion only for gases at very low pressures, so that
molecules escape through an orifice individually, not as a jet of gas.
Also, the orifice must be tiny so that no collisions occur as molecules pass through.
When effusion takes place under the restrictions described above, equation (6.23) can be
used to determine which of two gases effuses faster,
𝑟𝑎𝑡𝑖𝑜 𝑜𝑓
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑𝑠
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒𝑠
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑡𝑖𝑚𝑒𝑠
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑔𝑎𝑠 𝑒𝑓𝑓𝑢𝑠𝑒𝑑
=
𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡𝑤𝑜 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠𝑒𝑠
6.25
Applications of Diffusion
Addition of mercaptants (CH3SH) to liquefied petroleum gas
(LPG) to detect leaks. Liquefied petroleum gas (LPG) are odorless.
The mercaptan has an odor that can be detected in parts per billion
(ppb) or less. When a leak occurs, which can lead to asphyxiation
or an explosion, we rely on the diffusion of this odorous compound
for a warning.
Separation of 235U from 238 U. the Manhattan Project. The method
is based on the fact that uranium hexafluoride is one of the few
compounds of uranium that can be obtained as a gas at moderate
temperatures
Nonideal (Real) Gases
The measure of how much a gas deviates
from ideal gas behavior is found in its
compressibility factor.
The compressibility factor of a gas is the
Ratio PV/nRT. For an ideal gas it is = 1
At 300 K and 10 bar, He, H2, CO, N2 and O2
and behave almost ideally PV/nRT ≈ 1
But NH3 and SF6 do not PV/nRT ≈ 0.88
Conclusion from this plot is that all gases behave ideally at sufficiently low pressures,
say, below 1 atm, but that deviations set in at increased pressures.
The van der Waals Equation
For modeling the behavior of real gases, the van der Waals equation, equation is the
simplest to use and interpret
𝑎𝑛2
𝑃+ 2
𝑉
𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇
…
6.26
The equation incorporates two molecular parameters, a and b, whose values vary from
molecule to molecule
𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑐𝑡𝑜𝑟 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑛𝑅𝑇
The van der Waals equation uses
A modified pressure factor
A modified volume factor
P + an2/V2
V-nb
To explain the significance of the term an2/V2 in the modified pressure factor,
𝑛𝑅𝑇
𝑎𝑛2
𝑃=
−
𝑉 − 𝑛𝑏
𝑉2
Provided V is not too small, the first term in the equation above is approximately equal to
the pressure exerted by an ideal gas,
𝑛𝑅𝑇
𝑛𝑅𝑇
≈
= 𝑃𝑖𝑑𝑒𝑎𝑙
𝑉 − 𝑛𝑏
𝑉
The equation given above for P predicts that the pressure exerted by a real gas will be
less than that of an ideal gas.
The term an2/V2 takes into account the decrease in pressure caused by intermolecular
attractions
The proportionality constant, a, provides a measure of how strongly the molecules attract
each other.
the values of both a and b increase as the sizes of the molecules increase.
The smaller the values of a and b, the more closely the gas resembles an ideal gas.
Deviations from ideality, as measured by the compressibility factor, become more
pronounced as the values of a and b increase.
Using the van der Waals Equation to Calculate the
Pressure of a Gas
Example 6.17
Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 (g)
confined to a volume of 2.00 L at 273 K. The value of and that of a = 6.49 L2 atm mol-2
and that of b= 0.0562 L mol-1
Solution
𝑛𝑅𝑇
𝑎𝑛2
𝑃=
−
𝑉 − 𝑛𝑏
𝑉2
Then substitute the following values into the equation.
n = 1.00 mol; V = 2.00 L; T = 273 K; R = 0.08206 atm L mol-1 K-1
n2a = (1.00)2 mol2 * 6.49 L2 atm /mol2 = 6.49 L2 atm
nb = 1.00 mol * 0.0562 L mol-1 = 0.0562 L
−1
−1
𝐾
1.00 𝑚𝑜𝑙 𝑥 0.8206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙
𝑃 =
2.00 − 0.0562 𝐿
𝑥 273 𝐾
P = 11.5 atm - 1.62 atm = 9.9 atm
6.49 𝐿2 𝑎𝑡𝑚
−
(2.00)2 𝐿2