1) The document discusses concepts related to gas pressure, temperature, volume and chemical reactions involving gases. 2) Key gas laws described include Boyle's law relating pressure and volume, Charles' law relating volume and temperature, Avogadro's law relating amount of gas and volume, and the ideal gas law combining these relationships. 3) The ideal gas constant R is derived from the ideal gas law and standard temperature and pressure conditions. 4) Gas pressure, density and molar mass can be calculated using the ideal gas law and data on mass, volume, temperature and pressure of gases.

1. Chemical Reactions in Gases
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. Pressure
Pressure is defined as a force per unit area. 𝑃 =
𝐹
𝐴
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)
𝑃 𝑃𝑎 =
𝐹 (𝑁)
𝐴 (𝑚2)
… 6.1
it is difficult to measure the total force exerted by gas molecules

3. 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.
𝑃 =
𝐸
𝐴
=
𝑊
𝐴
=
𝑔 𝑥 𝑚
𝐴
=
𝑔 𝑥 𝑉 𝑥 𝑑
𝐴
=
𝑔 𝑥 𝐴 𝑥 ℎ 𝑥 𝑑
𝐴
= 𝑔 𝑥 ℎ 𝑥 𝑑
because g is a constant,
liquid pressure is directly proportional to the liquid density and the height of the
liquid column.

4. Barometric Pressure
To measure the pressure exerted by the atmosphere the device is called a barometer
The pressure exerted by a column of mercury that is exactly 1000 mm in height when the
density of mercury is equal to and the acceleration due to gravity, g, is equal to 9.80655
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. Typically, the pressure exerted by the
atmosphere can support a column of mercury that is about 760 mm high and thus,
atmospheric pressure is typically about 760 mmHg. L
The density of mercury is, d 13.5951 g/cm3 = 1.35951 104 kg/m3 and g 9.80655 m/s2

5. 𝑃 = 𝑔 𝑥 ℎ 𝑥 𝑑
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)

6. 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 can be compared with
barometric pressure by using a manometer
In manometer A, the pressure inside the
flask is equal to atmospheric pressure
In manometer B, the pressure inside the flask balances the atmospheric pressure plus an
additional pressure of 300. mm of Hg. If the actual atmospheric pressure is 750. mm of
Hg, then the pressure in the flask is 1050 mm of Hg
In the manometer C is less than atmospheric pressure, so the excess mercury is in the inside
arm of the manometer

7. 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
The gas pressure is less than the barometric pressure. Therefore, we subtract 8.6 mmHg
from the barometric pressure to obtain the gas pressure.
𝑃𝑔𝑎𝑠 = 748.2 𝑚𝑚𝐻𝑔 − 8.6 𝑚𝑚 𝐻𝑔 = 739.6 𝑚𝑚𝐻𝑔

8. Boyle s law
For a fixed amount of gas at a constant temperature,
the gas volume is inversely proportional to the gas pressure
𝑃 ∝
1
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.

9. 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 ( 𝑜 𝐶) + 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
𝑉𝑖
𝑇𝑖
=
𝑉𝑓
𝑉𝑓
𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑃 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The equation above is often used to relate volume and temperature changes

10. 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 = 100 K Pa = 105 Pa.
Standard conditions of temperature and pressure are usually abbreviated as STP

11. Avogadro’s Law
At a fixed temperature and pressure, the volume of a gas is directly proportional to the
amount of gas.
If the number of moles of gas (n) is doubled, the volume doubles, and so on. A
mathematical statement of this fact is
𝑉 ∝ 𝑛
𝑉 = 𝑐 𝑥 𝑛
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.

12. 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,
𝑉 ∝
𝑛 𝑇
𝑃
𝑉 = 𝑅
𝑛 𝑇
𝑃
=
𝑛𝑅𝑇
𝑃
𝑃𝑉 = 𝑛𝑅𝑇 6.11
A gas whose behavior conforms to the ideal gas equation is called an ideal, or perfect, gas.

13. 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
𝑅 =
𝑃𝑉
𝑛𝑇
=
1 𝑎𝑡𝑚 𝑥 22.4140 𝐿
1 𝑚𝑜𝑙 𝑥 273.15 𝐾
= 0.082057 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙−1 𝐾−1
Using the SI units of for volume and Pa for pressure gives
𝑅 =
𝑃𝑉
𝑛𝑇
=
101325 𝑃𝑎𝑥 22.4140 𝑥 10−2 𝑚3
1 𝑚𝑜𝑙 𝑥 273.15 𝐾
= 8.3145 𝑃𝑎 𝑚3 𝑚𝑜𝑙−1 𝐾−1
𝑅 = 8.3145 𝐽𝑚𝑜𝑙−1
𝐾−1

14. Calculating a Gas Pressure with the Ideal Gas
Equation
What is the pressure, in kilopascals, exerted by 1.00 x 1020 molecules of in a 305 mL flask
at 175 oC
The first step is to convert from molecules to moles of a gas, n.
𝑛 = 1.00 𝑥 1020 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁2 𝑥
1 𝑚𝑜𝑙 𝑁2
6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁2
= 0.000166 𝑚𝑜𝑙 𝑁2
Convert from milliliters to liters and then to cubic meters
𝑉 = 305 𝑚𝐿 𝑥
1𝐿
1000 𝑚𝐿
𝑥
1 𝑚3
1000 𝐿
= 305 𝑥 10−4 𝑚3

15. Rearrange the ideal gas equation to the form and substitute the above data.
𝑃 =
𝑛𝑅𝑇
𝑉
=
=
0.000166 𝑚𝑜𝑙 𝑥 8.3145 𝑃𝑎 𝑚2 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
𝑃 = 2.03 𝑥 103 𝑃𝑎 𝑥
1 𝑘 𝑃𝑎
1000 𝑃𝑎
= 2.03 𝑘𝑃𝑎

16. 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.
𝑛 =
𝑚
𝑀
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑃𝑉 =
𝑚𝑅𝑇
𝑀
Propylene is an important commercial chemical (about ninth 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?

17. Solution
Determine the mass of water required to fill the vessel
= 138.241 𝑔 − 40.1305 𝑔 = 98.1105
Use the density of water in a conversion factor to obtain the volume of water (and hence,
the volume of the glass vessel).
= 98.1105 𝑥
1 𝑚 𝐿 𝐻2 𝑂
0.9970 𝑔 𝑜𝑓 𝐻2 𝑂
= 98.41 𝑚𝐿 = 0.09841 𝐿
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
The values of temperature = 𝑃 = 740.3 𝑚𝑚𝐻𝑔 𝑥
1 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
= 09741 𝑎𝑡𝑚

18. 𝑀 =
𝑚𝑅𝑇
𝑃𝑉
𝑀 =
0.1654 𝑔 𝑥 0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙−1 𝐾−1 𝑥 297.2 𝐾
0.9741 𝑎𝑡𝑚 𝑥 0.09841 𝐿
𝑀= 42.08 g mol -1

19. Gas Densities
To determine the density of a gas, we can start with the density equation,
𝑑 =
𝑚
𝑉
=
𝑛 𝑥 𝑀
𝑉
=
𝑛
𝑉
𝑥 𝑀
with the ideal gas equation, we can replace n/V by its equivalent, P/RT
𝑑 =
𝑚
𝑉
=
𝑀𝑃
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

20. Using the Ideal Gas Equation in Reaction
Stoichiometry Calculations
What volume of measured at 735 mmHg and 26 oC is produced when 75.0 g NaN3 is
decomposed?
𝑚𝑜𝑙 𝑁2 = 75.0 𝑁𝑎 𝑁3 𝑥
1 𝑚𝑜𝑙 𝑁𝑎𝑁3
65.01 𝑔 𝑁𝑎𝑁4
𝑥
3 𝑚𝑜𝑙 𝑁2
2 𝑚𝑜𝑙 𝑁𝑎𝑁3
= 1.73 𝑚𝑜𝑙 𝑁2
𝑃 = 735 𝑚𝑚𝐻𝑔 𝑥
1 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔
= 0.967 𝑎𝑡𝑚
𝑉 = ?
n = 1.73 mmol
R = 0.08206 atm L mol-1 K-1
𝑇 = 26 𝑜 𝐶 + 273 = 299 𝐾
𝑉 =
𝑛𝑅𝑇
𝑃
=
1.73 𝑚𝑜𝑙 𝑥 0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙−1
𝐾−1
𝑥 299 𝐾
0.967 𝑎𝑡𝑚
= 43.9 𝐿

21. Mixtures of Gases
Dalton s law of partial pressures states that t he 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 =
𝑛 𝐴 𝑅𝑇
𝑃𝑡𝑜𝑡
; 𝑉𝐵 =
𝑛 𝐵 𝑅𝑇
𝑃𝑡𝑜𝑡
𝑉𝑡𝑜𝑡 = 𝑉𝐴 + 𝑉𝐵 + …
We can derive a particularly useful expression
𝑃𝐴
𝑃𝑡𝑜𝑡
=
𝑛 𝐴 𝑅𝑇 𝑉𝑡𝑜𝑡
𝑛 𝑏 𝑅𝑇 𝑉𝑡𝑜𝑡
=
𝑛 𝐴
𝑛 𝑡𝑜𝑡
𝑎𝑛𝑑
𝑉𝐴
𝑉𝑡𝑜𝑡
=
𝑛 𝐴 𝑅𝑇 𝑃𝑡𝑜𝑡
𝑛 𝑏 𝑅𝑇 𝑃𝑡𝑜𝑡
=
𝑛 𝐴
𝑛 𝑡𝑜𝑡
𝑛 𝐴
𝑛 𝐵
=
𝑃𝐴
𝑃𝑡𝑜𝑡
=
𝑉𝐴
𝑉𝑡𝑜𝑡
= 𝑥 𝐴which means that

22. 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 and 5.00 g He when the
mixture is confined to a volume of 5.0 L at 20 oC
𝑛 𝑡𝑜𝑡 = 1.0 𝑔𝐻2 𝑔 𝑥
1 𝑚𝑜𝑙 𝐻2
2.02 𝑔 𝐻2
+ 5.00 𝑔 𝐻𝑒 𝑥
1 𝑚𝑜𝑙 𝐻𝑒
4.003 𝑔 𝐻𝑒
= 0.50 𝑚𝑜𝑙 𝐻2 + 1.25 𝑚𝑜𝑙 𝐻𝑒 = 1.75 𝑚𝑜𝑙 𝑔𝑎𝑠
𝑃 =
1.75 𝑚𝑜𝑙 𝑥 0.0831 𝑏𝑎𝑟 𝐿 𝑚𝑜𝑙−1 𝐾−1 𝑥 293 𝐾
5.0 𝐿
= 8.5 𝑏𝑎𝑟
It is also possible to solve this problem by starting from equation (6.12).
𝑃𝑖 𝑉𝑖
𝑛𝑖 𝑇𝑖
=
𝑃𝑓 𝑉𝑓
𝑛 𝑓 𝑇𝑓

23. 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
+ 𝑃 𝐻2 𝑂 to calculate and then use the ideal gas equation to calculate the
number of moles of
𝑃𝑂2
= 𝑃𝑏𝑎𝑟 − 𝑃 𝐻2 𝑂
= 751 𝑚𝑚𝐻𝑔 − 21,1 𝑚𝑚𝐻𝑔 = 730 𝑚𝑚𝐻𝑔
𝑃𝑂2
= 730 𝑚𝑚 𝐻𝑔 𝑥
1 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
= 0.961 𝑎𝑡𝑚

24. 𝑉 = 81.2 𝑚𝐿 = 0.0812 𝐿
n = ?
R = 0.08206 atm L mol -1 K-1
𝑇 = 23 𝑜 𝐶 + 273 = 296 𝐾
𝑛 =
𝑃𝑉
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.
𝐴𝑔2 𝑂 𝑔 = 0.00321 𝑚𝑜𝑙 𝑂2 𝑥
2 𝑚𝑜𝑙 𝐴𝑔2 𝑂
1 𝑚𝑜𝑙 𝑂2
𝑥
231.7 𝑔 𝐴𝑔2 𝑂
1 𝑚𝑜𝑙 𝐴𝑔2 𝑂
= 1.49 𝑔 𝐴𝑔2 𝑂