10.
Gases
10.1
Pressure
Pressure is defined as the force across a unit area.
P=
Force
Area
N
m2
Pascal , Pa
In chemistry, the SI unit for pressure, the Pascal (Pa), is typically too small to be of practical use. Typically
the pressure in represented in units of kiloPascals, kPa.
There are several devices used to measure pressure. One such device, the barometer, was invented in
1643 by Evangelista Torrecelli. A schematic diagram of the barometer is presented below.
Vacuum
h
P= g h
Hg
External Pressure
The height of the mercury in the column is proportional to the pressure exerted on the bowl of mercury. The
atmospheric pressure varies according to climatic conditions and also to altitude (height above the earth’s
surface). At sea level, on the average, the height of a column of mercury will be 760.0 mmHg. While this is
not an appropriate pressure unit, it is proportional to pressure according to the formula:
P = gh
= density of the liquid (i.e., Hg), 13.547 kg m–3 (at 20°C)
g = acceleration due to gravity, 9.80665 m s–2
h = height of the column of liquid, m
Since the density of mercury and gravity are constants under normal conditions, the height is linearly
proportional to pressure through known constants, we can (improperly) refer to pressure by the height of the
column of mercury. Thus,
page 87
P = (1.3547×104 kg m–3) × ( 9.80665 m s–2) × (0.7600 m) = 1.013×105 kg m–1 s–2 = 1.013×105 Pa = 101.3 kPa
Therefore a column of mercury 760.0 mmHg high is equivalent to a pressure of 101.3 kPa.
In the British System of units, pressure is measured in pounds per square inch, psi. The standard
atmosphere then is:
Standard atmosphere: 1.000 atm = 101.3 kPa = 14.70 psi = 760.0 mmHg = 760.0 torr.
As an exercise, show that in the conversion of SI to British units, 101.3 kPa = 14.70 psi. The unit “mmHg” is
also referred to as a “torr” in honour of Torricelli.
The various expressions for the standard atmosphere can be used as conversion factors.
Example:
10.2
If the atmospheric pressure in the lab is found to be 753.3 mmHg, what is the pressure in units
of atm and in units of kPa?
P = 753.3 mmHg
atm
= 0.9912 atm
760.0 mmHg
P = 753.3 mmHg
101.3 kPa
= 100.4 kPa
760.0 mmHg
The Gas Laws
The systematic study of gases began in around the mid-seventeenth century and continues today. This is in
part because so many reactions studied in chemistry occur in the gas phase and because much of our
understandings of fields such as thermodynamics (the study of energy transformations) have their origins in
gas theory. Finally gas theory was instrumental in establishing the existence of atoms.
10.2.1 BOYLE’S LAW
In 1662, the British chemist, Robert Boyle, published a manuscript describing his studies on the relationship
between the pressure and the volume of a gas. In summary, Boyle found that the volume of a sample of
gas varied inversely with an externally applied pressure as long as the temperature is held constant. That
is:
V
1
P
which means that the product of pressure and volume is invariant as long as the temperature and number
of moles are held constant:
P1V1 = P2V2
This formula means if we know the pressure and volume of a sample of gas at a given temperature, then if
the pressure is changed (to P2) we can predict the volume (V2) that the gas would occupy at this new
pressure.
10.2.2 CHARLES’S LAW
In 1787, Jacques Charles studied the effect of temperature on the volume of a gas. Fifteen years later, in
1802, Joseph Gay-Lussac, carried out careful similar studies and published his work. What these scientists
found was that volume increases in proportion to temperature:
V
T
Similarly to Boyles Law, given the relation between the volume and temperature of a sample of gas at
constant pressure, if the temperature of the gas is changed (to T2), we can predict the volume the gas (V2)
that the gas would occupy at this new temperature:
page 88
V1 V2
=
T1 T2
Note that the absolute Kelvin temperature is always used in gas law calculations because in a plot of
Celsius temperature gives the following formula:
V = a TC + b
Gas Volume
It turns out that when experimental data for gas volume is plotted as a function of Celsius temperature,
regardless of the gas, the temperature intercept is always –273.15°C.
–273.15°C
0°C
Temperature (°C)
10.2.3 AVOGADRO’S LAW
In 1811, the Italian scientist, Amadeo Avogadro proposed that for a gas at constant pressure and
temperature, the volume of gas is directly proportional to the number of moles of gas present.
V
n
Thus if the number of moles of gas is changed from n1 to n2 , the volume will correspondingly be changed
from a volume V1 to V2 ,
V1 V2
=
n1 n2
10.3
The Ideal Gas Law
We have seen that over a period of 150 years various relationship between the volume of a gas and
variables such pressure, temperature and the number of moles have been developed:
V
1
P
T and n are constant
Charles and Gay-Lussac: V
T
P and n are constant
Avogadro’s Law:
n
P and T are constant
Boyle’Law:
V
Combining these laws yields the Combined Gas Law:
V
nT
P
The proportionality can be removed as usual:
page 89
PV
PV
1 1
= 2 2
n1T1 n2T2
What this means is that for any gas the ratio of PV to nT is invariant (this means that the ratio is a constant).
The constant is conventionally given the symbol R.
PV
=R
nT
or more conventionally
PV = nRT
Depending on the choice of units for pressure, the universal gas constant:
R = 8.3143 kPa L K–1 mol–1
= 0.08206 L atm K–1 mol–1
Note that these at the same number expressed in different systems of units. In calculations, if the pressure
is represented in units of kPa, the appropriate form of R to use is 8.3143 kPa L K–1 mol–1; if the pressure is
represented in units of atm, the appropriate form of R to use is 0.08206 L atm K–1 mol–1.
Example:
A certain sample of gas occupies 10.0 L at 1.50 atm at a given temperature. What volume
would the same sample of gas occupy if the pressure is increased to 2.0 atm when the
temperature is held constant?
Initial State (P1 , V1 , n1 , T1)
P1V1= n1R T1
Final State (P2 , V2 , n2 , T2)
P2V2= n2R T2
Since the sample of gas is the same n1 = n2 . Since the temperature is held constant,
T1 = T2 .
Thus:
V2 = V1
Example:
P1V1= P2V2
P1
1.5 atm
= 10.0 L
= 7.50 L
P2
2.0 atm
Calculate the molecular mass of a gas if a 14.275 g sample of the gas occupies 4.84 L at 753.3
torr and 25.0°C.
PV = nRT
101.3 kPa
= 100.4 kPa
760.0 atm
T = 25.0°C + 273.15 = 298.2 K
PV
(100.4 kPa)(4.84 L)
n=
=
= 0.196 mol
RT (8.3142 kPa L K 1 mol 1 )(298.2 K )
14.275 g
MM =
= 72.8 g mol 1
0.196 mol
P = 753.3 torr
page 90
Example:
The density of a gas was found to be 1.95 g L
Calculate its molecular mass.
STP
–1
at STP (Standard Temperature and Pressure).
1.000 atm and 0°C.
Recall that
=
m
V
PV = nRT
m
PV =
RT
MM
RT
P=m
MM V
m RT
RT
=
P=
V MM
MM
RT (1.95 g × mL-1 )(0.08206 L × atm × K -1 × mol -1 )(273.15 K )
MM =
=
= 43.7 g mol
P
1.000 atm
10.4
1
Gas Stoichiometry
Example: In a reaction used for making chlorine gas, 25.4 mL of 3.42 M hydrochloric acid are added to
2.50 g MnO2. Chlorine gas is produced according to the balanced chemical equation:
MnO2 (s) + 4 HCl (aq)
MnCl2 (aq) + Cl2 (g) + 2 H2O (g)
Determine the volume of chlorine gas formed at 542 mmHg and 36.0°C.
Atomic Mass:
Mn = 54.94
O = 16.00
H = 1.008
Cl = 35.453
Step 1: Write the balanced chemical equation
The balanced equation is given above.
Step 2: Calculate the number of moles of given compounds
i)
Molecular masses: MnO2 = (54.94) + 2 (16.00) = 86.94
HCl = 1.008 + 35.453 = 36.461
nMnO = 2.50 g
2
mol
= 0.0288 mol
86.94 g
nHCl = (3.42mol L 1 )(0.0254 L) = 0.0869 mol
Step 3: Determine the Limiting Reactant.
Assume MnO2 is the Limiting Reactant
nHCl = 0.0288 mol MnO2
4 mol HCl
= 0.115 mol HCl are required
mol MnO2
Since the actual moles of HCl = 0.0869 mol < 0.115 mol HCl required our assumption that MnO2 is
the limiting reactant is not correct. Thus HCl is the limiting reactant.
Step 4: Calculate the number of moles chlorine in the chemical equation produced based on the Limiting
Reactant, HCl.
page 91
1 mol Cl2
= 0.0217 mol
4 mol HCl
nHCl = 0.0869 mol Cl2
Step 5: Calculate the volume of chlorine gas formed at 542 mmHg and 36.0°C..
PV = nRT
P = 542 mmHg
atm
= 0.713 atm
760.0 mmHg
T = 36.0 + 273.15 = 309.2 K
(0.0217 mol )(0.08206 L atm K
0.713 atm
V=
Example:
1
mol 1 )(309.2 K )
= 0.773 L
If a 3.25 g sample of acetylene, C2H2 , is burned in 27.0 g oxygen, what volume of gas will be
present at 761.7 mmHg and 21.5°C? What volume will the gas occupy at STP?
Atomic Mass:
C = 12.011
H = 1.008
2 C2H2 (g) + 5 O2 (g)
O = 16.00
4 CO2 (g) + 2 H2O (l)
Step 2: Calculate the number of moles glucose
ii)
Molecular masses: C2H2 =2 (12.011) + 2 (1.008) = 26.038
O2 = 2 (16.00) = 32.00
nC H = 3.25 g
2
iii)
2
Moles:
nO = 27.0 g
2
mol
= 0.125 mol
26.038 g
mol
= 0.844 mol
32.00 g
Step 3: Determine the Limiting Reactant.
Assume C2H2 is the Limiting Reactant
5 mol O2
nO2 = 0.125 mol C2 H 2
= 0.312 mol O2 are required
2 mol C2 H 2
Since the Actual moles of O2 = 0.844 mol > 0.312 mol O2 required, our assumption that acetylene is
the limiting reactant is correct.
Step 4: Calculate the number of moles of the other compounds in the chemical equation at the end of the
reaction based on the Limiting Reactant.
At the end of the reaction:
nC H = 0.00 mol
2
2
nO = 0.844 mol 0.312 mol = 0.532 mol
2
nCO = 0.125 mol C2 H 2
2
4 mol CO2
= 0.250 mol
2 mol C2 H 2
Since at
P = 761.7 mmHg
atm
= 1.002atm
760.0 mmHg
page 92
and 21.5°C, water is essentially a liquid. Thus the total moles of gas will be due to the contributions
of CO2 and O2 .
ntot = nO + nCO = 0.532 mol + 0.250 mol = 0.782 mol
2
2
Step 5: Calculate the volume of gas at the end of the reaction at 761.7 mmHg and 21.5°C.
PV = nRT
V =
nRT (0.782 mol )(0.08206 L atm K 1 mol 1 )(21.5 + 273.15) K
=
= 18.9 L
P
1.002 atm
The volume of gas at STP is:
V =
nRT (0.782 mol )(0.08206 L atm K
=
P
1.000 atm
1
mol 1 )(273.15 K )
= 17.5 L
page 93