Chapter 11
Properties of Gases
In gases, the intermolecular forces of attraction are very weak to the extent that the
physical behavior of gases is independent on their composition. The behavior of gases
is dependent on volume, pressure, temperature, as well as the number of moles. The
movement of gas molecules is very rapid and independent on other molecules.
Pressure
The pressure is defined as the force per surface area. The standard atmospheric
pressure (1 atmosphere) is equal to the pressure that supports a 760 mm column of
mercury at 0 oC at sea level. A common unit for pressure is the psi (pound per square
inch). However, other units are usually used including atmosphere (atm), pascal (Pa),
and the torr. The following conversion factors are routinely used:
Unit
1 atm
1 atm
1 atm
1 atm
Equivalent
760 mm Hg
14.7 psi
101.325 kPa
760 torr
Unit
psi
1 Pa
1 torr
Bar
Equivalent
lb/in2
1 N/m2
1 mm Hg
100 kPa
Atmospheric pressures can be measured using a mercury barometer like the one in the
photo below:
Pressures of Trapped Gases
An open end manometer is usually used for measurement of pressures of gases in
chemical reactions. A U shaped mercury tube is connected to the outlet of the reaction
container where the pressure inside the container can do one of the following:
 Pushing the mercury toward the open end of the tube indicating that P gas>Patm.
In this case we have:
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Pgas = Patm + PHg
 Atmospheric pressure pushes the mercury downward the open end of the tube
indicating that Pgas<Patm. In this case we have:
Pgas = Patm - PHg
 The position of mercury in the U tube does not change indicating that P gas =
Patm.
Another type of manometer used for measurement of low pressures is the closed end
manometer where it is formed from a U tube closed at one end while the other end is
connected to the gas container. The tube is filled with mercury and the gas at low
pressure will draw the mercury column down the closed end arm; leaving vacuum
above. The pressure of the gas is then equals the pressure of the mercury.
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Using Liquids Other than Mercury in Manometers
The high density of mercury makes it of the best practical materials to be used in
manometers. 1 torr is equivalent to 1 mm height of mercury. However, if water is
used instead, 13.6 mm height of water will be equivalent to 1 torr of pressure. The
heights of fluids, in barometers and manometers, are related to their densities by the
relation:
ha da = hb db
Where: ha is the height of fluid a and da is the density of fluid a. Also, hb is the height
of fluid b and db is the density of fluid b.
Example
In an experiment, the barometric pressure was measured to be 756.0 torr. Using a
liquid with a density of 1.15 g/mL in an open end manometer resulted in 14.7 mm
decrease in the height of the fluid in the arm connected to an apparatus containing
trapped gas. What is the pressure of the trapped gas?
Solution
Since the height of the fluid connected to the container was lower than the open end
arm, the pressure of the trapped gas is higher than the barometric pressure where:
Pgas = Patm + Pliquid
From the relation connecting heights to density, we can calculate equivalent mercury
height where:
ha da = hHg dHg
14.7*1.15 = hHg * 13.6
hHg = 1.24 mm Hg = 1.24 torr
Pgas = 756.0 + 1.24 = 757.2 torr
Gas Laws
Some basic observations are unambiguous; like:
 The decrease in the volume of a gas as its pressure increases
 The increase in the volume of a gas as its temperature increases
 The increase in the pressure of a confined gas as its temperature increases
 The increase in the pressure of a confined gas as its amount (number of moles)
increases

Boyle’s Law
When the temperature of a gas is held constant, its volume is inversely proportional to
pressure.
V  1/P (at constant temperature), or:
PV = constant
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
Charles’s Law
Charles studied the variation of the volume of a gas as its temperature was varied at
constant pressure. A plot of the results of Charles experiments at different pressures
shows that the relationship between volume and temperature was linear and at a
hypothetical zero volume the temperature was always - 273.15 oC (zero Kelvin).
The assumed zero volume at –273.15 K makes this temperature the coldest possible
temperature since negative volumes are impossible. This temperature is called the
absolute zero; which corresponds to the zero point on the Kelvin scale.
TK = toC + 273.15 or simply:
TK = toC + 273
The figure above also suggests that at constant pressure, the volume of the gas is
proportional to its temperature.
V  T and thus V/T = constant
Gay-Lussac Law
Gay-Lussac studied the variation of the pressure of the gas as its temperature was
varied at constant volume. He arrived at the conclusion that at constant volume the
pressure of a gas is proportional to its temperature.
P  T and thus P/T = constant
The combined Gas Law
Combining Boyle’s, Charles’s, and Gay-Lussac’s laws result in the combined
relation:
PV/T = constant; therefore we can write the following relation for a specific gas if it
was changed from initial conditions (i) to any other final condition (f)
PiVi/Ti = PfVf/Tf
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Where Pi,Vi, and Ti are initial pressure, volume, and temperature while Pf, Vf, and Tf
are final pressure, volume, and temperature, respectively
Standard Temperature and Pressure (STP)
The state of a gas at STP is that where the temperature of the gas is 273 K and its
pressure is 1 atm (760 torr).
Example
A sample of a gas exerts a pressure of 625 torr in a 300 mL vessel at 25 oC. What
pressure would this gas sample exert if it were placed in a 500 mL vessel at 50 oC.
Solution
Ti = 25 + 273 = 298 K
Tf = 50 + 273 = 323 K
PiVi/Ti = PfVf/Tf
625 torr * 300 mL/ 298 K = Pf * 500 mL/323 K
Pf = 406 torr
Example
What would the volume of a gas be at STP if it were found to occupy 255 mL at 25 oC
and 650 torr.
Solution
PiVi/Ti = PfVf/Tf
650 torr * 255 mL/298 K = 760 torr * VSTP/273 K
VSTP = 200 mL
Example
A sample of a gas occupies 250 mL at 27 oC. What volume would it occupy at 35 oC
if the pressure does not change?
Solution
Pi = Pf = P
PVi/Ti = PVf/Tf
P * 250 mL/300 K = P * Vf/308 K
Vf = 257 mL
The Ideal Gas Law
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The combined gas law suggests that PV/T = constant. However, the pressure of a gas
depends on its number of moles (n) when the volume and temperature are kept
constant. Also, the volume of the gas depends on its number of moles when the
pressure and temperature are kept constant. Therefore, one can write
PV/T = n * constant
This constant in the above relation is called the universal gas constant R and the
relation becomes:
PV = nRT
This relation is called the ideal gas law or the equation of state for an ideal gas.
Molar Volume
The volume of 1 mol of a gas at STP (1 atm and 273 K) is 22.4 L. This is called the
molar volume of the gas at STP. It should be realized that this value (22.4 L) is just an
average value for what we refer to as the ideal gas. However, since gas molecules
have finite volume (non ideal), this value will change according to the gas of interest.
R = PV/nT
R = 1 atm * 22.4 L/1 mol * 273 K = 0.0821 L atm mol-1 K-1
Example
What volume will 25.0 g O2 (FW = 32.0) occupy at 20 oC and a pressure of 0.880
atm?
Solution
PV = nRT
n = Wt/FW = 25.0 g/(32.0 g/mol) = 0.781 mol
0.88 atm * V = 0.781 mol * 0.0821 L atm mol-1 K-1* 293 K
V = 21.3 L
Example
A 250 mL gas sample at 25 oC has a pressure of 550 torr. If the gas weighs 0.118 g,
find its molecular mass.
Solution
The ideal gas law uses the unit of atm rather than torr; therefore convert torr to atm:
P = 550 torr/(760 torr/1 atm) = 0.724 atm
PV = nRT
0.724 atm * 0.250 L = n * 0.0821 L atm mol-1 K-1 * 298 K
n = 0.00740
n = Wt/FW
0.00740 mol = 0.118 g/FW
FW = 0.118g/0.00740 mol = 15.9 g/mol
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Example
The density of a gas is 1.34 g/L at 25 oC and 760 torr. If the gas is composed of 79.8%
carbon and 20.2% hydrogen: Find the molecular mass and molecular formula of the
gas.
Solution
 Mol C = 79.8 g C * mol C/12.0 g C = 6.65 mol C
Mol O = 20.2 g H * mol O/1.0 g H = 20.0 mol H
Empirical formula = CH3
 Assume we have 1 L of the gas; the mass of the gas is thus 1.34 g
PV = nRT
1 atm * 1 L = n * 0.0821 L atm mol-1 K-1 * 298 K
n = 0.0409 mol
n = Wt/FW
0.0409 mol = 1.34 g/ FW
FW = 1.34 g/0.0409 mol = 32.8 g/mol
 Molecular formula can be obtained by getting the actual number of atoms in
the formula. This can be done by dividing the molecular mass by the empirical
formula to get the factor:
Factor = 32.8/15.0 = 2.19 ~ 2
Therefore, the molecular formula is C2H6
Calculation of Density of Gases
The ideal gas equation could be more conveniently written as:
PV = (Wt/FW) RT
FW = (Wt/V) RT/P
FW = d * RT/P
Where d is the density of the gas. Therefore, both the formula weight and density of
the gas can be calculated provided that the other is known, a very easy way to
calculate molecular weights of gases. Doing the calculation above using this relation:
FW = 1.34 g/L * 0.0821 L atm mol-1 K-1 * 298 K/ 1atm = 32.8 g/mol
Avogadro’s Law
Avogadro's law can be stated in either of two ways:
 Equal volumes of different gases at the same temperature and pressure contain
equal numbers of molecules.
 Equal numbers of molecules of different gases at the same temperature and
pressure occupy equal volumes.
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Dalton’s Law of Partial Pressures
When two or more gases that do not react with each other are placed in a container,
the resulting pressure is the sum of partial pressures of all gases.
PT = PA + PB + PC + ….
Where PT is the total pressure and PA, PB, and PC are partial pressures of gases A, B,
and C.
Example
If 200 mL of N2 at 25 oC and a pressure of 250 torr are mixed with 350 mL of O2 at
25 oC and a pressure of 300 torr, so that the resulting volume is 300 mL. What would
be the final pressure of the mixture at 25 oC?
Solution
The final volume is 300 mL and the final temperature is 25 oC. Therefore, we can find
the final pressure of each gas (partial pressure) and the sum will give the final
pressure of the mixture:
 For N2
PiVi/Ti = PiVf/Tf
250 torr * 200 mL/298 K = PN2 * 300 mL/298 K
PN2 = 167 torr
 For O2
PiVi/Ti = PiVf/Tf
300 torr * 350 mL/298 K = PN2 * 300 mL/298 K
PN2 = 350 torr
 PT = PN2 + PO2
PT = 167 + 350 = 517 torr
Practice Problem
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What is the pressure exerted by a mixture of 1.0 g H2 (FW = 2.02 g/mol) and 5.00 g
He (AtWt = 4.003 g/mol) when the mixture is confined to a volume of 5.0 L at 20 °C?
(Clue: find the number of moles of each gas and then find the partial pressure of each
gas from the ideal gas equation. Answer = 8.4 atm).
Calculation of Pressures of Gases Over Water
On doing so, one should remember that water has a vapor pressure. The total pressure
of the gas collected over water is the sum of the partial pressure of the gas plus the
vapor pressure of water.
PT = Pgas + Pwater
When the levels of water inside and outside the inverted container are the same, the
pressure (PT) inside the container equals the atmospheric pressure.
Example
In an experiment conducted at 25 oC, generated oxygen was collected over water until
the levels inside and outside the container were equal. If the volume of the gas was
245 mL and the atmospheric pressure was 758 torr, find the pressure of oxygen at 25
o
C and the volume of dry oxygen at STP. The vapor pressure of water at 25 oC is 23.8
torr.
Solution
PT = Pgas + Pwater
758 torr = Pgas + 23.8 torr
PO2 = 758 – 23.8 = 734 torr
At STP, T = 273 K and P = 760 torr
PiVi/Ti = PiVf/Tf
734 torr * 245 mL/298 K = 760 torr * VO2/273
VO2 = 217 mL
The partial pressure of a gas in a mixture is related to its mole fraction. The mole
fraction of a substance is equal to the number of moles of that substance divided by
the total number of moles.
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XA = Number of moles of A/Total number of moles
Example
Assume a gas mixture containing 1 mole of O2 and 4 moles of N2. If the pressure of
the mixture is 500 torr, find the partial pressure of each gas.
Solution
XO2 = 1/5 = 0.2
XN2 = 4/5 = 0.8
PO2 = 0.2 * 500 = 100 torr
PN2 = 0.8 * 500 = 400 torr
Graham’s Law of Effusion
The process whereby a gas under pressure escapes through a very small opening is
called effusion. Graham found that the rate of effusion is inversely proportional to the
square root of the densities of the gases:
Effusion  {1/d}1/2
Rate of effusion of gas A/Rate of effusion of gas B = {dB/dA}1/2 since density of the
gas is related to its molecular mass (M = d * RT/P), one can also write:
Rate of effusion of gas A/Rate of effusion of gas B = {dB/dA}1/2 = {MB/MA}1/2
Example
Which will effuse faster NH3 (FW = 17 g/mol) or CO2 (FW = 44 g/mol). What are
their relative rates of effusion?
Solution
Ammonia will effuse faster as it has a smaller molecular mass.
Rate of effusion of gas A/Rate of effusion of gas B = {MB/MA}1/2
Rate of effusion of NH3/Rate of effusion of CO2 = {44/17}1/2 = 1.6
Therefore, ammonia will effuse 1.6 times faster than carbon dioxide.
Real Gases
Ideal gases are characterized by the following:
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 Molecules of an ideal gas have no volume
 Molecules of an ideal gas have no attractive or repulsive forces. Ideal gases
will not condense into liquids even at absolute zero
However, an ideal gas is definitely not existent except at very high temperatures and
very low pressures. Real gases can be liquefied by cooling under pressure and will
even freeze into a solid at lower temperatures. Definitely molecules of real gases do
have finite volume. In addition, when a compressed gas is allowed to expand into a
vacuum, the gas will cool which is a manifestation of attractive forces. This is because
the potential energy of molecules will increase when they get apart from each other.
This results in a lowering in their kinetic energy since the sum will always be the
same (law of conservation of energy). Therefore, one can simply write:
Vmeasured = Videal + nb
Videal = Vmeasured - nb
Where n is the number of moles and b is a correction factor for a specific gas.
In addition, molecules of real gases do not travel in straight lines to hit the walls and
cause the pressure. They can change their path due to attraction forces as well as
collisions; which means that real gas molecules will travel longer distances to reach
the walls. Therefore, the pressure of an ideal gas will be larger than that of a real gas
and one can write:
Pideal = Pmeasured + n2a/V2
Where (a) is a measure of the attractive forces.
Van der Waals Equation of State
This equation takes the attraction forces between molecules of gases as well as their
volume into account. The ideal gas equation can thus be modified to give the equation
of state for real gases as below:
(Pmeasured + n2a/V2)(Vmeasured – nb) = nRT
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