Properties of Gases
If a gas is sufficiently dilute it obeys the ideal gas law
The ideal gas law can also be written
PV = nRT
A molar quantity is indicated by the bar across the top. The ideal gas law is an equation of state.
PV = RT where V = V/n
An equation of state relates the pressure, volume, and temperature of the gas given a quantity of n
moles of gas. The properties of a gas are of two type.Extensive variables are proportional to the size
of the system.
Intensive variables do not depend on the size of the system.
Extensive variables: volume, mass, energy
Intensive variables: pressure, temperature, density
If we divide an extensive quantity by the number of moles (or number of particles), we obtain an
intensive quantity. For example, volume V (L) is an extensive quantity, but molar volume V/n
(L/mole) is an intensive quantity.
Pressure in the ideal gas law has units of N/m2 which is corresponds to force per unit area. When
thinking about the unit of atmospheres which corresponds to the pressure at sea level, the force is that
due to the weight of the atmosphere above the surface of the earth. We look at some problems below.
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Problem: what is the mass of the atmosphere above 1 m2 of the earth surface?
Solution: The pressure is a force per unit area. The force in this case is the weight of the atmosphere
(F = mg). If we assume that g is a constant for the entire column of the atmosphere above 1 m2 we
have
m = PA/g = (1.0325 x 105 N/m2)(1 m2)/(9.8 m/s2)
m = 1.05 x 104 kg.
SI units of pressure refer to the N/m2 as the Pascal (Pa). There are 1.0325 x 105 Pa per atm. A new
standard unit of pressure is the bar, where 1 bar = 105 Pa.
If m is the mass of the liquid and g is the gravitational
acceleration, the force is F = mg. The pressure is
where ρ is the density, ρ = m/V. Note that the area of the object cancels exactly as in our example
mg ρhAg
F
P= =
=
= ρhg
A
A
A
above for the mass of the atmosphere. The application of P = ρhg is most common in liquids.
Problem: What is the pressure due to water at a depth of 100 m? (You can solve it).
We can also consider the pressure of a gas in gravitational field using the fluid formula above
and the ideal gas law. For example, we can solve the following problem.
Problem: What is the pressure of the atmosphere at an elevation of 6000 m (the top of Mt.
Kilimanjaro)? You may assume that T = 273 K.
We must recognize that the the pressure is a function of height P(h) = ρgh. The change in
pressure dP is related to the change in height dh by dP = - ρg dh. The minus arises from the fact that
the pressure decreases as the height above the earth's surface increases. We need to know how the
density changes with height. We obtain this from the ideal gas law. The density is ρ = nM/V where n
is the number of moles, M is the molar mass and V is the volume. We can obtain n/V from the ideal
gas law: n/V = P/RT, thus ρ = MP/RT. We can now substitute the density into the pressure vs. height
formula.
MPg
dP = –
dh
RT
P
1
dP = – Mg
P
RT
ln P = –
h
dh
0
Mgh
RT
Mgh
RT
At this point you can plug in numbers. We will use M = 29 grams/mole as an average molar mass for
the atmosphere.
P = exp –
0.029 kg/ mol 9.8 m/s 2 6000 m
P = 1 atmospere exp –
8.31 J / mol – K 273 K
P = exp – 0.751 = 0.471 atmospheres
The big mistake you can make in a problem like this one is to get the units wrong. For example, if you
forget to convert the molar mass to kilograms and plug in the number 29 instead of 0.029 then you get
an answer of P = exp{ - 751}, which is essentially zero pressure.
--------------------------------------------------Temperature is perhaps the most difficult quantity to conceptualize. We shall see that
temperature depends on microscopic motions of molecules. However, the temperature can be defined
macroscopically based on the ideal gas law. Since the PV product cannot be less than zero, T cannot
be less than zero. This implies an absolute zero of temperature.
We can consider a definition of a thermometer based on the following theorem. We consider
object A that is in thermal equilibrium with object B. Further we can consider object B in equilibrium
with C. The zeroth law of thermodynamics states that A is in equilibrium with C. This law implies
that object C can act as a thermometer for other objects. The definition of a temperature scale based
on the properties of water is known as the Kelvin scale. The triple point of water is defined to be at
273.16 K. K represents the degrees in the Kelvin temperature scale. The triple point is the unique
point in the phase diagram where solid, liquid, and vapor coexist. The boiling point of water at 1bar of
pressure is defined as the being 100 degrees higher (373.16 K).
Ideal vs. Real Gases
All gases obey the ideal gas equation of state provided they are sufficiently dilute. The
ratio Z = PV/nRT = 1 at all pressures for an ideal gas. For a real gas Z is deviates from one as the
pressure increases. First Z < 1 at intermediate pressures of tens to several hundreds of bar due to the
attractive forces between molecules. At higher pressure the repulsive forces begin to dominate
resulting in Z > 1.
Microscopic view of the deviation of gases from ideal behavior
Real gases differ from ideal gases in two ways. First, they have finite
size. Secondly, there are forces acting between the particles or molecules
in a real gas. The microscopic description of these forces has a
consequence for the macroscopic equation of state.
Hard sphere gas
We can begin with the finite size. If we think of the gas a collection of
solid spheres of diameter d then the potential energy between them has
the form shown in the Figure below.
The hard sphere diameter is often called σ. The volume of the sphere is
4π(σ/2)3/3 = πσ3/6. In the hard sphere model of a gas we call the molar
volume of the hard spheres b. It appears in the gas law as follows:
P(V-nb) = nRT. In this equation nb is the "excluded volume". This
means that it is the volume occupied by the gas molecules themselves.
Conceptual problem: what is the maximum possible density of a hard
sphere gas?
Answer: A hard sphere gas differs from an ideal gas in that it has a finite
size. As you increase the pressure on a hard sphere gas you will
eventually start packing the spheres into a solid. Since the spheres are
"hard" you will reach a point where you cannot compact them any
further. To find the formula for that point we solve for the density of a
hard sphere gas.At infinite pressure RT/P Æ 0 and so ρ = M/b is the
maximum density.
P V – nb = nRT
P V
n – b = RT
V = RT + b
n
P
ρ = nM = M
RT + b
V
P
Van der Waal's gas: microscopic considerations
The van der Waal's gas includes both the finite extent of a hard sphere
gas and intermolecular attractions due to induced dipole-induced dipole
interactions. Next we consider the origin of attractive forces between
molecules. These are the so-called London dispersion forces and they
arise due to the fluctuating electron clouds in molecules.
London dispersion interactions or induced ipole-induced dipole interactions are always attractive.
μ 21α 2
μ 22α 1
u ind(r) = –
–
4πε 0 r 6
4πε 0 r 6
where α is the molecular polarizability.
Induced-dipole-induced-dipole interactions are dominant if the dipole moment is zero.
where I is the ionization potential. This term is also known as the dispersion term (or the London
dispersion attraction). Surprisingly, this is the dominant term in the attractive part of the potential
I 1I 2
α 1α 2
u disp(r) = – 3
2 I 1 + I 2 4πε 2r 6
0
energy surface.
If the molecule has a permanent dipole moment as well there are dipoledipole interactions.
where μ is the ground state dipole moment. Note that the factor of kT in the denominator signifies that
u d,d(r) = –
2μ 1μ 2 1
6
4πε 0 3kT r
thermal fluctuations tend to disrupt interactions between the dipole moments and to reduce the
magnitude of this term.
Potential Energy Surface
The attractive force is a mutual electrostatic interaction.
The repulsive force is due to the finite dimension of molecules.
One common model for the potential energy surface is the LennardJones potential.
c 12 c 6
u(r) = 12 – 6
r
r
Repulsive
Attractive
Energy
u(r)
Internuclear Distance , r
-12
The r term accounts for repulsion due to molecular radii. This term
replaces the hard sphere model. The r-6 term accounts for attraction at
larger distances.
Contributions to the r-6 term include both dipole-dipole and induceddipole-induced-dipole interactions.
Van der Waal's gas: macroscopic equation
d
Finite
Size
Electrostatic
Attraction
There are non-ideal equations of state designed to account for the deviations of gases from ideal
behavior. The best known of these is the van der Waals equation
Where V bar designates the molar volume. The constants a and b are called van der Waals constants,
and they depend on the gas being described. The parameter a depends on the strength of
intermolecular interactions. The parameter b depends on the size of the molecules.
P + a2 V – b = RT
V
The van der Waals equation can be rewritten a number of ways:
(a) in terms of pressure
P=
RT – a
2
V
V–b
(b)
in terms of the compressibility factor Z
Z=
V – a
RTV
V–b
(c)
in terms of the molar volume
3
2
V – b + RT V + a V – ab = 0
P
P
P
The last form shows clearly that the van der Waals equation of state is a cubic polynomial. This is
important since it implies that below a critical temperature there will be a phase transition. In other
words this simple equation also serves as an equation of state for the liquid!
Equations of State: Virial and van der Waal's
The ideal gas law (PV = nRT) assumes that atoms and molecules have no extent (no finite size) and
there are no interactions between particles.
The virial equation of state shows a natural connection between the microscopic and macroscopic
view.
The coefficients B(T) are called virial coefficients.
B (T) B3V(T)
Z = PV = 1 + 2V
+
+ ...
2
RT
V
V
Z is called the compressibility. The second virial coefficient is negative at low temperature.
A negative B2V represents the dominance of intermolecular attractions. B2V can be related to the
intermolecular potential through
∞
e – u(r) / kT – 1 r 2dr
B2V(T) = – 2πN A
0
NA is Avagadro’s number. kB is the Boltzmann constant.
kB = R/NA. kB is a microscopic constant that corresponds to the macroscopic universal gas constant, R.
If u(r) is the Lennard-Jones potential the equation is not analytic, but has been solved
numerically. This is an important equation since it establishes a connection between a microscopic
potential energy function and macroscopic term in the compressibility expansion.
The second virial coefficient can be expressed in terms of molecular volume analytically if we use a
hard sphere potential. The hard sphere potential is u(r) = ∞ for r < σ and u(r) = 0 for r > σ.
∞
e – u(r) / kT – 1 r 2dr
B2V(T) = – 2πN A
0
σ
∞
2
= 2πN A
e – 1 r 2dr
0 – 1 r dr 2πN A
0
σ
= 2πN A
0
0
0
3
NA
2πσ
2
r dr =
3
which is equal to four times the volume of NA hard spheres.
The van der Waals equation has a natural interpretation in terms of finite size since b is the volume
reduction due to the size of the gas molecules and a is a pressure reduction due to interactions between
molecules.
P = RT – a2
V–b V
The interpretation the van der Waals parameters can be obtained by writing a virial expansion
2
b
PV
1
a
Z=
=1+ b–
+
+ ...
RT
RT V V 2
Microscopic Interpretation of the van der Waals Equation
Since
B2V(T) = b – a
RT
we can use a hard sphere 1/r6 potential,
u(r) = ∞ for r < σ and u(r) = -c6/r-6 for r > σ.
∞
e – u(r) / kT – 1 r 2dr
B2V(T) = – 2πN A
0
σ
∞
– 1 r 2dr – 2πN A
= –2πN A
0
σ
≈
∞
2
2πN A
r dr – 2πN A
0
σ
e – c 6 / kTr 6 – 1 r 2dr
σ
r 2dr
r6
2πσ 3 N A 2πN Ac 6
=
–
3
3kTσ 3
2πσ 3 N A
2πN A2c 6
Thus b =
and a =
3
3σ 3
Isotherms
Isotherms are constant temperature curves in pressure-volume space (PV). Using the ideal gas
law we can plot a number of isotherms. To evaluate simple expression and plot functions while at the
computer we will use the program Maple. To plot an ideal gas isotherm (P = RT/V) we go to the
Maple prompt and type in the command
¾ with(plots):plot(0.082*300/v,v=0.1..1);
where we have used R = 0.082 L-atm/mol-K and T = 300 K. To see three isotherms on a plot
type
¾ plot({0.082*300/v,0.082*200/v,0.082*100/v},v=0.05..0.3);
Your plot should look like this
where the red, yellow, green curves are isotherms.
The {} bracket allows more than one function to be
plotted on the same graph as {function1, function2,..}.
We can make the same plots for a van der Waals gas. Let’s look at ammonia (a = 4.25
2
L atm/mol2 and b = 0.037 L/mol). At 298 K we have
¾ with(plots):plot(0.082*298/(v-0.037)-4.248/(v*v),v=0.05..1);
You will note that this isotherm has a strange shape. The reason is that the cubic polynomial is
giving rise to inflection. At high enough temperature this will no longer be important (and the
van der Waals equation will begin to approach the ideal gas law).
¾
¾
¾
plot({0.082*298/(v-0.037)-4.248/(v*v),0.082*400/(v-0.037)-4.248/(v*v),0.082*500/(v-0.037)4.248/(v*v)},v=0.05..0.3);
Note that we have changed the range on the abcissa (the molar volume axis) to better see how the
shape of the curves is changing. You can compare the ideal gas law to the van der Waals
equation at 300 K
with(plots):plot({0.082*300/v,0.082*300/(v-0.037)-4.248/(v*v)},v=0.05..0.3);
and 500 K
with(plots):plot({0.082*500/v,0.082*500/(v-0.037)-4.248/(v*v)},v=0.05..0.3);
The plot at 500 K shows more similarity than the plot at 300 K. Above a critical temperature there is
no inflection point and the curves for real and ideal equations of state begin to have the same shape.
This behavior corresponds to the known behavior of substances. They all have a critical temperature
(and pressure) beyond which there is no distinction between liquid and gas. Below the critical
temperature there will be a phase transition for a given temperature and pressure. Since the molar
volume changes (think about liquid water turning to vapor) there should be an abrupt change in the
molar volume as shown in Figure 16.7 in your text for isotherms below the critical point.
The critical point in a van der Waal's gas
The critical point is defined mathematically as an inflection point. At the critical temperature the
curve will no longer turn down. For ammonia the critical point must be between 300 K where the
isotherm shows a change in the sign of curvature and 500 K where it does not. At the inflection point
the slope of the curve will be zero and the curvature will be zero.
In pressure-volume space this means
2
∂P = 0 and ∂ P = 0
2
∂V
∂V
There is an elegant and simple way of solving this. They note that at the critical point there is only one
root. Therefore, the cubic equation can be written
3
3
2
2
3
V – Vc = 0 or V – 3VcV + 3Vc V – Vc = 0
The coefficient Vc bar can be equated with the coefficients above for the van der Waals equation of
state expressed as a cubic polynomial. A single point in P-V space is solved for, thus, there is a critical
temperature, Tc, pressure, Pc, and volume, Vc. To find these we equate the terms in the van der Waal's
expression to the above cubic polynomial. At the critical point:
3
V – b+
RTc 2 a
V + V – ab = 0
Pc
Pc
Pc
Equating the coefficients we have the following equations:
3V c = b +
2
3V c = a
Pc
3
V c = ab
Pc
RTc
Pc
Vc = 3b , Pc =
a , and T = 8a
c
27bR
27b 2
The critical values are:
in terms of the van der Waals parameters.
For example, we can now calculate the critical temperature of ammonia,
Tc = 8(4.25)/(27(0.039)(0.082)) = 393.7 K.
I did not explicity write out the units. I used R = 0.082 L-atm/mol-K in this case since we are working
in the L-atm units of P-V space.
In practice, experimental critical data are used to obtain the parameters a and b. We will see
further that the virial equation of state can be related to the van der Waals equation of state and to
parameters that describe molecular interactions.
The Law of Corresponding States
We can define reduced quantities
PR = P/Pc, VR = V/Vc, TR = T/Tc
By substitution into the van der Waals equation we find
PR + 32 VR – 1 = 8 TR
3
3
VR
This equation is a universal equation for all gases. Although the actual pressures and volumes may
differ, two gases are said to be in corresponding states if their reduced pressure, volume, and
temperature are the same.
The compressibility factor Z can also be cast into the form of corresponding states showing that
Z also can be expressed as a universal function of VR and TR or any other two reduced quantities.
Z =
VR
– 9
VR – 1/3 8V RTR