Lecture 2
CHM3400
• Homework????
• Aims of the class:
Prosperities of real gases
Kinetic theory of gases
The Maxwell distribution of speed
• What is the equation of state for perfect
gases?
• What is the difference between perfect gas
and ideal gas?
• Define compression factor
What is van der Waals equation of state?
What is virial equation of state.
• Real gases
Perfect gas law – holds for real gases at low p (p<10 atm) and high T
At high pressure – deviation from ideal behavior
For a perfect gas: pVm = RT and thus pVm/RT = 1
Real gases:
Compressibility factor Z
Z  1 Z = pVm/RT = Vm/Vm with Vom = molar volume of perfect gas
Z > 1 Vm of gas is larger than Vm of perfect gas due to the repulsion between molecules
Z < 1 attractive forces are dominant
• Equation of state for real gas
Z = 1 + B/Vm + C/Vm2 + ….
B – second virial coefficient
C- third virial coefficient
(Virial: comes from that Latin word
vis, viris, meaning force)
Determined experimentally from
the plot of p V T plots for real gases
pVm/RT = 1 + B/Vm + C/Vm2 +…
p= RT/Vm (1 + B/Vm + C/Vm2 +…)
p = nRT/V (1 + B/Vm + C/Vm2 +…)
B/Vm >> C/V2m
Virial equation of state
• Van der Waal equation of state
p= nRT/(V-nb) – a(n/V)2
Gas
a (Pa m3) b(m3/mol)
nb term - is the very small approximate volume
occupied by
the molecules themselves
Helium
3.46 x 10-3 23.71 x 10-6
Neon
2.12 x 10-2 17.10 x 10-6
a(n/V)2 term - a is a positive proportionality
constant, taking into account that the pressure
is reduced due to the attractive forces
Hydrogen 2.45 x 10-2 26.61 x 10-6
(p + a(n/V)2)(V-nb) ) = nRT
Van der Waals equation is valid over
the wider range of volume and
pressure
Provides molecular interpretation of
equation of state
Carbon
dioxide
Water
vapor
3.96 x 10-1 42.69 x 10-6
5.47 x 10-1 30.52 x 10-6
Value of a parameter correlates
with boiling point
Limitation of perfect gases : ideal gas law does not predict that real gas can be liquidified
Curve T1: high T: the gas behavior corresponds to perfect gas (p  1/V)
Curve T2: low T: from point A to B:decrease in V leads to and increase in P, gas phase only
from B to C: decrease in V does not lead to increase in P, gas + liquid phase in
equilibrium, sharp interface between phases, change in V leads to change in
proportion between gas and liquid, pressure is called vapor pressure of the liquid
from point C to D: single phase system, only liquid phase, V nearly constant,
sharp increase in P
Increase in T- the region of
coexistance of two phases decreases
D
Curve T3: the isotherm shows a sharp
inflection point that corresponds to T,
P and V above which the gas can NOT
be liquidified. Critical P, V and T.
T1
T3 = Tc
Pc
C
B
A
Vc
T2
At Tc the density of liquid and gas are
equal
At T> Tc, the interface between the gas
and liquid can not be distinguished
Vapor: is the gaseous phase of substance below critical temperature
Gas is the gaseous phase of substance above critical temperature
Tc(H2O) = 374 oC
Tc (N2) = -147 oC
• Reduced temperature and pressure and volume
Type equation here.
Tr = T/Tc Pr= P/Pc, and Vr=V/Vc
Determination a and b parameters from Van der
Waals equation of state using Tc and Pc
𝑉𝑐 = 3𝑏
8𝑎
𝑇𝑐 =
27𝑅𝑏
𝑎
𝑃𝑐 =
27𝑏 2
• Liquefaction of gases - Joules –Thomson effect
Gases with low Tb can be liquefied by cooling
Gases with very low Tb are liquefied by expanding their
volume
Real gases – attractive forces between particles
- the gas is compressed and then allowed to expand
without heat exchange with it surrounding
increase in V leads to the expansion of gas, particles
have to overcome
attractive force
- the kinetic energy is converted into potential energy
- decrease of kinetic energy leads to decrease in T
- several cycles of expansion then lead to liquefy gas
• Kinetic model of gases
–
–
–
–
Gas particles are in constant motion
Particles have mass but volume is negligible
The molecules does not interact
Collisions are elastic (energy is transfer in form of
kinetic energy)
Derivation of pressure of gas
Pressure exerted by a gas of M enclosed in volume V
P = nMvrms2/3V (velocity – vector quantity, speed- scalar quantity)
v-root mean square speed
crms = v21/2 = (v12 + v22 + v32 +…+ vN2 /N)1/2
Pressure is proportional to the root mean square speed
and the kinetic energy of gas
• Molecules of the gas travel with kinetic energy
<Ek> = ½ mvrms2 and thus vrms = (2<Ek>/m)1/2
Mean speed, vmean = (v1 + v2 + v3 +…+ vN /N)
vmean = (8/3)1/2 vrms= 0.91vrms
p = nMv2/3V
pV = 1/3nM vrms 2 = nRT
rms speed of gas molecules is
proportional to T of the gas
Random motion of the gas
molecule = thermal motion
vrms = (3RT/M)1/2
<Ek> = 3/2 (RTm)/M = 3/2 RT/NA (using M=mN )
A
=
3/2 kBT (using R = kBNA; kB = 1.38x10-23 JK-1)
• Maxwell distribution of speed
– For large number of molecule, molecules do not
travel with the same speed
– Speed of molecules changes du to the individual
collisions
– Distribution of molecular speeds- expression that
determines what fraction of molecules travel with
speed in the range of c+dc
dN/N = f(c) dc ; f(c) is the Maxwell distribution of
speed
F(c) = 4c2(m/2 kBT)3/2e^(-mc2/2kBT)
100K
200K
300K
F(c) = 4c2(m/2 kBT)3/2e^(-mc2/2kBT)
As temperature increase, the distribution is wider, more fast moving molecule ;
Larger fraction of molecules have enough energy to overcome activation barrier
Heavy molecules move more slowly and have more narrow distribution of
speed, whereas lighter molecules move faster and thus have wider distribution
of speed
Most probable speed
mean speed
cmp = (2RT/M)1/2
cmean = (8RT/M)1/2
crms = (3RT/M)1/2
• Collision frequency: average rate of collisions made by
molecule (z)
z = (21/2 NAcp)/RT
• Mean free path () path between 2 collisions
 =c/z = RT/ (21/2 NAp)
•  - collision cross section – a molecule will hit another
molecule if the molecules is situated within the circle
of radii d
Diameter
of
molecule
 = d2
Z increases with P
Z increases with c
Since c  1/M2, heavy molecules
have lower frequency of collisions
Homework
Textbook: 1.12 (not in previous edition); 1.19 (1.21) ; 1.21 (not in the previous edition)
); 1.37a (1.23), 1.37b (1.24); 1.37c (1.35).
1. The van der Waals constants of the gas can be obtained from its critical constants
where:
a = (27R2Tc2/64Pc) and b = (RTc/8Pc). Given Tc= 562 K and Pc = 48.0 atm for benzene,
calculate it’s a and b value.
2. Consider the vitriol equation Z = 1+B’P +C’P2, which describe behavior of a gas at a
certain temperature. From the following plot of Z versus P, deduce the signs of B’ and
C’.
Z
1
P (atm)
3. A N2 molecule at 20 oC is released at sea level to travel upward. Assuming that the
temperature is constant and that the molecule does not collide with other molecules,
how far would it travel in meters) before it comes to the rest. Do the same calculation
for the He atom.
4. At certain temperature, the speeds of six gaseous molecules in a container are 2.0
ms-1; 2.2; 2.6; 2.7; 3.3; and 3.5 . Calculate root mean square speed, and the mean
speed of the molecules. These two values are close to each other, but the rootmean-square value is always the larger of the two. Why?
5. The Boyle temperature is the temperature at which the virial coefficient B is zero.
Therefore , a real gas behaves as an ideal at this temperature. Using your expression
for B in terms of van der Waals coefficients (a an b), homework #1; calculate the
Boyle temperature for argon, given a = 1.345 atmL2mol-2 and b = 3.22x10-2 L mol-1.
6. A sample of neon is heated from 300 K to 390 K. Calculate the percent in increase
in its kinetic energy.
7. Calculate the ratio of the number of O3 molecules with speed of 1300 ms-1 at 360
K to the number with that speed at 293 K.