Thermodynamics

Lecture 6

Ideal Gas Behavior

Non-ideal behavior

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Macroscopic variables P, T

Pressure is a force per unit area (P= F/A)

Temperature derives from molecular

motion (3/2RT = 1/2M<u2>)

The force arises from the

change in momentum as

particles hit an object and

change direction.

Greater average velocity

results in a higher

temperature. u is the velocity

M is molar

mass

Mass and molar mass

We can multiply the equation:

molar energy

by the number of moles, n, to obtain:

energy

32

RT = 12

M <u2>

32

nRT = 12

nM <u2>

Mass and molar mass

If m is the mass and M is the molar mass

of a particle then we can also write:

nM = Nm (N is the number of particles)

In other words nNA = N where NA is

Avagadro’s number.

32

nRT = 12

Nm <u2>

Kinetic Model of Gases Assumptions:

1. A gas consists of molecules that move randomly.

2. The size of the molecules is negligible.

3. There are no interactions between the gas molecules.

Because there are such large numbers of gas molecules

in any system we will interested in average quantities.

We have written average with an angle bracket.

For example, the average speed is:

<u2> = c2 =s1

2 + s22 + s3

2 + ... + sN2

N

c =s1 + s2 + s3 + ... + sN

N

The root-mean-square speed The ideal gas equation of state is consistent with an

interpretation of temperature as proportional to the kinetic

energy of a gas.

If we solve for <u2> we have the mean-square speed.

If we take the square root of both sides we have the r.m.s.

speed.

u2 = 3RTM

13

M u2 = RT

u2 1/2= 3RT

M

Example The r.m.s. speed of oxygen at 25 oC (298 K) is 482 m/s.

Note: M is converted to kg/mol!

u21/2

=3 8.31 J/mol–K 298 K

0.032 kg/mol= 481.8 m / s

The Maxwell Distribution Not all molecules have the same speed. Maxwell assumed

that the distribution of speeds was Gaussian.

As temperature increases the r.m.s. speed increases and

the width of the distribution increases. Moreover, the

functions is a normalized distribution. This just means

that the integral over the distribution function is equal to 1.

F(s) = 4 M2RT

3/2

s2exp –Ms2

RT

F(s)0

ds = 1

The Maxwell Distribution

Units of Pressure

Force has units of Newtons

F = ma (kg m/s2)

Pressure has units of Newtons/meter2

P= F/A = (kg m/s2/m2 = kg/s2/m)

These units are also called Pascals (Pa).

1 bar = 105 Pa = 105 N/m2.

1 atm = 1.01325 x 105 Pa

Units of Energy

Energy has units of Joules

1 J = 1 Nm

Work and energy have the same units.

Work is defined as the result of a force

acting through a distance.

We can also define chemical energy in

terms of the energy per mole.

1 kJ/mol

1 kcal/mol = 4.184 kJ/mol

Thermal Energy

Thermal energy can be defined as RT.

Its magnitude depends on temperature.

R = 8.31 J/mol-K or 1.98 cal/mol-K

At 298 K, RT = 2476 J/mol (2.476 kJ/mol)

Thermal energy can also be expressed on a

per molecule basis. The molecular

equivalent of R is the Boltzmann constant, kB.

R = NAkB

NA = 6.022 x 1023 molecules/mol

Extensive and Intensive Variables

Extensive variables are proportional to the

size of the system.

Extensive variables: volume, mass, energy

Intensive variables do not depend on the

size of the system.

Intensive variables: pressure, temperature,

density

Equation of state relates P, V and T

The ideal gas equation of state is

PV = nRT

An equation of state relates macroscopic

properties which result from the average

behavior of a large number of particles.

P

Macroscopic Microscopic

Microsopic view of momentum

A particle with velocity ux strikes a wall.

The momentum of the particle changes from mux

to –mux. The momentum change is Dp = 2mux.

a

b

c

ux

area = bc

Transit time

The time between collisions with one wall is

Dt = 2a/ux.

This is also the round trip time.

a

b

c

ux

area = bc

Transit time

The time between collision is Dt = 2a/ux.

velocity = distance/time.

time = distance/velocity.

a

b

c

ux

area = bc

Round trip

distance is 2a

The pressure on the wall

force = rate of change of momentum

The pressure is the force per unit area.

The area is A = bc and the

volume of the box is V = abc

F =Dp

Dt=

2mux

2a/ux

=mux

2

a

P = Fbc

=mux

2

abc=

mux2

V

Average properties

Pressure does not result from a single

particle striking the wall but from many

particles. Thus, the velocity is the average

velocity times the number of particles.

P =Nm ux

2

V

PV = Nm ux2

Average properties

There are three dimensions so the velocity along

the x-direction is 1/3 the total.

From the kinetic theory of gases

PV =Nm u2

3

ux2 = 1

3u2

12

Nm u2 = 32

nRT

Putting the results together

When we combine of microscopic view of

pressure with the kinetic theory of gases

result we find the ideal gas law.

This approach assumes that the molecules

have no size (take up no space) and that

they have no interactions.

PV = nRT

Pressure of a dense fluid

For a dense fluid (or a liquid) such as water

we can think of the pressure arising from

the weight of the column of fluid above the

point where the measurement is made.

The force is due to the mass of water m

(kg) accelerated by gravity (g = 9.8 m/s2).

P = = = = = rgh

where r is the density r = m/V.

F mg mgh mgh

A A Ah V

The dependence of atomspheric

pressure on altitude

We can think of the atmosphere is a fluid,

but it is not dense. Moreover, unlike water

the density of the atmosphere decreases

with altitude. Thus, at high elevations both

the pressure and the density are

decreased. To obtain the dependence of

pressure on height h above the earth’s

surface we use the ideal gas law to define

the density of an ideal gas.

The dependence of atomspheric

pressure on altitude

The density of an ideal gas is:

r = m/V = nM/V = MP/RT

The dependence of pressure on elevation

is:

We need to collect variables of integration on the

same side of the equation.

dPP

= –Mg

RTdh

dP = – rg dh = –MPg

RTdh

The barometric pressure formula

Then we integrate (assuming P0=1 at h=0):

dPPP0 = 1

P

= –Mg

RTdh

0

h

ln PP0

= –Mgh

RT

P = P0exp –Mgh

RTor P = exp –

Mgh

RTatm

Imperfect Gases

(Non-ideal or Real Gases)

NC State University

The Compression Factor

One way to represent the relationship between ideal and real

gases is to plot the deviation from ideality as the gas is

compressed (i.e. as the pressure is increased).

The compression factor is defined as:

Written in symbols this becomes:

Note that perfect gases are also called ideal gases.

Imperfect gases are sometimes called real gases.

CompressionFactor =Molar volumeof gas

Molar volumeof perfect gas

Z =Vm

Vm

perfect=

PVm

RT

The Compression Factor

A plot of the compression factor reveals that many gases

exhibit Z < 1 for low pressure. This indicates that attractive

forces dominate under these conditions.

As the pressure increases Z crosses 1 and eventually becomes

positive for all gases. This indicates that the finite molecular

volume leads to repulsions between closely packed gas

molecules. These repulsions are not including the ideal gas

model.

Attractive

Region

Repulsive

Region

The Virial Expansion

One way to represent the deviation of a gas from ideal (or

perfect) behavior is to expand the compression factor in

powers of the inverse molar volume. Such an expansion is

known as a virial expansion.

The coefficients B, C etc. are known as virial coefficients.

For example, B is the second virial coefficient.

Virial coefficients depend on temperature. From the preceding

considerations we see the B < 0 for ammonia, ethene, methane

and B > 0 for hydrogen.

Z = 1 + BVm

+ C

Vm2

+ ...

The Virial Equation of State

We write Z in complete form as:

An then solve for the pressure:

This expression is known as the virial equation of state.

Note that if B, C etc. are all equal to zero this is just the ideal

gas law. However, if these are not zero then this equation

contains corrections to ideal behavior.

PVm

RT= 1 + B

Vm

+ C

Vm2

+ ...

P = RTVm

1 + BVm

+ C

Vm2

+ ...

Relating the microscopic to

the macroscopic

Real gases differ from ideal gases in two ways.

First they have a real size (extent). The excluded volume

results in a repulsion between particles and larger pressure

than the corresponding ideal gas (positive contribution to

compressibility).

Secondly, they have attractive forces between molecules.

These are dispersive forces that arise from a potential energy

due to induced-dipole induced-dipole interactions.

We can relate the potential energy of a particle to the terms

in the virial expansion or other equation of state. While we

will not do this using math in this course we will consider the

graphical form of the potential energy functions.

Hard Sphere Potential

A hard sphere potential is the easiest potential to parameterize.

The hard sphere diameter corresponds to the interatomic

spacing in a closest packed geometry such as that shown

for the noble gas argon.

The diameter can be estimated

from the density of argon in

the solid state. The hard sphere

potential is widely used because

of its simplicity.

u(r) = r < s

u(r) =0 r > s

Ar Ar

Ar Ar

Ar Ar

Ar Ar

Ar Ar

Ar Ar

Ar Ar

Ar Ar

r

u(r

)

s

s

The Hard Sphere Equation of State

As a first correction to the ideal gas law

we can consider the fact that a gas has

finite extent. Thus, as we begin to decrease

the volume available to the gas the pressure

increases more than we would expect due

to the repulsions between the spheres of

finite molar volume, b, of the spheres.

P = nRTV – nb Gas molecule

of volume B

The Hard Sphere Model

Low density: These are ideal gas conditions

The Hard Sphere Model

Increasing density: the volume is V

b is the molar volume of the spheres.

The Hard Sphere Model

Increasing density

The Hard Sphere Model

Increasing density

The Hard Sphere Model

High density: At sufficiently high density

the gas becomes a high density fluid or

a liquid.

The Hard Sphere Model

Limiting density: at this density the hard

spheres have condensed into an ordered

lattice. They are a solid. The “gas” cannot

be compressed further.

If we think about the density in each of these cases we can

see that it increases to a maximum value.

The volume is nb

When the gas is

completely compressed.

The Lennard-Jones potential is a most commonly used

potential function for non-bonding interactions in atomistic

computer simulations.

V

LJ(R) = 4 s

R

12

– sR

6

The potential has a long-range attractive tail –1/r6, and

negative well depth , and a steeply rising repulsive wall

at R = s. Typically the parameter s is related to the

hard sphere diameter of the molecule. For a monoatomic

condensed phase s is determined either from the solid

state or from an estimate of the packing in dense liquids.

The well depth e is related to the heat of vaporization of

a monatomic fluid. For example, liquid argon boils at ~120K

at 1 atm. Thus, ~ kT or 1.38x10-23 J/K(120 K) = 1.65x10-21 J.

This also corresponds to 1.03 kJ/mol.

Lennard-Jones Potential Function

Graphical Representation L-J Potential

The L-J potential function has a steep rise when r < s.

This is the repulsive term in the potential that arises from

close contacts between molecules. The minimum is found

for Rmin = 21/6 s. The well depth is in units of energy.

Rmin

The van der Waal’s Equation of State

The microscopic terms and s in the L-J

potential can be related to the a and b

parameters in the van der Waal’s equation of

state below.

The significance of b is the same as for the

hard sphere potential. The parameter a is

related to the attractive force between

molecules. It tends to reduce the pressure

compared to an ideal gas.

P = nRTV – nb

– n2a

V2

The van der Waal’s Equation of State

in terms of molar volume

Recall that Vm = V/n so that the vdW equation

of state becomes:

We can plot this function for a variety of

different temperatures. As we saw for the

ideal gas these are isotherms. At sufficiently

high temperature the isotherms of the vdW

equation of state resemble those of the ideal

gas.

P = RTVm – b

– aVm

2

The argon phase diagram

Critical Point

For argon

Tc = 150.8 K

Pc = 4934.5 Pa

Vc = 74.9 cm3/mol

Significance of the critical point

Note that the vdW isotherms look very different

from those of the ideal gas below the critical

point. Below the critical point there are two

possible phases, liquid and gas. The liquid

phase is found at small molar volumes.

The gas phase is observed at larger molar

volumes. The shape of the isotherms is not

physically reasonable in the transition region

between the phases. Note that the implication

is that there is a sudden change in volume

for the phase transition from liquid to gas.

View of the liquid region

of the argon phase diagram

Phase Equilibrium Region

Liquid

Critical Parameters

The critical parameters can be derived in terms

of the vdW a and b parameters as well as the

gas constant R.

The derivation can use calculus since

the derivative of the vdW equation of

state is zero at the critical point.

Given that this is also an inflection point

the second derivative is also zero.

Pc = a27b

2

Tc = 8a27Rb

Vc = 3b

The critical point is a flat inflection point. This means

that both the first and second derivatives on the curve

vanish at that point. These derivatives are:

The solution of these equations (and the van der Waal;s

equation itself leads to the following values for the critical

constants:

The critical values, Pc, Vc and Tc, can be used to calculate a

critical compression factor.

The critical compression factor is predicted to be the same

for all gases. This fact leads to the principle of

corresponding states. This principle states that all gases

will have identical compression curves if they are

normalized to their critical constants. We define the

reduced, pressure, volume and temperature as:

The Law of Corresponding States