Dr.Salwa Al Saleh Salwams@ksu.edu.sa

Lecture 7Lecture 7

Deviations from IdealityDeviations from Ideality Van Der Waals Equation

Virial equations of state Compressibility factorCompressibility factor

Derivation of some laws from Derivation of some laws from kinetic theory of gases kinetic theory of gases

KINETIC THEORY

For a single molecule, the average force is:

For N molecules, the average force is:

root-mean-squarespeed

volume

Example : The Speed of Molecules in Air

Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0u) and oxygen O2 molecules (molecular mass 32.0u). Assumethat each behaves as an ideal gas and determine the rms speedsof the nitrogen and oxygen molecules when the temperature of the airis 293K.

For nitrogen…

The internal energy of monatomic ideal gas

Kinetic Energy

% of Molecules

High temp.

Low temp.

Kinetic Energy

% of Molecules

High temp.

Low temp.

Few molecules have very high kinetic energy

Kinetic Energy

% of Molecules

High temp.

Low temp.Average kinetic energies are temperatures

Kinetic Theory – SummaryKinetic Theory – Summary

Using Newtonian mechanics we have established:

◦ the relationship between p, N/V, T;

◦ the universality of the gas constant;

◦ the relationship between temperature and K.E.

◦ the internal energy of a monatomic ideal gas

Real GasesReal Gases

J. van der Waals, 1837-1923, J. van der Waals, 1837-1923, Professor of Physics, Professor of Physics, Amsterdam. Nobel Prize 1910.Amsterdam. Nobel Prize 1910.

Real Gases Real Gases General ObservationsGeneral Observations

Deviations from ideal gas law are particularly important at high pressures and low temperatures

Real gases differ from ideal gases in that there can be interactions between molecules in the gas state◦ Repulsive forces important only when molecules are

nearly in contact, i.e. very high pressures Gases at high pressures , gases less compressible

◦ Attractive forces operate at relatively long range (several molecular diameters) Gases at moderate pressures are more compressible

since attractive forces dominate◦ At low pressures, neither repulsive or attractive forces

dominate → ideal behavior

→

Real Gases Real Gases Deviations from IdealityDeviations from Ideality

J. van der Waals, 1837-1923, J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Professor of Physics, Amsterdam. Nobel Prize 1910.Nobel Prize 1910.

. The Ideal Gas Law ignores both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive

In reality, all gases have a volume and the molecules of real gases interact with one another.

For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis.

Real GasesReal GasesDeviations from IdealityDeviations from Ideality

For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis.

Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally.

The reasons for the deviations from ideality are:• The molecules are very close to

one another, thus their volume is important.

• The molecular interactions also become important.

Real Gases Real Gases Van Der Waals EquationVan Der Waals Equation

Real gases do not follow PV = nRT perfectly. The van der Waals equation corrects for the nonideal nature of real gases.

a corrects for interaction between atoms.

b corrects for volume occupied by atoms.

Real Gases Real Gases Van Der Waals EquationVan Der Waals Equation

A non-zero volume of molecules = “nb” (b is a constant depending on the type of gas, the 'excluded volume'). The molecules have less free space to move around in, so replace V in the ideal gas equation by V - nb

Very roughly, b 4/3 r3 where r is the molecular radius.

In the van der Waals equation ,

where “nb” represents the volume occupied by “n” moles of molecules.

Real Gases Real Gases Van Der Waals EquationVan Der Waals Equation

The attractive forces between real molecules, which reduce the pressure:

p wall collision frequency and p change in momentum at

each collision.

Both factors are proportional to concentration, n/V, and p is reduced by an amount a(n/V)2, where a depends on the type of gas.

[Note: a/V2 is called the internal pressure of the gas].

Also, in the van der Waals equation ,

where “n2a/V2” represents the effect on pressure to intermolecular attractions or repulsions.

Chapter 2 : 212 phys

So…

" p = nRT/V " becomes p = nRT/(V - nb) - a(n/V)2

i.e. [ p + an2/V2 ][ V - nb ] = nRT

This is what already could ″Van der Waals equation of state″

When V or T are very large i.e. at low p or high T, then this van der Waals equation of state becomes equivalent to the ideal equation.

a and b are empirical Van der Waals constants.

It is easy to solve for p given V.

To find V given p you need to solve a cubic equation (with Vm=V/n)

V.D.WV.D.W Cubic equations of state Cubic equations of state

Simple equation capable of representing both liquid and vapor behavior.

The van del Waals equation of state:◦ a and b are positive constants◦ unrealistic behavior in the two-phase region. In

reality, two, within the two-phase region, saturated liquid and saturated vapor coexist in varying proportions at the saturation or vapor pressure.

◦ Three volume roots, of which two may be complex.

◦ Physically meaningful values of V are always real, positive, and greater than constant b.

Real GasesReal GasesDeviations from IdealityDeviations from Ideality

• Problem • Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container

at 200. oC using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 L mol-1

n = 84.0g x 1mol/17 g T = 200 + 273P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2

5 L – (4.94 mol x3.71E-2 L mol-1) (5 L)2

P = 39.81 atm – 4.07 atm = 35.74P = 38.3 atm 7% error

Real GasesReal GasesNonideal

Conditions when gas gets close to conditions where it will liquify

◦ Lower Temperature

◦ Higher Pressure

Chapter 2 : 212 phys

Deviations from ideality can be described by the COMPRESSION FACTOR, Z (sometimes called the compressibility).

For ideal gases Z = 1 always.

Real GasesReal Gases Compressibility factor, ZCompressibility factor, Z

Compressibility factor, ZCompressibility factor, Z

Compression factor, Z, is ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same T & P◦ Z = Vm/ (Vm)id , where Vm = V/n

Using ideal gas law, p Vm = RTZ The compression factor of a gas is a

measure of its deviation from ideality◦ Depends on pressure (influence of

repulsive or attractive forces)◦ z = 1, ideal behavior◦ z < 1 attractive forces dominate,

moderate pressures◦ z > 1 repulsive forces dominate, high

pressures

…… Compressibility factor, ZCompressibility factor, Z

“compressibility factor”:Ideal gas: z = 1

z < 1: attractive intermolecular forces dominate

z > 1: repulsive intermolecular forces dominate

T=300K

EXAMPLE What is the molar volume Vm for an ideal gas at SATP?

V = nRT/p = 0.0248 m3 = 24.8 dm3 = 24.8 L = 24800 cm3

Z = PVm/RT =Vm/ (Vm)id

Chapter 2: 212 phys

Microscopic interpretation:

When p is very high, r is small so short-range repulsions are important. The gas is more difficult to compress than an ideal gas, so Z > 1.

When p is very low, r is large and intermolecular forces are negligible, so the gas acts close to ideally and Z 1.

At intermediate pressures attractive forces are important and often Z < 1.

Critical point Critical point of of Van Der Waals Van Der Waals EquationEquation

At fixed P and T, V is the solution of a cubic equation. There may be one or three real-valued solutions.

The set of parameters Pc, Vc, Tc for which the number of

solutions changes from one to three, is called the critical

point. The van der Waals equation has an inflection point at

Tc.

Critical point Critical point of of Van Der Waals Van Der Waals EquationEquation

Different gases have different values of p, V and T at their critical point

Comparing Different Gases

Chapter 2 : 212 phys

Comparing different gasesdifferent gases fall on the same curves.

Chapter 2: 212 phys

V

p

The isotherm at Tc has a horizontal inflection at the

critical point → dp/dV = 0 and d2p/dV2 = 0.

At the critical temperature the densities of the liquid and gas become equal - the boundary disappears. The material will fill the container so it is like a gas, but may be much denser than a typical gas, and is called a 'supercritical fluid'.}

The general Van der Waals pVT surface

}

Critical point Critical point of of Van Der Waals Van Der Waals EquationEquation

Chapter 2 : 212 phys

Consider 1 mol of gas, with molar volume V, at the critical point (Tc, pc, Vc)

0 = dp/dV = -RTc(Vc-b)-2 + 2aVc-3

0 = d2p/dV2 = 2RTc(Vc-b)-3 - 6aVc-4

The solution is Vc = 3b, pc = a/(27b2), Tc = 8a/(27Rb).

THE PRINCIPLE OF CORRESPONDING STATESDefine reduced variables pr = p/pc

Tr = T/Tc

Vr = Vm/Vm,c

Van der Waals hoped that different gases confined to the same Vr at the same Tr would have the same pr.

Chapter 2: 212 Phys

Proof:rewrite Van der Waals equation for 1 mol of gas, p = RT/(V-b)-a/V2, in terms of reduced variables:

Substitute for the critical values:

ThusThus all gases have the same reduced equation of state (within the Van der Waals approximation).

Chapter 2 : 212 phys

Chapter 1 :

Slide 38

VAN DER WAALS LOOPSAn artifact of the equation of state below Tc.

V

p

The general Van der Waals pVT surface

ProblemProblem

If sulfur dioxide were an “ideal” gas, the pressure at 0°C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure.

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

SolutionSolution

First, let’s rearrange the van der Waals equation to solve for pressure.

R= 0.0821 L. atm/mol. KT = 273.2 KV = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

ProblemProblem

The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the

“ideal” pressure.

Virial equations of Virial equations of statestatePV along an isotherm:

◦ The limiting value of PV as P →0 for all the gases: ◦ ◦ , with R as the proportionally

constant.◦ Assign the value of 273.16 K to the temperature of

the triple point of water: Ideal gas:

◦ the pressure ~ 0; the molecules are separated by infinite distance; the intermolecular forces approaches zero.

◦

Chapter 2 : 212 phys

VIRIAL COEFFICIENTS and the Boyle temperature

We can consider the perfect gas law as the first term of a general expression

p Vm = RT (1 + B’p + C’p2 + ...)

i.e. p Vm = RT (1 + B/Vm + C/Vm2 + ...)

This is the virial equation of state and B and C are the second and third virial coefficients. The first is 1.

B and C are themselves functions of temperature, B(T) and C(T).

Usually B/Vm >> C/Vm2

Chapter 2 : 212 phys

Consider dZ/dp. This is zero for a perfect gas since Z is constant.

dZ/dp = B’ + 2pC’ + ....In the limit as p tends to zero, dZ/dp = B’ which is zero only when B = 0.

The temperature at which this occurs is the Boyle temperature, TB, and then the gas behaves ideally over a wider range of p than at other temperatures.

Each gas has a characteristic TB, e.g. 23 K for He, 347 K for air, 715 K for CO2.

Real Gases - Real Gases - Other Equations of Other Equations of StateState

Virial equation is phenomenolgical, i.e., constants depend on the particular gas and must be determined experimentally

Other equations of state based on models for real gases as well as cumulative data on gases◦ Berthelot (1898)

Better than van der Waals at pressures not much above 1 atm

a is a constant

◦ Dieterici (1899)

Chapter 2 : 212 phys

SUMMARY

• The STATE of a system is specified by a few variables, that may be linked by an EQUATION OF STATE e.g. for IDEAL GASES pV = nRT. ISOTHERMS and ISOBARS.

• Dalton's Law of partial pressures.

• The VIRIAL EQUATION and the BOYLE TEMPERATURE.

• REAL GASES: the COMPRESSION FACTOR and INTERMOLECULAR FORCES. pV diagrams: the CRITICAL POINT.

Question timeQuestion time!!

Question1

Consider a fixed volume of gas. When N or T is doubled the pressure doubles since pV=NkT

T is doubled: what happens to the rate at which a molecule hits a wall?

(a) 1 (b) 2 (c) 2

N is doubled: what happens to the rate at which a molecule hits a wall?

(a) 1 (b) 2 (c) 2

Question 2Question 2

Container A contains 1 l of helium at 10 °C, container B contains 1 l of argon at 10 °C.

a) A and B have the same internal energy

b) A has more internal energy than Bc) A has less internal energy than B

Question 3Question 3

Container A contains 1 l of helium at 10 °C, container B contains 1 l of argon at 10 °C.

a) The argon and helium atoms have the same average velocity

b) The argon atoms are on average faster than the helium atoms

c) The argon atoms are on average slower than the helium atoms

Question 4Question 4

Container A contains 1 l of helium at 10 °C, container B contains 1 l of helium at 20 °C.

a) The average speeds are the sameb) The average speed in A is only a little

higher c) The average speed in A is about 2 higherd) The average speed in A is about twice as

high