ERT 108/3PHYSICAL CHEMISTRY

INTRODUCTION

Prepared by:Pn. Hairul Nazirah Abdul Halim

Chemistry is the study of matter and the changes it undergoes

What is Physical Chemistry?

• is the study of macroscopic, atomic, subatomic, and particulate phenomena in chemical systems in terms of physical laws and concepts.

• It applies the principles, practices and concepts of physics such as motion, energy, force, time, thermodynamics, quantum chemistry, statistical chemistry and dynamic.

Subtopic

• The properties of gases

a) The perfect gas

b) Real Gases

THE PERFECT GAS

• The general form of an equation of state:

• If we know the values of T, V and n for a particular substance, then the pressure has a fixed value.

• Equation of state of a ‘perfect gas’:

V

nRTp

),,( nVTfp

a) PRESSURE

• Definition: A force divided by the area to which the force is applied.

• p = F/A

• SI unit of pressure: Pascal (Pa) = Nm-2 = kgm-1s-2

• Pressure is measured with a barometer

Units of Pressure

• Pascal (Pa) 1Pa = 1 Nm-2

• Bar 1 bar = 105 Pa• Atmosphere 1 atm = 1.01325 bar• Torr 760 Torr = 1 atm• mmHg 1mmHg = 133.322 Pa• Psi 1 psi = 6.894757 kPa

• When a region of high pressure is separated from a low pressure by a movable wall, the wall will be pushed into one region.

• If the two pressure is identical, the wall will not move.

Example 1.1 Calculating pressure

Suppose Isaac Newton weighed 65 kg. Calculate the pressure he exerted on the ground when wearing boots with soles of total area 250 cm2 in contact with the ground.

Solution

ap P10 x 6.2Nm10 x 6.2m10 x 2.50

)ms kg)(9.81 (65 42-422-

-2

b) Temperature

• Is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall.

• Two types of boundary:

a) Diathermic – if a change of state is observed when two objects at different temp. are bought into contact. Example: metal container

b) Adiabatic – if no change occurs even though the two objects have different temperatures.

• Thermal equilibrium: no change of state occurs when two objects are in contact through a diathermic boundary.

• Zeroth Law Thermodynamics:

If A is in thermal equilibrium with B,

and B is in thermal equilibrium with C,

then C is also in thermal equilibrium with A.

a) Boyle’s law

b) Charles’s Law

c) Avogadro’s Principle

d) The perfect gas law

e) Mixture of gases

f) Mole fractions and partial pressures

The Gas Laws

constantpV

Vp

1p

V1

a) Boyle’s Law• At constant temp., the pressure of a sample gas

is inversely proportional to its volume.

• The volume it occupies is inversely proportional to its pressure:

Fig 1.5

The pressure-volume dependence of a fixed amount of perfect gas at different temperatures.

Each curve is a hyperbola

(pV = constant) and is called an isotherm.

• Straight lines are obtained when the pressure is plotted against 1/V at constant temp.

b) Charles’s Law

at constant pressure

at constant volume

TV constant x

Tp constant x

c) Avogadro’s principle

At constant pressure and temp.

nV constant x

d) The perfect gas law• perfect gas = ideal gas

R = gas constant

nRTpV

nT

PVR

Molar volume (Vm) of a perfect gas under Standard Ambient Temperature Pressure (SATP):

• Temp. at 298.15 K• Pressure at 1 bar (105 Pa)

1-

1-1-2

mol L 789.24

bar) 1(

K) 15.298)(molKbar L1031447.8(

m

m

m

V

xV

p

RTV

• To calculate the change in conditions when constant amount of gas is subjected to different temp., pressures and volume:

• Combined gas equation:

nRT

VP

1

11 nRT

VP

2

22

2

22

1

11

T

VP

T

VP

Example 1.3 Using combined gas equation

In an industrial process, nitrogen is heated to 500K in a vessel of constant volume. If it enters the vessel at 100 atm and 300K, what pressure would it exert at the working temp., if it behaved as a perfect gas?

Solution

atm 167 atm 100 x K 300

K 500

x

2

11

22

2

2

1

1

P

PT

TP

T

P

T

P

e) Mixture of Gases

•Dalton’s Law: the pressure exerted by a mixture of gases is the sum of the partial pressures of the gases.

•Where p = total pressure

pA = partial pressure of perfect gas A

pB = partial pressure of perfect gas B

... BA ppp

Example 1.4 Using Dalton’s Law

A container of volume 10.0L holds 1.00 mol N2 and 3.0 mol H2 at 298 K. What is the total pressure in atmospheres if each component behaves as a perfect gas?

Solution

atm78.9L 10.0

K)298)(mol K atm L 10 x (8.206 x mol) 3.00 mol 00.1(

)(

1-1-2-

p

p

V

RTnnppp BABA

f) Mole fractions and partial pressures

Where;

XJ = mole fraction; amount of J expressed as a fraction of the total amount of molecules, n in the sample.

n

nX JJ

... BA nnn

1... BA xx

Example Mole fractions.

A mixture of 1.0 mol N2 and 3.0 mol H2 consist of:

Mol fraction of N2 = 1.0 mol / (1.0 + 3.0 mol) = 0.25

Mol fraction of H2= 3.0 mol / (1.0 + 3.0 mol) = 0.75

Partial Pressure, pJ of a gas J in a mixture

• Where p = total pressure

The sum of partial pressures is equal to the total pressure:

pxp JJ

ppxxpp BABA ...)(...

Example 1.5 Calculating Partial Pressures

The mass percentage composition of dry air at sea level is approximately N2 = 75.5; O2 = 23.2; Ar = 1.3. What is the partial pressure of each component when the total pressure is 1.00 atm?

Solution

Assume; total mass of the sample = 100g.

mol 3.45 mol Total

mol 0.033gmol 39.95

g 100 x 013.0)(

mol 725.0gmol 32.00

g 100 x 232.0)O(

mol 69.2gmol 28.02

g 100 x 755.0)N(

1-

1-2

1-2

Arn

n

n

N2 O2 Ar

Mole fraction 0.780 0.210 0.0096

Partial pressure (atm) 0.780 0.210 0.0096

p

px JJ

Real Gases

• Real gases do not obey the perfect gas law exactly.

1.3 Molecular Interactions

• Real gases show deviations from the perfect gas law because molecules interact with each other.

• Repulsive forces between molecules assist expansion.

• Attractive forces assist compression.

a) The compression factor

Compression factor, Z is the ratio of its molar volume, Vm to the molar volume of a perfect gas,

at the same pressure and temp.

• For perfect gas, Z = 1• Deviation of Z from 1 is a measure of departure

from perfect behavior.

om

m

V

VZ

omV

Fig 1.13

The variation of compression factor, Z, with pressure for several gases at 00C.

• At very low pressures, all gases have Z≈1 and behave nearly perfectly.

• At high pressure, all the gases have Z > 1, signifying they have a larger molar volume than a perfect gas. Repulsive forces are dominant.

• At intermediate pressure, most gases have Z < 1. Attractive forces are reducing the molar volume.

b) Virial Coefficients

Fig 1.14

Experimental isotherms of carbon dioxide at several temperatures.

b) Virial Coefficients• Refer to Figure 1.14• At large molar volumes and high temp., the real-

gas isotherms do not differ greatly from perfect gas isotherms.

• The small different suggest that the perfect gas law is in fact the first term in an expression of the form:

• Known as ‘Virial equation of state’

...)1( 2'' pCpBRTpVm

...1

2mm

m V

C

V

BRTpV

• A more convenient expansion for many applications is:

• The term in parentheses can be identified with the compression factor, Z.

• Known as ‘Virial equation of state’

• First virial coefficient = 1• Second virial coefficient = B• Third virial coefficient = C

1.4 The van der Waals equation

Eq. 1.25a

The equation is often written in terms of the molar volume Vm = V/n.

Eq. 1.25b

Constant a and b = van der Waals coefficients.

2

V

na

nbV

nRTp

2mm V

a

bV

RTp

Example 1.6 Using the van der Waals equation to estimate a molar volume

Estimate the molar volume of CO2 at 500K and 100 atm by treating it as a van der Waals gas.

Solution

1. Arrange equation 1.25b to become:

023

p

abV

p

aV

p

RTbV mmm

023

223

223

23

2

2

p

abV

p

aV

p

RTbV

p

abV

p

aV

p

RTbVV

abaVRTVbVVp

bVV

abaVRTVp

V

a

bV

RTp

mmm

mmmm

mmmm

mm

mm

mm

• Refer to Table 1.5,

For CO2, a = 3.610 L2 atm mol-2

b = 4.29 x 10-2 L mol-1

31-322

21-2

1-2

1--1-12

)mol L(10x 55.1100

)10x 29.4)(1061.3(

)mol L(1061.3100

61.3

mol L453.0410.010x 29.4

mol L 410.0atm 100

K500mol K atm L10 x 2057.8

x

p

ab

xp

a

p

RTb

p

RT

• Thefore, on writing x = Vm, the equation to solve is:

• The acceptable root is x = 0.366

• Hence Vm = 0.366 L mol-1.

0)355.1()261.3(453.0 23 ExExx