ERT 108 Physical ChemistryINTRODUCTION-Part 2

by

Miss Anis Atikah binti Ahmad

Thermodynamic- Basic concepts (cont.) Equilibrium:

Variable (eg: pressure, temperature, & concentration) does not change with time

Has the same value in all parts of the system and surroundings.

Thermal equilibrium: No change of temperature occurs when two objects A and B are in contact through a diathermic boundary (thermally conducting wall).

Mechanical equilibrium: No change of pressure occurs when two objects A and B are in contact through a movable wall.

Example: Thermal Equilibrium

Both pressures change.

Reach the same value after some time.

Wall is diathermal

In thermal equilibrium

(T1=T2)

Example

No pressure change.

P1≠ P2.

Wall is adiabatic

Not in thermal equilibrium

Thermodynamic- Basic concepts (cont.) Zeroth Law of thermodynamics:

Two systems that are each found to be in thermal equilibrium with a third system will be found to be in thermal equilibrium with each other.

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

C

Thermal equilibrium

Thermal equilibrium

Thermal equilibrium

B

High pressure

Example: Mechanical equilibrium

Low pressure

Equal pressure

Equal pressure

Low pressure

High pressure

In mechanical equilibrium

(P1=P2)

Movable wall

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

There will come a stage when two pressures are equal and the wall has no tendency to move.

Pressure

The greater the force acting on a given area, the greater the pressure

A

FP

P= pressure, PaF= Force, NA=Area, m2

Exercise: Calculate the pressure exerted by a mass of 1.0

kg pressing through the point of a pin of area 1.0 x 10-2 mm2at the surface of the Earth. The force exerted by a mass m due to gravity at the surface of the Earth is mg, where g is the acceleration of free fall.

Solution: Calculate the pressure exerted by a mass of 1.0

kg pressing through the point of a pin of area 1.0 x 10-2 mm2at the surface of the Earth. The force exerted by a mass m due to gravity at the surface of the Earth is mg, where g is the acceleration of free fall.

A

FP mgF 28.9 msg

22 8.98.91 kgmsmskgF

292

26

222/1098.0

10

100.1

8.9mskg

m

mm

smm

kgmP

GPa98.0

Gas laws Boyle’s law

at constant mass and temperature

A decrease in volume causes the molecules to hit the wall more often, thereby increasing the pressure.

kPV is a constant

P and V are inversely

proportional.

Gas laws Charle’s law

at constant mass and pressure

at constant mass and volume

kTV / constant

P and T are directly proportional.

kTP /

Gas laws Avogadro’s principle;

Equal volumes of gases at the same temperature and pressure

contain the same numbers of molecules.

knV / at constant pressure and temperature

Boyle’s and Charle’s law are examples of a limiting law that are strictly true only in a certain limit, p0

Reliable at normal pressure (P≈1 bar) and used widely throughout chemistry.

Ideal Gas Ideal gas is a gas that obeys ideal gas law:

nRTPV

Gas Constant

Ideal gas law

Exercise In industrial process, nitrogen is heated to 500

K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as an ideal gas?

11 nRTVP

22 nRTVP 2

2

1

1

T

P

T

P

21

12 TT

PP

atmKK

atmP 167500

300

1002

Ideal Gas Mixture Dalton’s law:

The pressure exerted by a mixture of gases is the sum of the pressure that each one would exert if it occupied the container alone.

VRTnVRTnVRTnP tot /..// 21

RTnPV totIdeal gas mixture

Partial pressure, Pi of gas i in a gas mixture:

Where

For an ideal gas mixture:

PxP ii

totii nnx /

any gas mixture

VRTnnnPxP tottotiii //

VRTnP ii /

Exercise The mass percentage composition of dry air at

sea level is approximately N2= 75.5, O2=23.2, Ar= 1.3What is the partial pressure of each component

when the total pressure is 1.20 atm?

Real gas Real gas do not obey ideal gas law except in

the limit of p0 (where the intermolecular forces can be neligible)

Why real gases deviate from ideal gas law?

Because molecules interact with one another. (there are attractive and repulsive forces)

Real gas- molecular interaction At low P, when the sample occupies at large

volume, the molecules are so apart for most time that the intermolecular forces play no significant role, and behaves virtually perfectly/ideally.

At moderate P, when the average separation of the molecules is only a few molecular diameters, the ATTRACTIVE force dominate the repulsive forces. The gas can be expected to be more compressible than a perfect gas because the forces help to draw the molecules together.

Real gas- molecular interaction At high pressure, when the average

separation of molecules is small, the repulsive force dominate, and the gas can be expected to be less compressible because now the forces help to drive molecules apart.

Real gas Compression factor, Z

The extent of deviation from ideal gas behaviour is calculate using compression factor, Z

idealm

m

V

VZ

,

PRTnVVm //

RTpVZ m /

ZRTpVm

At very low pressures, Z ≈ 1At high pressures, Z>1At intermediate pressure, Z<1

Real gas equations Virial equation of state:

van der Waals equation:

2

2

V

na

nbV

nRTP

...1

2mm

m V

C

V

BRTPV

Compression factor, Z