Download Introduction to Physical Chemistry-Part 2

ERT 108 Physical Chemistry
INTRODUCTION-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
Wall is diathermal
Both pressures change.
Reach the same value
after some time.
In thermal
equilibrium (T1=T2)
Example
Wall is adiabatic
No pressure change.
P1≠ P2.
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.
A
Thermal
equilibrium
B



Thermal
equilibrium
Thermal
equilibrium
C
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.

Example: Mechanical equilibrium
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:
Movable wall
High
pressure
Low
pressure
Low
pressure
High
pressure
There will come a stage when two pressures are equal and the wall has no
tendency to move.
Equal
pressure
Equal
pressure
In mechanical
equilibrium (P1=P2)

Pressure
F
P
A

P= pressure, Pa
F= Force, N
A=Area, m2
The greater the force acting on a given area, the greater the
pressure
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.
F
P
A
g  9.8ms2
F  mg


F  1kg 9.8ms 2  9.8kgms2
9.8kgm
106 mm2
9
2
P


0
.
98

10
kg
/
ms
1.0 102 mm2 s 2
m2
 0.98GPa
Gas laws

Boyle’s law
PV  k
is a constant
at constant mass and temperature
P and V are inversely
proportional.
A
decrease in volume causes the
molecules to hit the wall more often,
thereby increasing the pressure.
P and T are directly
proportional.
Gas laws

Charle’s law
V /T  k
constant
at constant mass and pressure
P /T  k
at constant mass and volume
Gas laws

Avogadro’s principle;
V /nk
at constant pressure and temperature

Equal volumes of gases at the same temperature and pressure
contain the same numbers of molecules.

Boyle’s and Charle’s law are examples of a limiting law
that are strictly true only in a certain limit, p0

Reliable at normal pressure (P≈1 bar) and used widely
throughout chemistry.
Ideal Gas

Ideal gas is a gas that obeys ideal gas law:
PV  nRT
Ideal gas law
Gas Constant
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?
P1V  nRT1
P2V  nRT2
P1 P2

T1 T2
P1
P2   T2
T1
100atm
P2 
 500 K  167atm
300 K
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.
P  n1RT / V  n2 RT / V  ..  ntot RT / V
PV  ntot RT
Ideal gas mixture

Partial pressure, Pi of gas i in a gas mixture:
Pi  xi P
Where

any gas mixture
xi  ni / ntot
For an ideal gas mixture:
Pi  xi P  ni / ntot ntot RT / V 
Pi  ni RT / V
Exercise

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 1.20 atm?
Real gas


Real gas do not obey ideal gas law except in the limit of
p0 (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
Z
Vm
Vm  V / n  RT / P
Vm,ideal
Z  pVm / RT
pVm  ZRT
At very low pressures, Z ≈ 1
At high pressures, Z>1
At intermediate pressure, Z<1
Real gas equations

Virial equation of state:


B C
PVm  RT 1 
 2  ...
 Vm Vm


van der Waals equation:
nRT
n2
P
a 2
V  nb
V
Compression
factor, Z