Post on 16-Dec-2015
transcript
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