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# ERT 108 Physical Chemistry INTRODUCTION-Part 2 - PowerPoint PPT Presentation

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.

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ERT 108 Physical Chemistry INTRODUCTION-Part 2

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## ERT 108 Physical ChemistryINTRODUCTION-Part 2

by

### 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

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.

• 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

Thermal equilibrium

Thermal equilibrium

Thermal equilibrium

B

C

• 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:

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

Movable wall

High pressure

Low pressure

Low pressure

High pressure

Equal pressure

Equal pressure

In mechanical equilibrium (P1=P2)

• Pressure

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

P= pressure, Pa

F= Force, N

A=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.

### 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.

is a constant

P and V are inversely proportional.

### Gas laws

P and T are directly proportional.

• Charle’s law

at constant mass and pressure

at constant mass and volume

constant

### Gas laws

• Equal volumes of gases at the same temperature and pressure

contain the same numbers of molecules.

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, 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:

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 atmand 300 K, what pressure would it exert at the working temperature if it behaved as an ideal gas?

### 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.

Ideal gas mixture

• Partial pressure, Pi of gas iin a gas mixture:

Where

• For an ideal gas mixture:

any gas mixture

### 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

At very low pressures, Z ≈ 1

At high pressures, Z>1

At intermediate pressure, Z<1

### Real gas equations

• Virial equation of state:

• van der Waals equation:

Compression factor, Z