Presentation is loading. Please wait.

Presentation is loading. Please wait.

ERT 108 Physical Chemistry INTRODUCTION-Part 2 by Miss Anis Atikah binti Ahmad.

Similar presentations


Presentation on theme: "ERT 108 Physical Chemistry INTRODUCTION-Part 2 by Miss Anis Atikah binti Ahmad."— Presentation transcript:

1 ERT 108 Physical Chemistry INTRODUCTION-Part 2 by Miss Anis Atikah binti Ahmad

2 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.

3 Example: Thermal Equilibrium Both pressures change. Reach the same value after some time. Wall is diathermal In thermal equilibrium (T 1 =T 2 )

4 Example No pressure change. P 1 ≠ P 2. Wall is adiabatic Not in thermal equilibrium

5 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 B

6 High pressure  Example: Mechanical equilibrium Low pressure Equal pressure Low pressure High pressure In mechanical equilibrium (P 1 =P 2 ) 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.

7  Pressure  The greater the force acting on a given area, the greater the pressure P= pressure, Pa F= Force, N A=Area, m 2

8 Exercise:  Calculate the pressure exerted by a mass of 1.0 kg pressing through the point of a pin of area 1.0 x mm 2 at 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.

9 Solution:  Calculate the pressure exerted by a mass of 1.0 kg pressing through the point of a pin of area 1.0 x mm 2 at 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.

10 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.

11 Gas laws  Charle’s law at constant mass and pressure at constant mass and volume constant P and T are directly proportional.

12 Gas laws  Avogadro’s principle;  Equal volumes of gases at the same temperature and pressure contain the same numbers of molecules. at constant pressure and temperature

13  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.

14 Ideal Gas  Ideal gas is a gas that obeys ideal gas law: Gas Constant Ideal gas law

15 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?

16 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

17  Partial pressure, P i of gas i in a gas mixture: Where  For an ideal gas mixture: any gas mixture

18 Exercise  The mass percentage composition of dry air at sea level is approximately N 2 = 75.5, O 2 =23.2, Ar= 1.3 What is the partial pressure of each component when the total pressure is 1.20 atm?

19 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)

20 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.

21 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.

22 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

23 Real gas equations  Virial equation of state:  van der Waals equation: Compression factor, Z


Download ppt "ERT 108 Physical Chemistry INTRODUCTION-Part 2 by Miss Anis Atikah binti Ahmad."

Similar presentations


Ads by Google