Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 9.8ms2 F mg F 1kg 9.8ms 2 9.8kgms2 9.8kgm 106 mm2 9 2 P 0 . 98 10 kg / ms 1.0 102 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 /nk 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, 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: 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 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 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