ENGINEERING PHYSICS-I

1. ENGINEERING PHYSICS-I

3. EVALUATION • Pre mid term Snap Test : 05 • Mid term exam : 20 • Post mid term Snap Test : 05 • Final Exam : 70 (Theory:50,Lab:20) • Total : 100

4. Engineering Physics by R.K. Gaur and S.L Gupta, (Dhanpat Rai Pub., New Delhi) Heat and Thermodynamics by Brijlal and N Subramaniyam (S. Chand & Co.) Fundamentals of Physics by Haliday Resnik, Walker (Wiley, 6th edition)

6. MECHANICS X THERMODYNAMICS MECHANICS THERMODYNAMICS Deals about the transfer of heat and its conversion to other forms of energy • Deals only about the motion.

7. KINETIC THEORY OF GASES

8. KINETIC THEORY • Derives all Thermodynamic properties of the system using particle dynamics. • This is the bridge between Mechanics and Thermodyanmics.

9. POSTULATES: • Smallest indivisible particles (Molecules) • Distance between molecules is large • The molecules are continously in motion • Size of molecules is negligible • Time of impact is negligible • Molecules are perfectly hard elastic spheres

10. Pressure of a Gas M- Total mass of Gas. V- Volume.

11. MEAN KINETIC ENERGY

12. RMS velocity m- mass of a molecule M- mass of one mole of gas.

13. Think over it! GasVrms Velocity (m/s) • Hydrogen 1920 • Helium 1370 • Water Vapour 645 • Nitrogen 517 • Oxygen 483 • Carbon dioxide 412 • Sulphur dioxide 342 Why does it takes less than a minute before you smell perfume , when someone opens a bottle across room?

14. DISTRIBUTION OF VELOCITY: MAXWELL’S LAW All the molecules are not in same velocity. Maxwell studied this problem statistically and he plotted a graph between Number of molecules and Velocity. From the graph it is clear that for a particular velocity number of molecules is very large. That velocity is known as most probable velocity.

15. MEAN FREE PATH It is defined as average distance travelled by a molecule between two successive collisions.

16. d- diameter of molecule All molecule will have a sphere of influence around it and the radius of this sphere is equal to the diameter of the molecule. Only the molecule under consideration is assumed to be in motion and all other molecules are at rest.

17. As the molecule moves, sphere of influence will also move and it will sweep a cylinder of radius d. Volume of the cylinder swept out by sphere of influence in one second, The number of collisions made by molecule in one second is given by, n- number of molecule per volume.

18. Therefore average time taken by molecule between two collisions Mean free path According to Maxwell's law, all the molecule will have a certain velocity. So the corrected mean free path is given by,

19. TRANSPORT PHENOMENA • A molecule in motion can be considered as a transporter of energy or momentum. • In equilibrium state there would be no net transport. • In non equilibrium state, the energy and momentum will be transferred from one region to other. Such phenomena is known as transport phenomena. • Example: Thermal conduction of Heat.

20. VISCOSITY OF GASES • It is the property of the fluids due to which they oppose relative motion between the adjacent layers. • Newton found that viscous force is directly proportional to velocity gradient and surface area of the layer. • Maxwell explained the phenomena on basis of Kinetic theory of Gases.

21. Consider a gas in motion along a horizontal surface. Assume gas to be consisting of layers over layers. Velocity of gas in contact with wall is Zero. Velocity increases with increase in distance from the fixed layer.

22. Velocity of layer at height x, Velocity of layer at height x- , Velocity of layer at height x+ ,

23. Molecule moves in all possible random direction 1/3 of total no . of molecules moved along any one axis , out of these 1/3 half will move in one direction and remaining half in other direction Number of molecules crossing unit area in one second should be “nC/6” Thus mass of the total molecules crossing unit area in one second M = mnC/6. Momentum in forward direction = Momentum in opposite direction =

24. The rate of change of momentum, Thus viscosity of gas is, If A is unity

25. We know that product of is constant. And Therefore,

26. VANDER WALL’S EQUATION OF STATE • Equation of state of ideal gas is, PV=nRT • The following assumptions are used to derive the above equation of state. • The molecules are point particles and occupy no space. • There is no force of attraction or repulsion between the two molecules.

27. In real world, molecules in the gas occupy some volume and also attract each other. So, PV=nRT is not strictly true in the case of real gases. Thus Vander Waal corrected the above equation of state by applying correction for the size and attraction force of the gas molecules.

28. Correction for intermolecular attraction The identical molecule will attract each other. If the molecule is well inside the vessel, it will experience an attractive force in all possible directions, as a result net force will be zero. If the molecule is very near to the wall, there will be a net attraction towards the interior. So the velocity of molecule which strikes the wall of the vessel is reduced due to mutual attraction, thus decreasing Pressure on the walls.

29. The magnitude of this decrease in pressure is proportional to the number molecules present per unit volume and to the number of molecules striking the unit area of the walls of the vessel per second. Both of these proportional to density. Decrease in pressure proportional to (density)2

30. Correction for the size of Molecule If the molecules have a definite volume, the effective volume available to the molecule for the movement will be less than the actual volume V of gas. Let b is the decrease in volume, then * Substituting corrected values of pressure and volume, the perfect gas equation is, This is Vander Waal’s equation of state

31. Vander Waal’s Equation and Critical Constants Vander Waal’s equation of state can be written as, At the point X & Thus point X is known as Critical point and pressure, volume and temperature corresponding to this point are called critical pressure, critical volume and critical temperature and the curve is known as critical isotherm.

32. Critical Temperature Critical Pressure Critical Volume Critical coefficient

33. JOUL’S EXPERIMENT (b) Joul’s experiment with different calorimeter

34. Two vessels v1 and v2 were connected with a tube having stop cock C V2 was exhausted and V1 was filled with dry air at a pressure of 22 atm Case I: Apparatus was placed in single calorimeter and stop cock C was opened No change in temperature of water in calorimeter was observed , hence no internal work done .

35. Case II: V1 , V2 stopcock C placed in separate calorimeter When C is opened, Air enters into V2 and temp of V1 drops and for V2 and C temperature rises. Heat loss by V1 is equal to heat gain by V2 & C . Again total change in internal energy of the gas during expansion is Zero. Where, U is the internal energy , V is the volume. NO INTERNAL WORK IS DONE BY GAS IN EXPANDIND OR CHANGE IN U WITH VOLUME AT CONSTANT TEMPERATURE IS ZERO. THIS IS CALLED JOULE’S LAW OR MAYER’S HYPOTHESIS.

36. POROUS PLUG EXPERIMENT Lord Kelvin along with Joule devised a modified Joule’ experiment to test Joules law. This is called porous plug experiment. Highly compressed gas is being continuously forced at constant pressure through a narrow nozzle or porous plug. The porous plug is cotton or wool etc, having number of fine holes.

37. Consider mass of gas traversing by plug C from left to right . Let PA , VA , TA , UA and PB , VB , TB , UB be the pressure, volume, temperature and internal energy of one gram of gas before and after traversing the orifice respectively. Flow of gas is by forward motion of piston A. Thus the gas exerts a constant pressure PA at M . Gas after transverse the plug pushes forward the imaginary piston B Let us consider steady flow, Net work done PBVB - PAVA Adiabatic process, dQ = 0 Work done is equal to change in internal energy , PBVB - PAVA = UA – UB UA + PAVA = UB + PBVB Hence, U + PV is constant

38. For a perfect gas, PV= constant (Boyle’s Law) U = C T (Joule’s Law) Therefore U+PV depends only on temperature. But we know that U+PV= constant. Thus perfect gases temp should remains constant on both sides. But , for O2,N2 ,CO2 a cooling effect is observed. for H2 , heating effect is observed . Hence, for actual gases there is deviation in Boyle’s law or Joule’s law or in both…

39. Case I: Deviation from Boyle’s law • Gas is more compressible at low pressure, therefore • and • Hence, the gas will show ‘cooling effect’ • (b) Gas is less compressible at low pressure ,then • and • hence, ‘Heating Effect’ • Case II: Deviation from Joule’s law • Cohesive force will be present between the molecules, work will be done in order to overcome inter molecular attraction and the gas will become cooled. ie, cooling effect

40. Cooling or Heating of gas due to free expansion through a porous plug from high pressure to low pressure side will depends on deviation from Boyle’s Law and the work done in overcoming Inter molecular attraction. This is known as Joule Thomson effect. Cooling or Heating of gas due to free expansion through a porous plug from high pressure to low pressure side.

41. Temperature of inversion Ti • Let Ti and T is the temperature of inversion and vessel. Then, • If T> Ti, the gas will be heated when it passed through porous plug. • If T< Ti, the gas will be cooled when it passed through porous plug. • If T= Ti, there will be no change in temperature when it passed through porous plug. • According to Vander Waal’s equation, temperature of inversion is given by,

42. RELATION BETWEEN BOYLE TEMPERATURE, TEMPERATURE OF INVERSION AND CRITICAL TEMPERATURE. Temperature of inversion Boyle Temperature Critical Temperature