Physics 160 Thermodynamics and Statistical Physics: Lecture 2. Dr. Rengachary Parthasarathy Jan 28, 2013

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1 Physics 160 Thermodynamics and Statistical Physics: Lecture 2 Dr. Rengachary Parthasarathy Jan 28, 2013

2 Chapter 1: Energy in Thermal Physics Due Date Section /3 Section 1.2: 1.12, 1.14, 1.16, /3 Section 1.3: /10 Section 1.4: /10 Section 1.5: 1.32, /10 Section 1.6: 1.41, 1.45, 1.47,1.50 2/17 Section 1.7: 1.56, 1.63, 1.66, /17

3 State Variables: 1.Are fully determined by the values at present and do not depend on the history of the system. 2. Not all state variable have to be specified to define the state of the system. Why? Because the variables are interdependent and only a small number can be varied independently.

4 The small number of variables that can be varied independently are called the independent variable; the others are dependent variables.. The number of independent variables to describe a macroscopic system thermodynamically is very small, about half a dozen.

5 The Zeroeth Law-Temperature (Maxwell, Fowler, Sommerfeld). If two bodies are in thermal equilibrium with a third, they are in thermal equilibrium with each other. The formulation of this law has a long history.

6 The concept of temperature. As a natural generalization of experience we introduce the postulate: if two assemblies are each in thermal equilibrium with a third assembly, they are in thermal equilibrium with each other.

7 From this it may be shown to follow that the condition for thermal equilibrium between several assemblies is the equality of a certain single-valued function of the thermodynamic states of the assemblies, which may be called the temperature t, any one of the assemblies being used as a thermometer reading the temperature t on a suitable scale. This postulate of the existence of temperature could with advantage be known as the zeroth law of thermodynamics. Fowler and Guggenheim

8 What is thermal equilibrium? Thermal equilibrium refers to systems in thermal contact that do not change with time. Thermal contact refers to systems in contact via a diathermal wall (is one in which the state of the system can be changed by means other than moving the boundary.

9 independent. How to get this single-valued function of the thermodynamic states? Consider three systems A, B and C. Let P A, P B,P C, V A, V B, and V C be their mechanical variables. P = pressure, V = molar volume. If A and C are in thermal contact across a rigid wall and in equilibrium, not all four variables are

10 The variables are connected by an equation of state. This can be written as (P A, V A ; P C, V C ) = 0 or P C = Φ 1 (V A, V C ; P A ). Similarly, when B and C are in equilibrium, as (P B, V B ; P C, V C ) = 0 or P C = Φ 2 (V B, V C ; P B ). By Zeroeth law, A and B are in thermal equilibrium.

11 This requires that (P A, V A ; P B, V B ) = 0 Eqn (1) implying that P A, V A, P B and V B are interdependent. We also know that P C = Φ 1 (V A, V C ; P A ) = Φ 2 (V B, V C ; P B ) implying a functional relationship exits between P A, P B, V A, V B, and V C

12 Implying (P A, V A ; P B, V B; V C ) = 0. (Eqn 2.) How to reconcile the two functions and, one implying that P A, V A ; P B, V B; are interdependent but not dependent on V C and the other that they are dependent on V C?) To Summarize: 1. P C = Φ 1 (V A, V C ; P A ) 2. P C = Φ 2 (V B, V C ; P B ) 3. (P A, V A ; P B, V B; V C ) = 0.

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14 The problem is V C How to develop a function Φ that depends on V C but also satisfies 1 and 2 that does not involve V C? Make the dependence on V C a product type and the same for Φ 1 and Φ 2. This way when we write Φ 1 = Φ 2 the common function that depends on V C gets factored out and the dependence on V C gets eliminated.

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16 In addition, we can also add a common function of V C to Φ 1 and Φ 2 that also gets eliminated when we write Φ 1 = Φ 2. This is possible if the functions Φ 1 and Φ 2 are of this form: Φ 1 (V A, V C ; P A ) = θ 1 (V A, P A )ε (V C ) + η (V C ) and

17 Φ 2 (V B, V C ; P B ) = θ 2 (V B, P B )ε (V C ) + η (V C ) Φ 1 = Φ 2 implies the validity of equation 1. Looking at the right hand side, it is seen that θ 1 (V A, P A ) = θ 2 (V B, P B ) = θ

18 We call this θ empirical temperature. Thus, for every system, it is possible to find a function θ of mechanical variables P and V such that when the systems are in equilibrium, they have the same value of θ.

19 Does it make sense to say that one object is twice as hot as another? In the Celsius or Fahrenheit scales, it makes no sense since the zero is arbitrary. Is 2 0 C twice as hot as 1 0 C? But in using the kelvin scale, it makes perfect sense. 2K is twice as hot as 1K.

20 The Ideal Gas: What is an ideal gas? Two important characteristics. One, the gas molecules are point molecules; second, there is no intermolecular attraction or repulsion. At low pressures, gases behave as ideal ones.

21 The ideal gas law: PV = nrt P = pressure, V = volume, n = number of moles of gas, R is a universal gas constant, T = temperature in kelvin. What is the value of R? To calculate it, take STP (0 0 C= K; 1 atm= 1.013x10 5 Pa)

22 Solve for R R = PV/nT = (1atm)( L) (1mol) (273.15) = L.(atm)/mol.K The calculated value is usually rounded to L.atm/mol.K The numerical value of R changes when other units for P, V, n, T are used.

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24 A mole of molecules is Avagadro s number of them, N A =6.02 x N A is more useful in chemistry; here, we use the number of molecules denoted by N=n x N A Using the above in the ideal gas equation, we have PV = NkT; k is Boltzmann s constant = R/ N A = x J/K

25 It is best to memorize the following relationship: nr = Nk Nearly all laws of physics is an approximation including the ideal gas law. At RT and 1atm, the average distance between gas molecules is about 10 times the size of the molecule.

26 What is the volume of 1mol of air at RT and 1atm? V = nrt/p = (1mol)(8.31 J/mol.K) (300K)/10 5 N/m 2 = m 3 = 25 L.

27 Calculate the average volume per molecule for an ideal gas at RT and 1atm. V/N = kt/p = (1.381 x J/K) x (300K) / 10 5 N/m 2 = 41 nm 3 V = (4/3)π r 3 R =( (3/4)41 nm 3 ) 1/3 = 3.129nm

28 Molecular size is a few angstroms, 2 x m = 0.2nm versus the distance of about 6nm. The distance is quite larger molecular size in general. WE will recognize this fact to discuss the ideal gas

29 Thermal equilibrium arises due to exchange of energy between two systems. How exactly is temperature related to energy? The answer is not simple except for ideal gas.

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32 Assumptions: 1.The collisions are elastic; the speed never changes. 2. As the molecule bounces off the wall, its path is symmetric. 3. As time passes, the velocity ( not speed ) of the molecule changes.

33 The only thing we know for perfect gas is PV = NkT. First, we have to relate pressure P to the kinetic energy; then pressure to temperature What is the pressure on the piston due to this molecule bouncing all over?

34 The average pressure on the piston Is

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49 Difficulties with Equipartition Theorem: 1. U Thermal is never the total energy of a system. 2. Applicable only to changes in energy due to changes in temperature. 3. Not applicable to phase transitions where bonds may be broken or formed.

50 4. How to count the degreees of freedom of a system? For monatomic molecules, only translational motion counts giving rise to f = 3. For diatomic molecules, in addition to 3 translations, three rotational motions occur about three axes of the molecule..but

51 For a single atom, whether isolated or part of a molecule, any motion can be resolved into components along the axes of a Cartesian coordinate system Every atom has three degrees of freedom.. For a molecule composed of N atoms, there are thus 3N degrees of freedom

52 Molecular Translations Simultaneous motion of all of the atoms in the same direction, resulting in a linear movement of the entire molecule Molecular Rotations Circular movement of the atoms in the molecule around the molecule s center of mass

53 Normal Modes of Vibration An oscillating motion of the atoms, or vibration, without a change in the center of mass of the molecule

54 Translational Degrees of Freedom All molecular translations can be resolved into components along the three axes of a Cartesian coordinate system, Tx, Ty and Tz. All molecules have three translational degrees of freedom

55 Rotational Degrees of Freedom of Linear Molecules Molecular rotations of linear molecules can be resolved into rotations about the two Cartesian axes perpendicular to the molecular axis, Rx and Ry, but not the molecular axis itself. Linear molecules have two rotational degrees of freedom

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