Introduction to Statistical Mechanics

Solutions to Problems

John Dirk Walecka

• Verlag: WSPC
• Jahr: 2016
• ISBN:
  9789813148154
• Thema: Naturwissenschaften
• Thema: Mathematische & Computerphysik

Statistical mechanics is concerned with defining the thermodynamic properties of a macroscopic sample in terms of the properties of the microscopic systems of which it is composed. The previous book Introduction to Statistical Mechanics provided

Chapter 1

Introduction

Problem 1.1 Prove from Eq. (1.1) that the integral in Eq. (1.3) is independent of path.

Solution to Problem 1.1

Equation (1.1) states the following integral around a closed cycle vanishes

Pick two points on the cycle (A, B) with specified thermodynamic variables. Let C₁ denote that part of the cycle running from A → B, and C₂ that part of the cycle returning from B → A. Then the above states

where the arrows indicate the direction during the cycle. Since both the heat flow and work are algebraic and change sign if the change occurs in the opposite direction, the integral changes sign if the trajectory is traversed in the opposite direction¹

Hence the previous relation can be re-written as

This states that the integral is the same whether we go from A to B along C₁ or along C₂ in the opposite direction. Since the cycle containing the points (A, B) is arbitrary, the integral

is independent of the path from A → B.

Problem 1.2 Start from either statement of the second law, and see how far you can get in verifying the statements leading to Eqs. (1.5) and (1.6); then compare with [Zemansky (1968)].

Solution to Problem 1.2

The two equivalent statements of the second law of thermodynamics are given at the start of section 1.1.2:

(1)Kelvin: It is impossible to construct an engine that, operating in a cycle, will produce no effect other that extraction of heat from a reservoir and performance of an equivalent amount of work.

(2)Clausius: It is impossible to construct a device that, operating in a cycle, will produce no effect other than the transfer of heat from a cooler to a hotter body.

We shall not provide a rigorous derivation of the consequences, but leave that to a basic course in thermodynamics.² The key element is to establish that all reversible engines operating between two heat baths at distinct temperatures T₁ > T₂ have the same efficiency, equal to that of a Carnot engine, from which Eqs. (1.5)–(1.7) follow. The argument goes something like this. Consider two such engines operating between the two heat baths, and let the second operate in the reverse direction. Suppose the first absorbs heat Q₁ at T₁, produces work W, and expels heat Q₂ at T₂. Now put the heat Q₂ and work W into the second engine. If it does not expel exactly the same heat Q₁ into the bath at T₁, then the combined engines constitute a device that can be arranged to violate the second statement of the second law.

Problem 1.3 (a) Why does each point on the dotted curve in Fig. 1.2 in the text correspond to a given T?³

(b) How could one carry out the Carnot cycles shown in Fig. 1.2 in the text in a continuous manner with each segment being covered only once?

(c) Why is it unnecessary for the construction of the entropy to actually traverse opposing segments of the adiabats in (b)?

(d) Show that the total heat input and total work output in (b) satisfy Q = W.

Solution to Problem 1.3

(a) A perfect gas obeys the equation of state PV = nRT (see, for example, Prob. 1.4). Thus if we specify (P, V), we also specify T;

(b) If we simply take the assembly around the outer segments of all the Carnot cycles, omitting all the common segments traversed in opposite directions, we move around the loop in a clockwise direction, in a continuous manner, with each segment being covered only once;

(c) The opposing segments along the adiabats in Fig. 1.2 in the text are traversed in a reversible manner, and along each of them the reversible heat flow vanishes, dQ_R = 0. Hence there is no change in entropy along those segments;

(d) Since the gas retu...