Exam (elaborations) TEST BANK FOR Statistical Physics of Particles By

Exam (elaborations) TEST BANK FOR Statistical Physics of Particles By Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 II. Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 III. Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . 38 IV. Classical Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . 72 V. Interacting Particles . . . . . . . . . . . . . . . . . . . . . . . . . . 93 VI. Quantum Statistical Mechanics . . . . . . . . . . . . . . . . . . . . 121 VII. Ideal Quantum Gases . . . . . . . . . . . . . . . . . . . . . . . . 138 Problems for Chapter I - Thermodynamics 1. Surface tension: Thermodynamic properties of the interface between two phases are described by a state function called the surface tension S. It is defined in terms of the work required to increase the surface area by an amount dA through dW¯ = SdA. (a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius R is larger than outside pressure by 2S/R. What is the air pressure inside a soap bubble of radius R? • The work done by a water droplet on the outside world, needed to increase the radius from R to R + ∆R is ∆W = (P − Po) · 4πR2 · ∆R, where P is the pressure inside the drop and Po is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy S∆A = S · 8πR · ∆R, where S is the surface tension, and ∆Wtotal = 0, =⇒ ∆Wpressure = −∆Wsurface , resulting in (P − Po) · 4πR2 · ∆R = S · 8πR · ∆R, =⇒ (P − Po) = 2S R . In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and Pfilm − Po = Pinterior − Pfilm = 2S R , leading to Pinterior − Po = 4S R . Hence, the air pressure inside the bubble is larger than atmospheric pressure by 4S/R. (b) A water droplet condenses on a solid surface. There are three surface tensions involved Saw, Ssw, and Ssa, where a, s, and w refer to air, solid and water respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting). • When steam condenses on a solid surface, water either forms a droplet, or spreads on the surface. There are two ways to consider this problem: Method 1: Energy associated with the interfaces 1 In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore dE = SawdAaw + SasdAas + SwsdAws = 0. Since the total surface area of the solid is constant, dAas + dAws = 0. From geometrical considerations (see proof below), we obtain dAws cos θ = dAaw. From these equations, we obtain dE = (Saw cos θ − Sas + Sws) dAws = 0, =⇒ cos θ = Sas − Sws Saw . Proof of dAws cos θ = dAaw: Consider a droplet which is part of a sphere of radius R, which is cut by the substrate at an angle θ. The areas of the involved surfaces are Aws = π(R sin θ) 2 , and Aaw = 2πR2 (1 − cos θ). Let us consider a small change in shape, accompanied by changes in R and θ. These variations should preserve the volume of water, i.e. constrained by V = πR3 3