Statistical Physics

52 4 Ideal GasesFor N molecules, the phase space is 6N-dimensional, and the states witha total energy less than E occupy a nonzero volume in this space. The numberof states Ω 0 (E) can be calculated by dividing the volume of phase spaceby h 3N .We begin the calculation with the simplest case. We consider a systemof a single molecule where the motion of the molecule is restricted to a onedimensionalspace of le ngth L x . The phase space is two-dimensional, as shownin Fig. 4.5a. If the energy of the system is E, the momentum of the moleculemust satisfy |p x | = √ 2mE, wherem is the mass of the molecule. Therefore,the region of phase space for states with an energy less than E is the shadedarea in the figure, and the number of states isΩ 0 = 2 × √ 2mEL x. (4.8)hNext, the momentum subspace for a molecule in a two-dimensional space ofarea L x × L y is shown in Fig. 4.5b, and the region corresponding to energiesless than E is the shaded circle, the area of which is π( √ 2mE) 2 .Thus,Ω 0 (E) =( √2mE ) 2π Lx L yh 2 . (4.9)Similarly, in a three-dimensional space of volume V , the region of the momentumsubspace where the energy is less than E is a sphere of radius √ 2mE,and Ω 0 is given byΩ 0 (E) = 4π (2mE) 3/23 h 3 V. (4.10)Fig. 4.5. The shaded area shows the region in phase space in which the energy of thesystem is less than E. (a) One-dimensional system. (b) Two-dimensional system,where only the momentum subspace is shown