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Entropy of an Ideal Monatomic Gas 1

Entropy of an Ideal Monatomic Gas 1. We wish to find a general expression ω (U,V,N) for a system of N weakly-interacting particles of an ideal monatomic gas, confined to a volume V, with the total energy in the range U to U + U. Since U = p 2 /2m, the total momentum lies in the ranges

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Entropy of an Ideal Monatomic Gas 1

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  1. Entropy of an Ideal Monatomic Gas 1 We wish to find a general expression ω(U,V,N) for a system of N weakly-interacting particles of an ideal monatomic gas, confined to a volume V, with the total energy in the range U to U + U. Since U = p2/2m, the total momentum lies in the ranges ± p to ± (p + p). In order to count microstates, we imagine the phase-space to be divided into cells of “area” ΔpiΔqi ≈ h, per degree of freedom. N particles moving in 3 dimensions is represented by a point in 6N-dimensional phase space, while the number of degrees of freedom of the system is f = 3N. Note that f may be more than 3N if a particle is vibrating in simple harmonic motion, since degrees of freedom are associated both with kinetic energy and with vibrational potential energy.

  2. Entropy of an Ideal Monatomic Gas 2 A state describing N particles moving in 3 dimensions is a point in 6N-dimensional phase space, The number of degrees of freedom of the system is f = 3N. For f = 1 (one particle moving in one dimension), the relation U = p2/2m means that p lies between the values ± p to ± (p + p). Thus ω(U,V,N) = 2 p L/h, where is the length of the container, so that ω L = V1/3and ω p0 U0 .

  3. Entropy of an Ideal Monatomic Gas 3 • f = 1 (1 atom in 1 dimension) ω is the number of squares contained in the two rectangles; i.e. ω= 2 p L/h  p0L  U0V1/3. • f = 2 (1 atom in 2 dimensions or 2 atoms in 1 dimension) ω is the number of squares contained in the area between the circles; i.e. ω= 2πp p L2/h2  p1L2  U1/2V2/3. • f = 3 (1 atom in 3 dimensions or 3 atoms in 1 dimension) ω is the number of squares contained in the area between the spheres; i.e. ω= 4πp2p L3/h3  p2L3  U1V.

  4. Entropy of an Ideal Monatomic Gas 4 • General result for f degrees of freedom ω p(f – 1)Lf or p(f – 1)Vf/3. ω (mU)(f – 1)/2Vf/3 = (mU)(3N – 1)/2VN ≈(mU)3N/2VN for large N. Thus ω(U,V,N) = A(N) m3N/2 U3N/2 VN For a single species of mass m, we may write B(N) = A(N) m3N/2. Thusω(U,V,N) = B(N) VN U3N/2, so that S(U,V,N) = k [C(N) + N lnV + (3N/2) lnU]. Assuming that S is an extensive function, we have S(U,V,N) = N s(U,V) = Nk [K + lnv + (3/2) ln u]. Using the results u =(3/2)kT and v = V/N, we obtain S(U,V,N) = Nk [D + (3/2)lnT + ln(V/N)].

  5. Mixing of gases 1: Gibbs’ Paradox If N molecules of an ideal monatomic gas make a free expansion from a V to 2V, ΔU = 0, so that ΔS = Sf – Si = Nk ln(2V/V) = Nk ln 2. If the container of volume 2V is now redivided into equal parts, so that (N/2) molecules are in each half, then ΔS’ = Sf’ – Si’ = 2(N/2)k ln(V/2V) = – Nk ln 2. This is Gibbs’ paradox, which is removed by assuming that the molecules are indistiguishable; i.e. that ω(U,V,N) = B(N) U3N/2 VN/N!. The term Nk lnV in the expression for S is replaced by Nk ln(V/N!), where (by Stirling’s theorem), Nk ln(V/N!) ≈ Nk [ln(V/N) + 1].

  6. Entropy of an Ideal Monatomic Gas 5 • If S is assumed to be an extensive quantity, Gibbs’ paradox may be avoided. • Letting S = Ns, with s(U,V) = k [lnC + lnv +(3/2) lnu], where C = B(N)/N, v = V/N, and u = U/N. • Thus, S(U,V,N) = Nk [lnC + (3/2)ln(U/N) + ln(V/N)]. • Temperature is defined as 1/T  (S/U)V,N = (3/2)Nk/U, since U = (3/2)NkT, so that S(U,V,N) = Nk [D + (3/2)lnT + ln(V/N)]. • Note that P/T = (S/V)T,N = Nk/V, so that PV = NkT. Mixing gases • Gas A in the left compartment, a different gas B in the right compartment. ΔStot = ΔS1 + ΔS2. b. Gas A in the left compartment, the same gas B in the right compartment. ΔStot = Sfinal– S1initial– S2initial.

  7. Partition Function of an Ideal Monatomic Gas 1 z = iexp(– i) → d3x d3p exp(– )/h3(classical system).

  8. Partition Function of an Ideal Monatomic Gas 2

  9. Entropy of an Ideal Monatomic Gas 6 Z =zN/N!, where z = V(2πmkT/h2)3/2, and U = (3/2)N kT. Also S = k(lnZ + U) = k [ln zN – lnN! + (3/2)N], = Nk [lnz – lnN + 1 + 3/2]. Thus S = Nk [K + ln(V/N) + (3/2) ln T], with K = ln (2πmk/h2)3/2+ 5/2.

  10. Equipartition Theorem 1

  11. Equipartition Theorem 2

  12. Degrees of Freedom • Applying the equipartition theorem is simplified by using the concept of degrees of freedom f,which is the minimum number of independent coordinates needed to the specify the motion of a system of particles.* • For a single atom (assumed to be a point), there are 3 directions of motion so that f =3. • For a molecule consisting of n point-atoms, it is necessary to separate translational, rotational and vibrational motions; however the total number of degrees of freedom must be 3f. • For linear molecules, ftrans= 3, frot= 2 (since there can be no rotation about the axis for point-atoms), so that fvib= 3n – 5. • For non-linear molecules, ftrans= 3, frot= 3, so that fvib= 3n – 6. * Beware! Some texts, such as Carter, define f differently.

  13. Equipartition Theorem : Examples Harmonic oscillator in one dimension  = p2/2m + ½ k’ x2, where k’ is the spring constant. •  = ½ kT + ½ kT = kT. Ideal diatomic gas 50 Kvib,rotT trans, so that = (3/2)kT and CV = (3/2)nR. 500 KvibT trans,rot, so that = (5/2)kT and CV = (5/2)nR. CP = CV + nR = (7/2)nR(Mayer’s equation), so thatγ= 7/5 = 1.2. 5000 K(assuming no dissociation)T trans,rot,vib, so thatCV = (7/2)nR.

  14. Harmonic Oscillator in One Dimension 1 For a 1D harmonic oscillator, n = (n + ½) ħω, where ω = √(k/m).

  15. Harmonic Oscillator in One Dimension 2

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