1

Entropy of an Ideal Monatomic Gas 1

We wish to find a general expression ω(U,V,N) for a system of Nweakly-interacting particles of an ideal monatomic gas, confinedto 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 in6N-dimensional phase space, while the number of degrees offreedom of the system is f = 3N.

Note that f may be more than 3N if a particle is vibrating in simpleharmonic motion, since degrees of freedom are associated bothwith 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 = V1/3 and 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)/2 Vf/3 = (mU)(3N – 1)/2 VN ≈ (mU)3N/2 VN 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.

Thus ω(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 expansionfrom 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 themolecules 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

a. 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 = i exp(– 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

Degrees of Freedom• Applying the equipartition theorem is simplified by using the

concept of degrees of freedom 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 : ExamplesHarmonic 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 that γ = 7/5 = 1.2.

5000 K (assuming no dissociation) T trans ,rot ,vib ,

so that CV = (7/2)nR.

Motion f c

(for T c)

Translation 3 10–16 K (3/2) kT

Rotation 2 1 K kT

Vibration 1 103 K kT

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