Exercise Solutions 883

CHAPTER 3: Equations of State

3.2 For isolated hydrogen atoms the orbital electron radius isr = a0n2 wherea0 is theBohr radius andn is the principle quantum number. The average separation between atomsis d ∼ (3m/4πρ)1/3. If we assume that there are no states bound to a single atom ifr > d/2,then the requirementr < d/2 leads to

a0n2 <12

(

3m4πρ

)1/3

−→ n <

√

12a0

(

3m4πρ

)1/3

for bound states.

3.4 From Eq. (2.5),

logn+

n0= log

u+

u0+ 5

2 logT −Ei (eV)5040

T− logPe−0.179,

where for hydrogen gasu0 = 2, u+ = 1, Ei = 13.6 eV, and the electron pressurePe and thefraction ionizationf are given by

Pe =n+

2n+ + n0Pgas=

12+ n0/n+

Pgas f =n+

n+ + n0=

11+ n0/n+

.

Thus we must insert this expression forPe and solve the first equation forn+/n0. Thiscan be done numerically for the general case, but if we assumea significant fraction ofionization we can approximatePe ≃ 0.5Pgas and the first equation can be solved directly.Using a central solar temperature of 15.7 million K and a central gas pressurePgas≃ 2.3×1017 dyn cm−2 from Table 8.1, we find that at the center of the Sun the first equation predictsabout 75% ionization (justifying our approximation thatPe ∼

12Pgas). This is because the

Saha equation accounts for pressure effects through electron recombination, but not throughpressure ionization. In reality, the center of the Sun wouldbe 100% ionized because of thepressure.

3.10 For adiabatic processesδQ = 0 and the first law (3.7) reduces todU = −PdV . Butfor an ideal gasPV = NkT so

PdV +V dP = NkdT = NkdUCV

= −NkPdVCV

= (1− γ)PdV

wheredU = CVdT was used in the second step,dU = −PdV in the third step, and in thefinal stepγ = CP/CV andCP = CV + Nk for an ideal gas have been used. Rearranging thisexpression gives

γdVV

= −

dPP

.

884 Exercise Solutions

By substitution and application of the ideal gas law it is clear that

P1−γT γ = constant PV γ = constant TV γ−1 = constant

are solutions. The adiabatic speed of sound isvs= (B/ρ)1/2, whereB≡−V (∂P/∂V )adiabatic

is the bulk modulus. Taking the adiabatic equation of state in the formP = KV−γ whereKis a constant, and evaluating the partial derivative gives for the bulk modulusB = γP. Thusthe adiabatic speed of sound isvs = (γP/ρ)1/2.

3.18 We follow the solution of Phillips [11]. From Eqs. (3.68), (3.71), (3.58), and (3.50),the internal energy of the gas is

U =∫ ∞

0εp f (εp)g(p) dp g(p)dp = 4πgs(V/h3)p2dp ε2

p = p2c2 + m2c4.

The gas pressure may be defined by

P = −

∂U∂V

= −

∫ ∞

0

dεp

dVf (εp)g(p) dp = −

∫ ∞

0

dεp

dpdpdV

f (εp)g(p) dp.

where we considerεp to be a function ofp, andp to be a function ofV . By the uncertaintyprinciple, in a volumeV = L3 the momentum isp ∝ L ∝ V−1/3, so

dpdV

≃−13V−4/3 =

−V−1/3

3V≃−

p3V

,

and from the energy expression

dεp

dp=

12

(

2pc2

(p2c2 + m2c4)1/2

)

=pc2

εp≃ vp,

wherevp is the speed of a particle with momentump. Inserting these results in the aboveintegral forP gives

P =1

3V

∫ ∞

0vp p f (εp)g(p) dp.

Now insert a Maxwell distributionf (εp) = exp[−(εp − µ)/kT ] in the pressure integral togive

P =4π3h3 eµ/kT gs

∫ ∞

0p3e−εp/kT vp dp.

To evaluate the integral note that

d(e−εp/kT ) = −

1kT

e−εp/kT dεp = −

1kT

e−εp/kT vpdp,

Exercise Solutions 885

where we’ve useddεp = vpdp from earlier. Thus

∫ ∞

0p3e−εp/kT vp dp =

−1kT

∫ ∞

0p3 d(e−εp/kT )

= −kT p3e−εp/kT∣

∣

∣

p=∞

p=0+3kT

∫ ∞

0e−εp/kT p2 dp

= 3kT∫ ∞

0e−εp/kT p2 dp,

where an integration by parts and thatεp = ∞ if p = ∞ were used in the second line. Insert-ing this result in the earlier expression forP we obtain

P =kTV

eµ/kT∫ ∞

0e−εp/kT g(p) dp.

But the total particle number is

N =

∫ ∞

0f (εp)g(p) dp = eµ/kT

∫ ∞

0e−εp/kT g(p) dp.

Therefore, a comparison of the previous two expressions gives the ideal gas law

P =NkT

V= nkT,

wheren = N/V is the particle number density. Thus the ideal gas law is valid for arbitraryparticle velocity, provided that the gas has a Maxwell–Boltzmann distribution.