Solutions to Schroeder

(Dated: April 13, 2012)

SECTION 1

Schroeder 1.23

Helium has only 3 degrees of freedom (translational) per molecule. Using the equipartition theorem and the ideal gas law: U = 3N 1 kT = 3 PV. For P = 105 N/m2 and 2 2 V = 103 m3 PV = 100 J and U = 150 J. Air (a mixture of O2 and N2 )has 5 degrees of freedom (3 translational and 2 rotational), so U = 5 PV = 250 J. 2

Schroeder 1.24

A lead atom has 6 degrees of freedom: 3 translational (from the kinetic energy) and 3 vibrational (from the potential energy of vibration along 3 orthogonal directions). The atomic mass is 207 , so 1 gram of lead contains 1/207 moles. The total thermal energy is:
f U = 2 nRT = 6 1 (8.31 2 207

J/K)(300 K) = 36 J.

Schroeder 1.55 (a)-(c)

(a) Assume the particles rotate around a center of mass, in an orbit of radius r. As a result, they are at a distance 2r from each other. The kinetic and potential energies of the system are:
1 K = 2 2 mv2 ; V = gm . 2r
2

These energies are related by applying Newtons second law (applied to the motion of one of these particles): F = ma
gm2 (2r)2 mv2 r

2 When multiplying both sides of the equation by 2r, we obtain the following: V = 2mv2 = 2K. (b) The total energy of the system is: Utotal = K + V = K 2K = K = mv2 . Increasing the total energy will decrease the kinetic energy. (c) The total kinetic energy of the system is N 3 kT, so the total energy (kinetic and 2 potential) is: Utotal = K = 3 NkT. 2 The heat capacity is equal to: C=
dU dT

= 3 Nk. 2

The heat capacity is negative.

SECTION 2

Schroeder 2.3 (a)-(d)

(a) There are two possible states for the rst coin, and for each of these, two for the second state, and for each of these, two for the third state, and so on... The total number of microstates is 250 = 1.13 1015 . (b) The number of ways of getting 25 heads is: ( ) 50! (25) = 50 = (25!)2 = 1.26 1014 25 (c) The probability of getting 25 heads is: P(25) =
(25) 250

= 0.112.

(d) The probability of getting 30 heads is: P(30) =

1 50! 250 (30!)(20!)

= 0.042.

3
Schroeder 2.8 (a)-(d)

(a) From the total of 20 units of energy, solid A could take up between 0 to 20. Each of the 0 to 20 possibilities denes a macrostates. There will be 20 macrostates in total. (b) The combined system has 20 oscillators and 20 units of energy. The total number of microstates will be: ) ( (20, 20) = 20+201 = 20
39! (20!)(19!)

= 6.89 1010 .

(c) For the macrostate with all the energy in solid A(10,20), the multiplicity of solid A is: ( ) 29! (10, 20) = 20+101 = (20!)(9!) = 1 107 . The multiplicity of solid B is: 20 ( ) (10, 0) = 0+101 = 1. 0 If the system is in equilibrium, all the microstates are equally probable, so the probability of this macrostate is: Probability =
A B (total) 1.00107 6.891010

= 1.45 104 .

(d) For the macrostate with half the energy in each solid, the multiplicity of the combined system is: = A B = (10+101)(10+101)
10 10 19! = ( (10!)(9!) )2 = 1.454 .

The probability for this macrostate (at equilibrium) is: Probability =

= 0.124.

SECTION 3 [10 MARKS]

Schroeder 2.16 (a) [4 marks]

(a) For 1000 coins the total number of possible microstates is 21000 . [1 mark] The number of ways of getting exactly 500 heads and 500 tails is: ( ) 1000! (500) = 1000 = (500!)2 . [1 mark] 500 Using Stirlings approximation:

4 (500) =
10001000 e1000 21000 500 e500 2500)2 (500

21000 . 500

[1 mark]

The probability for this macrostate is: Probability =

1 500

= 0.025 = 25%. [1 mark]

Schroeder 2.17 [6 marks]

We start by using the same steps as in equations 2.17 and 2.18 (Schroeder, page 63) to obtain: ln (N + q) ln(N + q) q ln q N ln N. [1 mark] To simplify in the limit q N), we expand the rst logarithm in this limit:
q q q ln(N + q) = ln[N(1 + N )] = ln N + ln(1 + N ) ln N + N . [1 mark]

By plugging this result in the rst equation and cancelling the NlnN terms, we obtain: ln q ln N + q +
q2 N

q ln q. [2 marks]
q2 N

Next, we can neglect the

contribution, as it is much smaller than the others:

ln q ln N + q. [1 mark] q Exponentiating this equation will give: eq ln q eq = ( N )q eq = ( eN )q . [1 mark] q q

SECTION 4 [10 MARKS]

Schroeder 2.29 [6 marks]

As discussed in section 2.3 of the texbook, for a system of two Einstein solids, with NA = 300, NB = 200, and qtotal = 100, the most likely macrostate is the state with qA = 60 and a multiplicity 6.9 10114 and the least likely macrostate is the state with qA = 0 and a multiplicity 2.8 1081 . [1 mark] The entropies corresponding to these states are:

5 (qA = 60) (qA = 0)

= ln(6.9 10114 ) = ln 6.9 + 114 ln 10 = 264.4. [1 mark]

= ln(2.8 1081 ) = ln 2.8 + 81 ln 10 = 187.5. [1 mark]

Over long time scales, all microstates are accessible, so all macrostates will be allowed. [1 mark] The entropy in this case is:
S k

= ln total = ln(9.3 10115 ) = ln 9.3 + 115 ln 10 = 267.02. [2 marks]

Schroeder 2.37 [4 marks]

Gas B has xN molecules. When the gases start mixing, the molecules in B expand to ll a volume that is greater by a factor of 1/x. According to the Sackur-Tetrode equation, their entropy will increase by:
1 SB = (xN)k ln x = Nk ln x. [1 mark]

Analogously, the number of molecules in A is (1-x)N and they will expand in volume by a factor of 1/(1-x). Their entropy will increase by:
1 SA = [(1 x)N]k ln 1x = Nk(1 x) ln(1 x). [1 mark]

The total entropy increase due to mixing is: mixing = A + B = Nk[x ln x + (1 x) ln(1 x)]. [1 mark] For x=1/2, this expression is equal to:
1 1 Smixing = Nk[ 2 ln 2 + 1 ln 1 ] = Nk ln 1 = Nk ln 2. 2 2 2

The results are in agreement with equation 2.54, since N in this case is the total number of molecules (compared to 2N in equation 2.54). [1 mark]

6
SECTION 5

Schroeder 3.5

The result to 2.17 is: eqln q eq = ( eN )q . q The entropy in this limit (q N) is: S = kln( eN )q = kq ln( eN ) = kq[ln e + ln N ln q] = kq[ln N ln q + 1]. q q U = q, where is an energy unit. The entropy will become: S=
kU [ln N

ln U + ln + 1].

We can dierentiate with respect to U and use the following formula:

dS dU

= k [ln( N ) + 1] + U

kU 1 ( U )

ln( N ). Next, we solve for U: U

= ln( N ) e kT = U

N U

U = Ne kT .

The energy goes to zero, as T 0.

Schroeder 3.13

(a) We assume there is an average of eight hours of sunlight per day. The total energy that a square meter of Earths surface receives per year is: Q = W t = (1000 J/s)(3600 s/hr)(8 hrs/day)(365 days) = 1.05 1010 J. The entropy created in one year by the ow of solar heat is: SEarth =
Q T

1.051010 J 300 K

= 3.5 107 J/K.

(b) Let us assume that in one year we would be able to grow 1 kg of grass, which would roughly contain 1000 moles of carbon and other molecules. Even if the grass has zero entropy, the net reduction in entropy upon assembly from smaller molecules would be of the order: S Nk = nR (1000 moles)(8.3 J/mol K) 104 J/K.

7 That is 3000 times less than the entropy created by the ow of solar heat. Therefore, there is no violation of the second law of thermodynamics here, since the growth of the grass reduces the increase in entropy by a small percent.