ThermalPhysics_1_solns

Richard Thomas - FK4005 “Thermodynamik” - Solutions to Raknovningar 1

• 1.9: What is the volume of one mole of air, at room temperature and 1 atm pressure.

From the ideal gas law, we have PV = nRT . Therefore:

V =nRT

P=

(1 mol)(8.31 J/mol/K)(300 K)

(105 N/m2)

V ≈ 0.025 m3 ≈ 25 L

• 1.10: Estimate the number of air molecules in an average-sized room.

Consider that an average room is about 4 metres squared and 3 metres high. The numberof air molecules (at the standard room temperature and pressure) is

N =PV

kBT=

(105 N/m2)(4 × 4 × 3 m3)

(1.38 × 10−23 J/K)(300 K)

N = 1.2 × 1027 ≈ ×1027

which is about 2000 moles.

• 1.11: Rooms A and B are the same size, and are connected by an open door. Room A,however, is warmer (perhaps because its windows face the sun). Which room contains thegreater mass of air? Explain carefully.

Pressure is the same in both because they are connected by an open doorway; if the pressurewere different then a wind would blow from one room to the other. The volume of both roomsis also the same. Therefore PAVA = PBVB and, given the ideal gas law, NAkTA = NBkTB

as well. Thus the cooler temperature room has the higher number of molecules, and thusthe greater mass.

• 1.12: Calculate the average volume per molecule for an ideal gas at room temperature andatmospheric pressure. Then take the cube root to get an estimate of the average distancebetween molecules. How does this distance compare to the size of a small molecule like N2

or H2O?

The volume per molecule for an ideal gas at room temperature and atmospheric pressure is:

V

N=

kBT

P=

(1.38 × 10−23 J/K)(300 K)

105 N/m2

V

N= 4.1 × 10−26 m3 ≈ 41 nm3

The average distance between molecules can be estimated as

(

V

N

)1/3

≈ 3.5 nm3

The diameter of a molecule like N2 or O2 is only a few Angstroms, approximately 10×smaller than this average distance.

• 1.13: A mole is approximately the number of protons in a gram of protons. The mass of aneutron is about the same as the mass of a proton, while the mass of an electron is usuallynegligible in comparison, so if you know the total number of protons and neutrons in in amolecule (i.e. its “atomic mass”), you know the approximate mass (in grams) of a mole


of these molecules. Referring to the periodic table, find the mass of each of the following:water, molecular nitrogen, lead, and quartz (SiO2)

H2O = 2(1) + 1(16) = 18 grams/molN2 = 2(14) = 28 grams/molLead = Pb = 207 grams/molSiO2 = 1(28.08) + 2(16) = 60 grams/mol

• 1.14: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume),O2 (21%) and Ar (1%).

Use weighted average to calculate the mass of air:

f =

∑

i wifi∑

wi(wi = weights)

m =0.78mN2

+ 0.21mO2+ 0.01mAr

0.78 + 0, 21 + 0, 01

From the periodic table, mN2= 28.014 g/mol, mO2

= 32.00 g/mol, and mAr = 39.95 g/molTherefore, the mass of a mole of dry air:

m = 0.78(28.014) + 0.21(32) + 0.01(39)

m = 28.92 g/mol

• 1.15: Estimate the average temperature of the air in a hot air balloon (see Fig. 1.1 in book).Assume that the total mass of the unfilled balloon and payload is 500 kg. What is the massof the air inside the balloon?

The upward buoyant force on the balloon is equal to the weight of the air displaced. As-suming that this force is approximately in balance with gravity, we can write:

ρ0V g = (M + ρV )g or ρ0 − ρ = M/V

where ρ0 is the density of the surrounding air, V is the volume of the balloon, M is themass of the unfilled balloon and payload, and ρ is the density of the air inside the balloon.According to the ideal gas law, the density of air is:

ρ =mn

V=

mP

RT

where m is the mass of one mole of air (≈29 g). This formula applies either inside or outsidethe balloon, with the same pressure in both places but different temperatures. Therefore,teh balance of forces implies:

mP

RT0− mP

RT=

M

V

where T and T0 are the temperature inside and outside the balloon, respectively. Rearrangingyields:

1

T= − 1

T0− M

m

P

RV


Assume an outside air temperature of 290 K and atmospheric pressure. The volume of theballoon can be estimated from the figure. Comparing the heights of the people standingbeneath, a reasonable estimate for the diameter of the balloon in the foreground is ≈15metres. This gives a volume, assuming a spherical balloon (3

3πr3), of about 1770 m3. The

mass of the unfilled balloon and payload is assumed to be 500 kg, so the previous expressionevaluates to:

1

T= − 1

(290 K)− (500 kg)

(0.029 kg)

(8.31 J/K)

(105 N/m2)(1770 m3)1

T= − 1

(290 K)− 1

(1235 K)=

1

(379 K)

Thus, the temperature inside the balloon must be about 379 K or just over 100 C. Assumingthis temperature, the mass of the air inside the balloon should be roughly:

Mair = mn =mPV

RT

Mair =(0.029 kg)(105 N/m2)(1770 m3)

(8.31 J/K)(379 K)= 1600 kg

which s more than three times the mass of the unfilled ballon and payload.

• 1.16: The Exponential atmosphere.a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest,the pressure holding it up from below must balance both the pressure from above and theweight of the slab. Use this fact to find an expression for dP/dz, the variation of pressurewith altitude, in terms of the density of air.

This diagram shows the forces on the slab of air. P (z) is the atmospheric pressure as afunction of height z, and M is the total mass of the slab.

Because the slab of air is at rest, the net force on the slab must be zero. Thus

P (z + dz)A + Mg − P (z)A = 0 ⇒ [P (z + dz) − P (z)]A = −Mg ⇒ P (z + dz) − P (z) =−Mg

A

Now the volume of the slab is Adz, so if the density of air is ρ(z), then M = ρ(z)Adz, andso

P (z + dz) − P (z) = −ρ(z)Agdz

A= −ρ(z)gdz

⇒ P (z + dz) − P (z)

dz= −ρ(z)g


If dz is very small, then the left hand side is the definition of the derivative dP/dz. Thus,

dP

dz= −ρ(z)g

b. Use the ideal gas law to write the density of air in terms of pressure, temperature, andthe average mass m of the air molecules. Show, then, that the pressure obeys the differentialequation:

dP

dz= −mg

kTP,

called the barometric equation.

We can derive this from part a by finding an expression for ρ(z). The density of the slab isequal to the mass of the slab, M divided by its volume, V . The mass, in turn, is equal tothe mass per molecule, m, times the number of molecules, N . Thus:

ρ =mN

V

The ideal gas law says that PV = NkT , so that N/V = P/kT . Thus ρ = mP/kT , and

dP

dz= −mP

kTg = −mg

kTP (z)

c. Assuming that the temperature of the atmosphere is independent of height, solve thebarometric equation to obtain the pressure as a function of height: P (z) = P (0)e−mgz/kT .Show also that the density obeys a similar equation.

For convenience, let’s write α = mg/kT , so that

dP

dz= −αP (z)

This is one of the most basic differential equations, but if you’ve never seen it before, wecan solve it by separation of variables: put the pressure on one side, and the altitude z onthe other:

dP

dz= −αP (z)

dP

P= −αd(z)

∫ P (z)

P (0)

dP

P= −

∫ z

0αd(z)

ln P (z) − ln P (0) = −αz

ln

(

P (z)

P (0)

)

= −αz

P (z)

P (0)= e−αz

P (z) = P (0)e−αz

Substituting in the definition for α gives us P (z) = P (0)e−mgz/kT as expected.


We showed in part b that ρ = mP/kT , so

ρ(z) =m

kTP (0)e−αz,

we could also write mP (0)/kT as ρ(0), the density at z = 0. Note that the constant α hasunits of m−1; the constant z0 = 1/α is the length scale over which the pressure and densityvary.

d. Estimate the pressure, in atmospheres, at Mt. Everest. (Assume that the pressure at sealevel is 1 atm.)

We can write:

P (z) = P (0)e−z/z0

where z0 = kT/mg and it is convenient to calculate this quantity first. Assume that T =280K. Air is 80% N2, which has an atomic mass of 28, and 20% O2 which has an atomicmass of 32. An atomic mass unit is 1.67 × 10−27 kg (roughly the mass of a proton), so theaverage mass per molecule is:

m = 80%(28)(1.67 × 10−27 kg) + 20%(32)(1.67 × 10−27 kg) = 4.8 × 10−26 kg

and so

z0 =(1.38 × 10−23 J/K)(280K)

(4.8 × 10−26 kg)(9.8 m/s2)= 8200 m = 8.2 km

(Because this is so large, pressure and density does not vary appreciably inside a room).Now P (0) is the pressure at sea level, thus 1 atm, and so P (z) = (1atm)ez/8200m, and thusMt. Everest = 8850m, pressure = 0.34 atm

• 1.18: Calculate the rms speed of a nitrogen molecule at room temperature.

We know:

vrms =

√

3kT

m

One nitrogen atom has mass 14.00 u.1.66×10−27 kg1 u

= 2.3 × 10−26 kg, so a nitrogen moleculehas (approximately) twice that mass, or m = 4.6 × 10−26 kg. Thus

vrms =

√

3(1.38 × 10−23 J/K)(300K)

4.6 × 10−26 kg= 5200 m/s

• 1.19: Suppose you have a gas containing hydrogen and oxygen molecules, in thermal equi-librium. Which molecules are moving faster, on average? By what factor?

The rms speed of the molecules in an ideal gas is vrms =√

3kTm

. The ratio of the speeds ofhydrogen and oxygen at the same temperature is:

vH

vO=

√

mO

mH

Now oxygen molecules are sixteen times more massive than hydrogen molecules, so vH/vO =√16 = 4. So hydrogen molecules will be moving four times faster than oxygen molecules.


• 1.21: During a hailstorm, hailstones with an average mass of 2 g and a speed of 15 m/sstrike a windowpane at a 45 angle. The area of the window is 0.5 m2 and the hailstones hitit at a rate of 30 /s. What average pressure do they exert on the window? How does thiscompare to the pressure of the atmosphere?

The hailstones strike the window at intervals of 1/30 s on average. During this time period,the average force exerted by the window on the hailstone must be:

→

F x= m∆vx

∆t

where ∆vx is the change in the component of the hailstone’s velocity perpendicular to thewindow. Assuming elastic collisions and a velocity of 15m/s at 45, this change in velocityis 2vcos(45) = 21m/s. The average pressure is then

→

P=

→

F

A=

m∆vx

A∆t=

(0.002 kg)(21 m/s)

(0.5 m2)(0.033 s)= 2.5 N/m2 = 2.5 Pa

This is smaller than atmospheric pressure by a factor of 40,000. (However, the instantaneouspressure during a collision is much higher, as the force is localized in time and space; this iswhat might cause the window to break.)

• 1.22: If you poke a hole in a container full of gas, the gas will start leaking out. In thisproblem you will make a rough estimate of the rate at which gas escapes through a hole.

a. Consider a small portion (area A) of the inside wall of a container full of gas. Show that

the number of molecules colliding with this surface in a time interval ∆t is PA∆t/(2m→v x),

where P is the pressure, m the average molecular mass, and→v x is the average x velocity of

those molecules that collide with the wall.

As in Eq. 1.9, Newton’s Laws imply that each molecule colliding with the surface exerts anaverage pressure of:

→

P= −m∆vx

A∆t

For an elastic collision, ∆vx = −2vx. If there are N molecules, the total pressure is the sumof N terms of this form, one for each molecule; the sum over vx values can be written as Ntimes the average, or N

→v x. Therefore

P =m(2N

→v x)

A∆t⇒ N =

PA∆t

2m→v x

b. It’s not easy to calculate→v x, but a good enough approximation if (v2

x)0.5, where the bar

now represents an average over all molecules in the gas. Show that (v2x)

0.5 =√

kT/m.

This result follows directly from Eq. 1.15: kT = mv2x ⇒ v2

x =√

kT/m

c. If we now take away this small part of the wall, the molecules that would have collidedwith it will instead escape through the hole. Assuming that nothing enters through the hole,show that the number N of molecules inside the container as a function of time is governedby the differential equation:

dN

dt= − A

2V

√

kT

mN


Solve this equation (assuming constant T ) to obtain a formula of the form N(t) = N(0)e−t/τ

where τ is the “characteristic time” for N (and P ) to drop by a factor of e.

What we called N above is now −∆N , the change in the number of molecules in thecontainer. Substituting this into part (a), and bringing in part (b), gives us:

−∆N =PA∆t

2m

√

m

kT

Now use the ideal gas law to eliminate P , and divide through by ∆t, taking the limit ∆t → 0to get the derivative:

−δN =(NkT/V )A∆t

2m

√

m

kT

∆N

δt= −N

kTA

2mV

√

m

kT

dN

dt= − A

2V

√

kT

mN = −1

τN

where τ = (2V/A)√

m/kT . Thus N(t) is a function whose derivative is equal to −1/τ timesitself, which makes it an exponential:

N(t) = N0e−t/τ

d. Calculate the characteristic time for air at room temperature to escape from a 1-litrecontainer punctured by a 1mm2 hole.

To calculate τ we start with the quantity√

kT/m =√

RT/M , where M is the molar mass

of the gas (that is, the mass of a mole of air). Assuming the gas is near room temperature:

√

RT

M=

√

√

√

√

(8.3 J/mol/K)(300 K)

0.029 kg/mol= 293 m/s ≈ 300 m/s

A one liter bottle has volume 10−3m3, and 1 mm2 = 10−6 m2, so

τ =2V

A√

RT/M=

2(0.001 m3)

(10−6 m2)(300 m/s)= 7 s

e. The tire “going flat” I assume to mean that et/τ ≈ 0 in the shortest time. For example,e−3 ≈ .05, which is close to 0. That means:

− t

τ− 3 → t = 3τt = 3τ = 3

2V

A

√

m

kBT=

6V

A(3.42 × 10−3)

Again, I assume V = 1 L and 1 hour = 3600 s.

3600s =6(10−3 m3)

A(3.42 × 10−3 s/m

Calculate the area A;

A = 5.7 × 10−9 m2 = 5.7 × 10−3 mm2

f. Assuming that in time ∆t, all particles within the half sphere of radius vx∆t surroundingthe hole escape, since vx is on the order of several hundreds of meters per second on averageit seems unlikely that the travelers could toss the corpse out the window in a short enoughtime for an insignificant number of particles to escape. Of course, this hinges on the thedefinition of significant- if the space ship was very large, the percentage of air compared tothe total volume of the ship would be very small and perhaps termed “insignificant.”


• 1.26: A battery is connected in series to a resistor, which is immersed in water (to prepare anice hot cup of tea). Would you classify the flow of energy from the battery to the resistoras “heat” or “work”? What about the flow of energy from the resistor to the wire?

Energy flowing from the battery to the resistor should be classified as work, since it is aprocess that would not happen spontaneously. The flow of energy from the resistor to thewater is heat, since it is a spontaneous process.

• 1.27: Give an example of a process in which no heat is added to a system but its temperatureincreases. Then give an example of the opposite: a process in which heat is added to a systembut its temperature does not change.

An example of a process in which no heat is added to a system but the temperature increasesis pumping a tire full of air. A process in which heat is added to a system but its temperatureremains constant is a phase change, melting a piece of ice for example.

• 1.28: Estimate how long it would take to bring a cup of water to boiling temperature ina typical 600-watt microwave oven, assuming that all the energy ends up in the water.(Assume any reasonable initial temperature for the water). Explain why no heat is involvedin this process.

There are a few ways to approach this problem. ∆U = Q + W where, for this problem,Q = 0 so ∆U = W . Looking at the change in thermal energy (eqn 1.23) for a system of Nmolecules with f degrees of freedom:

∆U = Nf1

2k∆T

Assuming that the water is at 20 C at the start and it boils at 100 C, then ∆T = (100 -20) =80 C. H2O = 2H + O ⇒ 2 g/mol + 16 g/mol ⇒ 18 g/mol. A typical cup of water isabout 200g, and so:

N =(200 g)

(18 g/mol)

(6.02 × 1023 molecules)

(1 mol)

N = 6.7 × 1024 molecules

The number of degrees of freedom, f , is not so clearly defined in the problem. There arecertainly 3 translational degrees of freedom and there are 3 rotational. We do not know ifthe vibrational degrees of freedom are frozen out, but if we assume that they are then f = 6.

∆U ≈ 6.7 × 1024 6

21.38 × 10−23 73 J∆U ≈ 2.02 × 104 J

For a 600 W oven, the “heating” rate is 600 J/s, so:

∆t ≈ ∆U

600 J/s≈ 34 s

Alternatively, we can look at the definition of a calorie. 1 calorie = 4.186 J is the amountof energy required to raise the temperature of 1 g of water by 1 C. So, for 200 g of waterwe need:

U = (4.186 J/g/C)(200 g)(80 C)

U = 6.11 × 104 J


So,

∆t ≈ 6.11 × 104

600≈ 102 s

Clearly, f = 6 is not correct for liquid water and there must be a larger number of degreesof freedom.

Remember that flow of heat is a spontaneous process from higher T to lower T . Themicrowave overn is an example of electro-magnetic work.

• 1.31: Imagine some helium in a cylinder with an initial volume of 1 litre and an initialpressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 litres, insuch a way that the pressure rises in direct proportion to its volume. (a) Sketch a graph ofpressure vs volume for this process.

(b) Calculate the work done on the gas during this process, assuming that there are no“other” types of work being done.

The work done is just minus the area under the graph shown in (a). The easiest way tocompute the area is to note that the average pressure during this process is 2 atm and:

W = −P∆V = −(2 atm)(2 l)

W = −(2 × 105 Pa)(2 × 10−3 m3)

W = −400 J

The minus sign indicates that 400 J of work is done by the gas on its surroundings.

(c) Calculate the change in the helium’s energy content during this process

Each helium atom has three degrees of freedom, so at any point the thermal energy of thehelium is U = 3

2NkT = P

V. The change in energy during this process is:

∆U =3

2[PfVf − PiVi]

∆U =3

2[(3 atm)(3 l) − (1 atm)(1 l)]

∆U = 12 atm l = 1200J


(d) Calculate the amount of heat added to or removed from the helium during this process.

By the first law,

Q = ∆U − W

Q = 1200 − (−400) = 1600 J

This amount of heat enters the gas.

(e) Describe what you might do to cause the pressure to rise as the helium expands?

To cause such an increase in pressure (and temperature) as the gas expands, you mustprovide heat, for example, by holding a flame under the cylinder and letting the piston outslowly enough to allow the pressure to rise as desired.

• 1.33: An ideal gas is made to undergo the cyclic process shown in the figure. For each of thesteps A, B, and C, determine whether each of the following is positive, negative, or zero: (a)the work done on the gas, (b) the change in the energy content of the gas, and (c) the heatadded to the gas. Then determine the sign of each of these three quantities for the wholecycle. What does this process accomplish (in real-world terms)?

The work done on the gas depends entirely on the change in volume; decreasing volumemeans that work is positive, while increasing volume means that work is negative. Thethermal energy of the gas is directly proportional to the temperature, which is in turnproportional to PV . The heat must then be accounted for via the equation ∆U = W + Q.

For branch A, the volume increases, so W < 0. PV increases as well, so the thermal energyis increasing. Since the energy increases but work flows out, there must be a flow of heatinward to compensate, and Q > 0.

For branch B, there is no change in volume so W = 0. PV increases, so ∆U > 0. Becausethe energy is increasing, there must be a flow of heat inward.

For branch C, volume and pressure are both decreasing, so W > 0 and ∆U < 0. Becausework flows in and the energy still decreases, there must be a net flow of heat outward, soQ < 0. To summarize:

Branch energy W QA increases negative positiveB increases zero positiveC decreases positive negative

During one complete cycle, PV comes back to its original value, so the thermal energy doesnot change. The work done on this system is positive (because branch C has a greater area


under it than branch A, and thus has more work flowing). Because the system’s energydoesn’t change, Q must be negative: heat flows out of the system.

In short, this cycle transforms work into heat; it could, for instance, be some sort of spaceheater. If we note, however, that this system both absorbs heat (in steps A and B) and emitsheat, we can think of a more sophisticated use: in particular, we can absorb heat from oneroom and emit it into the other. Similar cycles can act as refrigerators, but this particularcycle would not make a practical refrigerator because the heat absorption stages are not ata lower temperature than the emission stages.

• 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangularcyclic process shown in the figure below. Assume that the temperature is always such thatthe rotational degrees of freedom are active but that the vibrational modes are “frozenout.” Also assume that the only type of work done on the gas is quasidiabatic compression-expansion work.

(a) For each of the four steps A through D, compute the work done om the gas, the heatadded to the gas, and the change in the energy content of the gas. Express all answers interms of P1, P2, V1 and V2 )Hint: compute ∆U before Q using the ideal gas law and theequipartition theorem.)

For an ideal diatomic gas at room temperature, U = 52NkBT = 5

2PV (since PV = NkBT )

Step A:

∆UA =5

2V ∆P =

5

2V1(P2 − P1)

WA = 0

QA = ∆UA − WA =5

2V1(P2 − P1)

Step B:

∆UB =5

2P∆V =

5

2P2(V2 − V1)


WB = −∫ V2

V1

PdV = −P2(V2 − V1) = P2(V1 − V2)

QB = ∆UB − WB =5

2P2(V2 − V1) + P2(V1 − V2) =

7

2P2(V2 − V1)

Step C:

∆UC =5

2V ∆P =

5

2V2(P1 − P2) = −5

2V2(P2 − P1)

WC = 0

QC = ∆UC − WC = −5

2V2(P2 − P1)

Step D:

∆UD =5

2P∆V = −5

2P1(V2 − V1)

WD = −∫ V1

V2

PdV = −P1(V1 − V2) = P1(V2 − V1)

QD = ∆UD − WD = −5

2P1(V2 − V1) − P2(V2 − V1) = −7

2P1(V2 − V1)

(b) Describe in words what is physically being done during each of the four steps; for example,during Step A, heat is added to the gas (from an external flame or something) while thepiston is held fixed.

In step A, the volume is held constant at V1, and heat is added to the gas in the cylinder,thus raising its temperature and pressure. In step B, the piston is moved out to increasethe volume from V1 to V2, and, at the same time, heat is added so as to maintain constantpressure at P2. In step C, the volume is held constant at V2, and heat is removed from thecylinder to lower the pressure from P2 to P1. Finally, in step D, the gas is compressed fromV2 to V1, while continuing to remove heat, so that the pressure is maintained at P1.

(c) Compute the net work done on the gas, the net heat added to the gas, and the net changein the energy of the gas during the entire cycle. Are the results as you expected? Explainbriefly.

For the cyclical process, we have:

∆Ucycle = ∆UA + ∆UB + ∆UC + ∆UD

∆UD =5

2V1(P2 − P1) +

7

2P2(V2 − V1) + −5

2V2(P2 − P1) + −7

2P1(V2 − V1)

∆Ucycle = 0

Similarly,

Wcycle = −P2(V2 − V1) + P1(V2 − V1)

Wcycle = −(P2 − P1)(V2 − V1)

and

Qcycle =5

2V1(P2 − P1) +

7

2P2(V2 − V1) −

5

2V2(P2 − P1) −

7

2P1(V2 − V1)

Qcycle =5

2(P2 − P1)(V1 − V2) +

7

2(P2 − P1)(V2 − V1)

Qcycle = (P2 − P1)(V2 − V1)


Yes, the results are as expected. For a complete cycle, the net change in the internal energyU should be zero, since U is a function of state. Furthermore, from the first law, we have∆U = Q + W ⇒ Q = −W if ∆U = 0.

• 1.38: Two identical bubbles of gas form at the bottom of a lake, then rise to the surface.Because the pressure is much lower at the surface than at the bottom, both bubbles expandas they rise. However, bubble A rises very quickly so that no heat is exchanged between itand the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so thatit always remains in thermal equilibrium with the water (which has the same temperatureeverywhere). Which of the two bubbles is larger by the time they reach the surface? Explainyour reasoning fully.

Both bubbles are identical to begin with, so both have the same number of air molecules Ninside. The pressure inside each bubble must be more-or-less equal to the pressure of thewater (we ignore the effects of surface tension here), so both bubbles end up at the samepressure as well. Therefore the volume of each bubble, V = NkT/P , depends entirely ontemperature: the warmer bubble will be bigger. That also means that the larger bubble willhave the larger thermal energy. Now both bubbles do work on the surrounding water as theyexpand (W < 0) Because bubble A does not exchange heat with its environment, it losesenergy and so cools down. Bubble B, on the other hand, remains at a constant temperature.Therefore, by our reasoning above, bubble B, the one that absorbs heat, will be larger.

• 1.40: In problem 1.16 you calculated the pressure of the earth’s atmosphere as a functionof altitude, assuming constant temperature. Ordinarily, however, the temperature of thebottommost 10-15 km of the atmosphere (called the troposphere) decreases with increasingaltitude, due to heating from the ground (which is warmed by sunlight). If the temperaturegradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-densityair will rise while cool, high-density air sinks. The decrease of pressure with altitude causesa rising air mass to expand adiabatically and thus to cool. The condition for convection tooccur is that the rising air mass must remain warmer than the surrounding air despite thisadiabatic cooling. (a) Show that when an ideal gas expands adiabatically, the temperatureand pressure are related by the differential equation:

dT

dP=

2

f + 2

T

P

From the book:

PV γ = const.

V T f/2 = const.

Using these and solving for V gives:

V =1

PT (f+2)/2 = const.

This can be written in terms of natural logarithms as follows:

ln T (f+2)/2 − ln P = const.f + 2

2ln T − ln P = const.


Now if we take a derivative:(

f + 2

2

)

dT

T− dP

P= 0

Simplifying this gives the desired differential equation

dT

dP=

2

f + 2

T

P

(b)Assume that dT/dz is just at the critical value fro convection to begin, so that thevertical forces on a convecting air mass are always approximately in balance. Use the resultof problem 1.16(b) to find a formula for dT/dz in this case. This result should be a constant,independent of temperature and pressure, which evaluates to approximately -10/km. Thisfundamental meteorological constant is known as the dry adiabatic lapse rate.

From Problem 1.16 (b):

dP

dz= − mg

kBT

From the answer in part (a), dT/dP can be decomposed into the product

dT

dP=

dT

dz

dz

dP

and therefore

dT

dz

dz

dP=

2

f + 2

T

P

Substituting in the expression for dP/dz:

dT

dz

(

−kBT

mg

)

1

P=

2

f + 2

T

P

The T s and P s cancel, and we are left with (after solving for dT/dz)

dT

dz=

mg

kB

(

2

f + 2

)

Using the mass of air, mair = 4.84 × 10−26 kg/molecule. Substituting in all the numbers:

dT

dz=

(4.84 × 10−26 kg/molecule)(9.8 m/s2)

1.38 × 10−23 J/K

2

5 + 2

dT

dz= .0098

K

m≈ .01

K

m

To change to K/km we just have to mutliply by 1000 m/km, and we get

dT

dz= 10

K

km

Which was to be shown.


• 2.1: Suppose you flip four fair coins.

(a) Make a list of all the possible outcomes, as in Table 2.1.

The number of outcomes is 24 = 16.

TTTT THTT HTTT HHTTTTTH THTH HTTH HHTHTTHT THHT HTHT HHHTTTHH THHH HTHH HHHH

(b) Make a list of all the different “macrostates” and their probabilities.

0 Heads Ω = 1 P = 1/161 Heads Ω = 4 P = 4/162 Heads Ω = 6 P = 5/163 Heads Ω = 4 P = 4/164 Heads Ω = 1 P = 1/16

(c) Compute the multiplicity of each macrostate using the combinatorial formula (2.6), andcheck that these results agree with what you got by brute force counting.

Ω(0, 4) = 4!0! 4!

Note : 0! = 1 So, Ω = 1, P=1/16.Ω(1, 4) = 4!

1! 3!So, Ω = 24

6, P=4/16.

Ω(2, 4) = 4!2! 2!

So, Ω = 244, P=6/16.

Ω(3, 4) = 4!3! 1!

So, Ω = 246, P=4/16.

Ω(4, 4) = 4!4! 0!

So, Ω = 1, P=1/16.

• 2.2: Suppose you flip 20 coins. (a) How many possible outcomes “macrostates” are there?

For only 2 possible states ⇒ 220 microstates. 220 = 1,048,576. Note that this is Ω(all) orΩ(total)

(b) What is the probability of getting the sequence HTHHTTTHTHHHTHHHHTHT (inexactly that order)?

Simply, this is only one macrostate out of 220. Since each state is equally probable, thepossibility of getting this particular one is 1/220 ⇒ less than one in a million.

(c) What is the probability of getting 12 heads and 8 tails (in any order)=

This is another macrostate:

P =Ω(20, 12)

Ω(all)=

20!

12! (20−12)!1048576

=125970

1048576= 0.12

So, the probability is about 12%

• 2.3: Suppose you flip 50 coins.

(a) How many possible outcomes (microstates) are there?

There are two outcomes for each coin, so total number of macrostates: Ω(all) = 250 =1.13 × 1015

(b) How many ways are there of getting exactly 25 heads and 25 tails?


To get exactly 25 heads:

Ω(q, N) =N !

(N − q)! q!

Ω(25, 50) =50!

(50 − 25)! 25!= 1.26 × 1014

(c) What is the probability of getting exactly 25 heads and 25 tails?

P (25) =Ω(25, 50)

Ω(all)

P (25) =1.26 × 1014

250= 0.112

So, not so large.

(d) What is the probability of getting exactly 30 heads and 20 tails?

P (30) =Ω(30, 50)

Ω(all)

P (30) =1

250

50!

30! 20!= 0.042

(e) What is the probability of getting exactly 40 heads and 10 tails?

P (40) =Ω(40, 50)

Ω(all)

P (40) =1

250

50!

40! 10!= 0.0000091 = 9.1 × 10−6

(f) What is the probability of getting exactly 50 heads and 0 tails?

P (50) =1

Ω(all)= 8.9 × 10−16

(g) Plot a graph of the probability of getting n heads as a function of n

• 2.4: Calculate the number of possible five-card poker hands, dealt from a deck of 52 cards.(The order of cards in a hand does not matter.) A royal flush consists of the five highest-ranking cards (ace, king, queen, jack, 10) of any one of the four suits. What is the probabilityof being dealt a royal flush (on the first deal)?

The number of hands:

N =52!

5! 47!= 2598960

The probability of getting a royal flush:

P =4!

5! 47!= 2598960


• 2.5: For an Einstein solid with each of the following values of N and q, list all of the possiblemicrostates, count them, and verify formula 2.9

(a) N = 3, q = 4 : Ω(N, q) = (q+N−1)!q! (N−1)!

= 15

Oscillator Oscillator1 2 3 1 2 34 0 0 0 1 30 4 0 2 2 00 0 4 2 0 23 1 0 2 1 13 0 1 0 2 21 3 0 1 2 10 3 1 1 1 21 0 3

15 states.

(b) N = 3, q = 5 : Ω(N, q) = (q+N−1)!q! (N−1)!

= 21

Oscillator Oscillator1 2 3 1 2 35 0 0 3 1 10 5 0 2 3 00 0 5 0 3 24 1 0 1 3 14 0 1 2 0 31 4 0 0 2 30 4 1 1 1 31 0 4 2 2 10 1 4 2 1 23 2 0 1 2 23 0 2

21 states.


(c) N = 3, q = 6 : Ω(N, q) = (q+N−1)!q! (N−1)!

= 28

Oscillator Oscillator1 2 3 1 2 36 0 0 1 4 10 6 0 2 0 40 0 6 0 2 45 1 0 1 1 45 0 1 3 3 01 5 0 3 0 30 5 1 3 2 11 0 5 3 1 20 3 5 3 3 04 2 0 0 3 34 0 2 2 3 14 1 1 1 3 22 4 0 2 1 30 4 2 1 2 3

28 states.

(d) N = 4, q = 2 : Ω(N, q) = (q+N−1)!q! (N−1)!

= 10

Oscillator1 2 3 42 0 0 00 2 0 00 0 2 00 0 0 21 1 0 01 0 1 01 0 0 10 1 1 00 1 0 10 0 1 1

10 states.

(e) N = 4, q = 3 : Ω(N, q) = (q+N−1)!q! (N−1)!

= 20

Oscillator Oscillator1 2 3 4 1 2 3 43 0 0 0 1 0 2 00 3 0 0 0 1 2 00 0 3 0 0 0 2 10 0 0 3 1 0 0 22 1 0 0 0 1 0 22 0 1 0 0 0 1 22 0 0 1 1 1 1 01 2 0 0 1 1 0 10 2 1 0 1 0 1 10 2 0 1 0 1 1 1

20 states.

(f) N = 1, q = anything

Ω(N, q) = (q+N−1)!q! (N−1)!

= q Oscillator 1: q


(g) N = anything, q = 1

Ω(N, q) = (q+N−1)!q! (N−1)!

= N

• 2.6: Calculate the multiplicity of an Einstein solid with 30 oscillators and 30 units of energy.(Do not attempt to list all the microstates)

Ω(30, 30) = (q+N−1)!q! (N−1)!

= 59!30! 29!

= 5.91 × 1016

• 2.7: For an Einstein solid with four oscillators and two units of energy, represent each possiblemicrostate as a series of dots and lines, as used in the text to prove equation 2.9

N = 4, q = 2, Ω(N, q) = (q+N−1)!q! (N−1)!

= 10

Oscillator Dots and lines1 2 3 4 1 2 3 42 0 0 0 ··0 2 0 0 ··0 0 2 0 ··0 0 0 2 ··1 1 0 0 · ·1 0 1 0 · ·1 0 0 1 · ·0 1 1 0 · ·0 1 0 1 · ·0 0 1 1 · ·

10 states.

• 2.8: Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharinga total of 20 units of energy. Assume that the solids are weakly coupled, and that the totalenergy is fixed.

(a) How many different macrostates are available to this system?

Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 differentmacrostates are available to the system of weakly coupled A and B solids sharing 20 unitsof energy.

(b) How many different microstates are available to this system?

The combined system has 20 oscillators and 20 units of energy. Therefore,

Ω(N, q) =(q + N − 1)!

q!(N − 1)!=

39!

20! 19!= 6.89 × 1010

(c) Assuming that this system is in thermal equilibrium, what is the probability of findingall of the energy in solid A?

For the macrostate with all energy in solid A:

Ω = ΩAΩB = ΩA(10, 20)ΩB(10, 0) =29!

20! 9!= 1 × 107

The probability of this macrostate is:

P =1 × 107

6.89 × 1010= 1.45 × 10−4

(d) What is the probability of finding exactly half the energy in solid A?


For the macrostate with exactly half the energy in solid A:

Ω = ΩAΩB = ΩA(10, 10)ΩB(10, 10) =19!

10! 9!= 8.53 × 109

P =8.53 × 109

6.89 × 1010= 0.124

(e) Under what circumstances would this system exhibit irreversible behaviour?

If the energy was initially all in A, the system would evolve toward half energy in A andhalf energy in B, and this distribution of energy would be irreversible. In other words, oncethe energy is evenly distributed, it is highly unlikely that the system would find itself withall the energy back in A.

• 2.16: Suppose you flip 1000 coins. (a) What is the probability of getting exactly 500 headsand 500 tails? (Hint: First write down a formula for the total number of possible outcomes.Then, to determine the “multiplicity” of the 500-500 “macrostate”, use Sterling’s approx-imation. If you have a fancy calculator that makes Stirling’s approximation unnecessary,multiply all the numbers in this problem by a sufficient factor that you have to manuallymake this approximation.

First, note that the number of possible outcomes (microstates) is 21000. The probability ofgetting 500 heads and 500 tails is (from Sterling’s approximation):

Ω(500) =

(

1000500

)

=1000!

(500!)2

Ω(500) ≈ 10001000e−1000√

2π 1000(

500500e−500√

2π 500)2 =

21000

√500π

The probability is Ω(500)/Ω(all) :

Ω(500) =21000

21000√

500π=

1√500π

= 0.025

So the chance of getting exactly 500 heads is about 2.5%, or 1 in 40.

(b)What is the probability of getting exactly 600 heads and 400 tails?

From Sterling’s approximation:

Ω(600) =

(

1000600

)

=1000!

600! 400!

Ω(600) ≈ 10001000e−1000√

2π 1000

600600e−600√

2π 600 400400e−400√

2π 400=

10001000

600600 400400√

480π

The probability is Ω(600)/Ω(all) :

Ω(600) =10001000

21000 600600 400400√

480π=

5001000

600600 400400√

480π=

500600 500400

600600 400400√

480π

Ω(600) =(

5

6

)600 (5

4

)400 1√480π

So the chance of getting exactly 600 heads is 4.6 × 10−11, much smaller than P (500)


• 2.21: Use a computer to plot formula 2.22 directly, as follows. Define z = qa/q such that(1 − z) = qb/q. Then, aside from a few constants that can be ignored, the multiplicityfunction is [4z(1 − z]N , where z ranges from 0 to 1 and the factor 4 ensures that the heightof the peak is equal to 1 for any N. Plot this function for N=1, 10, 100, 1000, and 10,000.Observe how the width of the peak decreases as N increases.

• 2.23: Consider a two-state paramagnet with 1023 elementary dipoles, with half the totalenergy fixed at zero so that exactly half the dipole point up and half point down.

(a) How many microstates are “accessible” to this system?

With N dipoles of which exactly N/2 point up, the multiplicity of the paramagnet is:

Ω =

(

NN2

)

⇒ N !(

N2

)

!(

N2

)

!

So, taking the Sterling approximation:

Ω =NNe−N

√2πN

[

(

N2

)N/2e−N/2

√πN

]2 = 2N

√

2

πN

For N = 1023, then, the multiplicity is roughly: 21023

/4 × 1011 Since the denominator ismerely “large,” we could just as well neglect it and say that Ω = 21023

, which is the numberof microstates if we allow any number of dipoles to be pointing up.

(b)Suppose that the microstate of this system changes a billion times per second. How manymicrostates will it explore in ten billion years (the age of the universe)?

A year is about 3 × 107 s, so 10 billion years is about 3 × 1017 s, or 3 × 1026 ns. If themicrostate of the system changes once every nanosecond, this is how many microstates thesystem will explore in the age of the universe. But this is a tiny fraction of the total numberof microstates. In fact, the fraction is so small that the ratio of states not explored to statesexplored is 21023

.


(c) Is it correct to say that, if you wait long enough, a system will eventually be found inev ery “accessible” microstate? Explain your answer, and discuss the meaning of the word“accessible.”

Even if we wait for the age of the universe, the fraction of all “accessible” microstates thatare actually explored by this system is so tiny that it might be more accurate to say that thesystem explores none of its “accessible” microstates. When we call a microstate “accessible,”therefore, we should not think that the system will ever actually be in that microstate. Sowhat do we mean? One of the best interpretations is in terms of our ignorance of whichmicrostates the system will actually explore in the future. Forallweknow, the system mightsoon be found in any of its “accessible” microstates, even though the probability of its beingfound in any of them is vanishingly small.

• 2.24: For a single large two-state paramagnet, the multiplicty function is very sharply peakedabout N↑ = N/2 (a) Use Sterling’s approximation to estimate the height of the peak in themultiplicity function.

If N = N↑, then N↑ = N↓ = N2

and the multiplicity is:

Ωmax =N !

N↑! N↓!=

N !(

N2!)2

Ωmax ≈ NNe−N√

2πN[

(

N2

)N/2e−N/2

√

2πN2

]2 = 2N

√

2

π N

Using Sterling’s approximation:

Ω =N !

N↑! N↓!≈ NNe−N

√2πN

NN↑

↑ e−N↑

√

2πN↑ NN↓

↓ e−N↓

√

2πN↓

=NN

NN↑

↑ NN↓

↓

√

N

2πN↑N↓

(b) Use the methods discussed in the lectures to derive a formula for the multiplicity functionin the vicinity of the peak, in terms of x ≡ N↑ − (N/2). Check that your formula agreeswith your answer to part (a) when x = 0

Using x ≡ N↑ − (N/2), we can rearrange for N↑:

N↑ =N

2+ x

and solve for N↓:

N↓ = N − N↑ ⇒ N↓ = N − N

2− x ⇒ N↓ =

N

2− x

Using the equation obtained from Sterlings approximation:

Ω =NN

NN↑

↑ NN↓

↓

√

N

2πN↑N↓

setting N↑ = (N/2) + x and N↓ = (N/2) − x:

Ω =NN

(

N2

+ x)N/2+x (

N2− x

)N/2−x

√

√

√

√

N

2π(

N2

+ x) (

N2− x

)


Rearranging . . .

Ω =N

[

(

N2

)2 − x2

]N/2 (N2

+ x)x (

N2− x

)−x

√

√

√

√

√

N

2π[

(

N2

)2 − x2

]

Apply the logarithmic approximation to this equation:

lnΩ ≈ N lnN − N

2ln

[

(

N

2

)2

− x2

]

− x ln(

N

2+ x

)

+ x ln(

N

2− x

)

+ ln

√

N

2π− 1

2ln

[

(

N

2

)2

− x2

]

Nothing yet has been assumed about the size of x relative to N but if x ≪ N , the logarithmscontaining two terms can be expanded:

ln

[

(

N

2

)2

− x2

]

= ln(

N

2

)2

+ ln

[

1 −(

2x

N

)2]

≈ 2 ln(

N

2

)

−(

2x

N

)2

ln(

N

2± x

)

= ln(

N

2

)

+ ln[

1 ± 2x

N

]

≈ ln(

N

2

)

± 2x

N

Using these expressions, we get:

lnΩ = N ln N − N lnN

2+

2x2

N− x ln

N

2− 2x2

N+ x ln

N

2− 2x2

N

+ ln

√

N

2π− ln

N

2+

2x2

N2

lnΩ = N ln 2 − 2x2

N+ ln

√

2

πN− 2x2

N2

The last term is clearly negligible and can be neglected. Taking the exponential:

eln Ω ≈ eN ln 2− 2x2

N+ln

√2

πN

Ω = 2N

√

2

πNe−2x2/N for x ≪ N

This is a Gaussian function, peaked at x = 0. Clearly, when x = 0:

Ω = 2N

√

2

πN

which is the same result obtained in part (a).

(c) How wide is the peak in the multiplicity function?

The Gaussian function falls off to 1/e of its peak value when:

2x2

N= 1 ⇒ x =

√N

2

But the width of the peak will be 2x so: width =√

2N


(d) If you flip one million coins. Would you be surprised to obtain 501,000 heads and 499,999tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

Here, N = 106 x = N2

+ 103. The half-width of the peak in the multiplicity function would

be√

500000 ≈ 700. So an excess of 1000 heads is only a little beyond the point where theGaussian has fallen off to 1/e of its maximum value. It would nit be too surprising to obtainapproximately this many heads, though it would be surprising to obtain an excess of exactly1000. Conversely, an excess of 10,000 heads lies fay outside the peak in the multiplicityfunction. At this point the Gaussian has fallen off to e200 ≈ 10−87 of its maximum value. Ifa result close to this was achieved, it’s likely that there is a problem with the coins!

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ThermalPhysics_1_solns