Thermodynamics | Custom PHD Thesis

Thermodynamics Lecture 2

Thermal equilibrium can be reached when two objects have been in contact for a long time.

Relaxation time is the time required for a system to come to thermal equilibrium/

Diffusive equilibrium between two substances (say , coffee-cream mixture) is reached when molecules of each substance are free to move

around but no longer has the tendency to one way or the other.

Mechanical equilibrium takes place when large-scale motions (expansion of a balloon) can take place but no longer do.

Keep in mind that energy is exchanged when two systems are in thermal equilibrium, volume is exchanged between two systems in

mechanical equilibrium, and particles are exchanged when two systems are in diffusive equilibrium.

Absolute zero is approached when the volume of a gas is held fixed, then its pressure will approach zero as the temperature approaches

– 273 oC. A absolute temperature scale or a Kelvin scale is of the same size as a degree Celsius, but the former is measured up from

absolute zero.

Q. 1. What is the absolute zero on the Fahrenheit scale?

Q. 2. Determine the Kelvin temperature for each of the following:

i) human body temperature

ii) the b. p. of water at standard pressure of 1 atm

iii) the coldest day you can remember

iv) -196 oC, the b. p. of liquid nitrogen

v) 327 oC, the m. p. of Pb.

Q. 3. Linear thermal expansion coefficient α= (∆L/L)/∆T

i) For steel α=1.1E-5 K-1, estimate the total variation in length of a 1-km steel bridge between a cold winter night and a hot summer

day. Ans 77 cm

Ideal Gas: Many of the properties of a low density gas can be summarized in the ideal gas equation.

PV= nRT

Where P = pressure, V= volume, n = number of moles of the gas, R is a universal constant= 8.31 J/mol.K (when you measure pressure in

N/m2 = Pa and volume in m3), and T is the temperature in kilvins
A mole of molecules is Avogadro’s number (NA = 6.02E23)of them.

Another form of ideal gas equation is

PV = NkT

whereN = n x NA

k is called the Boltzmann’s constant = 1.381E-23 J/K

Remember that nR = Nk
Q. 4. What is the volume of one mole of air at room temperature and 1 atm pressure?
N =n RT/P = 1mol x 8.31 J/mol.K x 300 K/(1E5 N/m2) Ans. 0.025 m3 = 25 liters

Q. 5. Estimate the number of molecule in your classroom.
N = PV/kT = (1E5 N/m2) x (Average room volume = sq area x height)/(1.38E-23 J/K x 300 K)
For 4 m x 4 m x 3m room size, it is 1.2E27 = !E27 = (2000 moles)

Q. 6. Calculate the mass of a mole of dry air which is a mixute on 78% N2, 21% O2, and 1% argon.
M = 0.78 x 28 g + 0.21 x 32 g + 0.01 x 40 g = 28.96 g = 29.0 g
Q. 7. Estimate the average temperature of the air inside a hot air balloon. Assume that the total mass of the unfilled balloon and the

payload is 500 kg. What is the mass of the air inside the balloon?
Let ρo = density of surrounding air, V = volume of the balloon, M = mass of the unfilled balloon and the payload, m = mas of 1 mole of

air, and ρ = density of air inside the balloon. Assume, outside temp T = 290 K, balloon volume = 1770 m3, mass of unfilled balloon and

payload = 500 kg

In flight, the buoyancy = weight of the displaced air is assumed to be equal to the weight of the payload, balloon, and the air inside

the balloon.
i) Mass of the displaced air = density x volume = ρo V = ? Weight of the displaced air = ρo V g. = ?
ii) Mass of air inside balloon = ρ V = ?
iii) Weight of this air = ρ V g = ?
iv) Weight of payload, balloon, and air = Mg + ρ V g = ?
v) From ρo V g = Mg + ρ V g , what is ρo – ρ = ?
vi) From gas equation, ρ = mn/V = mP/RT from ideal gas eqn., and ρo = mP/RTo. Substitute them in (v) above to get mP/RTo – mP/RT

Thermodynamics Lecture 3
Q. 4. What is the volume of one mole of air at room temperature and 1 atm pressure?
N =n RT/P = 1mol x 8.31 J/mol.K x 300 K/(1E5 N/m2) Ans. 0.025 m3 = 25 liters
Q. 5. Estimate the number of molecule in your classroom.
N = PV/kT = (1E5 N/m2) x (Average room volume = sq area x height)/(1.38E-23 J/K x 300 K)
For 4 m x 4 m x 3m room size, it is 1.2E27 = !E27 = (2000 moles)
Q. 6. Calculate the mass of a mole of dry air which is a mixute on 78% N2, 21% O2, and 1% argon.
M = 0.78 x 28 g + 0.21 x 32 g + 0.01 x 40 g = 28.96 g = 29.0 g
Q. 7. Estimate the average temperature of the air inside a hot air balloon. Assume that the total mass of the unfilled balloon and the

air, and ρ = density of air inside the balloon. Assume, outside temp T = 290 K, balloon volume = 1770 m3, mass of unfilled balloon and

payload = 500 kg
In flight, the buoyancy = weight of the displaced air is assumed to be equal to the weight of the payload, balloon, and the air inside

the balloon.
i) Mass of displaced air = density x volume = ρo V = ? Weight of the displaced air = ρo V g. = ?
ii) Mass of air inside balloon = ρ V = ?
iii) Weight of this air = ρ V g = ?
iv) Weight of payload, balloon, and air = Mg + ρ V g = ?
v) From ρo V g = Mg + ρ V g , what is ρo – ρ = ?
vi) From gas equation, ρ = mn/V = mP/RT from ideal gas eqn., and ρo = mP/RTo. Substitute them in (v) above to get mP/RTo – mP/RT

= M/V
vii) What is 1/T = ? (simplify to find)
viii) Substitute the known values to find the numerical answer, Ans. T = 370 K
ix) M air = mn = mPV/RT = 0.029 kg x 1E5 N/m2 x 1770 m3 / (8.31 J/K x 379 K) = 1600 kg
Microscopic Model of an Ideal Gas
Imagine a cylinder of length L, and piston cross-sectional area A containing just one gas molecule. Cylinder inside volume = ?
Let the velocity of the molecule be v having the horizontal component vx. The molecule bounces off the cylinder walls and changes its

velocity, but not the speed. Assume the collisions are elastic so that the molecule does not lose any energy.
Pressure = force/area. Ideal gas law gives PV = NkT.
The molecule periodically clashes into the piston and bounces off.
Average pressure on the pistion = average force on the piston/A

A
vx
v V = LA
L
= – average force on the molecule/A (justify how?)
= molecule mass x average acceleration/A
= m (∆vx/∆t)/A, where ∆t = time the molecule takes to undergo one round-trip from the left to the right and

back again.
 ∆t = total distance/velicity component = 2 L/vx
During the time the molecule undergoes exactly one collision with the piston, the change in its x-velocity = ∆vx = vx, final – vx,

initial = (- vx) – (vx) = -2vx
With this value of the velocity change, average pressure , P = – (m/A) (-2vx)/2L/vx) = mvx2/AL = mvx2/V
 PV = mvx2
 If there are N identical molecules with randon positions and directions of motion, PV = Nmvx2
 Putting NkT = PV, NkT = Nmvx2
 kT = mvx2
 1/2kT =1/2mvx2
The same kind of equation should hold for y and z directions also.
 Summing all of them, 1/2mvx2 + 1/2mvx2 + 1/2mvx2 = 1/kT + ½ kT, +!/2 kT =3/2 kT
Q. What is KT= how many Joules for an air molecule at room temperature?
= how many electrons volts? (1 eV = 1.6E-19 J)

Average speed of the molecules in a gas:
1/2mv2 =3/2 kT = > v2 average = 3kT/m.
Taking square roots of both sides, we get vrms = (3kT)1/2 = (3RT/M)1/2
1. Calculate rms speed on nitrogen molecule at room temp.
2. vrms = (3kT/m)1/2 = [(3T)(k/m)]1/2= [(3T)(k.NA/mNA)]1/2 = (3RT/M)1/2 = ? Ans. 517 m/s
M= molar mass = 0.028 kg
3. Suppose a container has hydrogen and oxygen molecules in thermal equilibrium. Which molecule moves faster, on average? By what

factor?
4. Uranium isotopes of atomic masses 238 and 235 form uranium hexafluoride UF6. Calculate the rms speed of each type of molecule

at room temperature, and compare them.
Uranium hexafluoride with U-238 has a molecular weight of 352 = 0.352 kg, and the other isotope will give a molecular weight of 349.
Use vrms = (3kT/m)1/2 = (3RT/M)1/2
Ans. heavier isotope has 145.8 m/s and lighter one has 146.4 m/s
THERMODYNAMICS LECTURE 4

Equipartition of energy. Equipartition theorem concerns all forms of energy for which the formula is a quadratic function of a

coordinate (viz. elastic potential energy 1/2ksx2 as a function of spring constant ks and the amount of displacement x from the

equilibrium) or velocity component (viz., translational motion in the x, y, and z directions – ½ mvx2, ½ mvy2, ½ mvz2, and rotational

and vibrational motions ½ Iωx2, ½ Iωx2). Each of these forms of energy is called a degree of freedom. Equipartition theorem says that

for each quadratic degree of freedom, the average energy will be ½ kT at temperature T. If a system has N molecules, each with f

degrees of freedom, and there are no other (non-quadratic) temperature-dependent forms of energy, then the total thermal energy is
Uthermal = N. f. 1/kT
This is just the average total thermal energy. For large N, fluctuations away from the average will be negligible. Uthermal is almost

never the total energy of a system because of the presence of some static energy like chemical bond energy or rest mass energy (mc2) of

the particles in the system.It is safest to apply equipartition theorem only to changes in energy following temperature variation to

avoid phase transformations and other reactions involving breaking bonds between particles.
Learning degrees of freedom in connection with equipartition theorem is important. In monatomic gas molecules like helium and argon,

only translational motion counts, so each molecule has three degrees of freedom, f=3. In diatomic gas molecules like O2 or N2, each

molecule can also rotate about two different axes at right angles to each other and perpendicular to the length of the molecule. The

same is true for CO2. Most polyatomic molecules can rotate about all three axes. Vibrations about the length-wise axis is prohibited by

quantum mechanics.

1.23 Calculate the total thermal energy of a liter of helium at room temperature and atmospheric pressure. Then repeat the calculation

for a liter of air
i) Is He monatomic or diatomic?
ii) What is the degrees of freedom per molecule?
iii) Write down the Uthermal formula =
iv) Put the value of f in (iii)
v) What is 3 N. (½)kT = ? in terms of PV
vi) P = 1 atmosphere = ? 1E5 N/m2
vii) V= 1 liter = ? m3
viii) What is the product PV = ?
x) What is the Uthermal = ?

Thermodynamics Lecture #5
Heat. Spontaneous flow of energy from one object to another because of difference of temperature.
Work. Any other transfer of energy, which is not heat, into or out of a system when an extensive mechanical variable changes.
Heat and work refers to energy in transit. Meaning ones are total energy inside a system, how much heat entered a system, and how much

work was done on a system. Meaningless ones are how much or how much work is in a system.
Symbols. U = Total energy inside a system, Q = amount of energy that entered a system as heat, W= work
Q + W = total energy that enters the system, and, by conservation of energy, this is the amount by which the system’s energy changes.

=> ∆U = Q + W which is known as the first law of thermodynamics.
= > Change in energy = heat added + work done. We have taken + sign for heat Q and work W entering the system. When they leave, we

will use – sign

∆U = Q + W W
Q

Compression work. Work = F∆x = PA∆x= -P∆V, the negative sign takes care of the fact that volume decreases as the piston moves in.

Piston area, A

Force, F
∆x ∆V = -A∆x
Fig. 1.8
If the pressure is constant, work done is just minus the area under a graph of pressure vs volume. If the pressure is not constant, we

divide the process into a bunch of tiny steps, compute the area

P P Area = ʃP

Area = P(V2-v1)

V2 V
V
V1 V2
Fig. 1.9
Under the graph for each step, then add up all the areas to get the total work. The work is still minus the total area under the graph

of P vs V. If the pressure is known as a function of volume, then
= > W = – ʃ P (V) dV (quasistatic) (1-29)
Evaluated between the limits of Vi to Vf.
Q. 1.31. Some helium in a cylinder with an initial volume of 1 L and an initial pressure of 1 atm is made to expand to a final volume

of 3 L in such a way that the pressure rises in direct proportion to its volume.
a) Sketch the graph of pressure vs volume for this process.
i) What is V i = ? ii) What is Vf = ? iii) What is Pi = ? iv) What is Pf = ?
v) Draw the graph with (Vi, Pi) and (Vf, Pf)
b) calculate the work done on the gas during the process, assuming that there are no “other” type of work done.
i) Write down the work done formula.
ii) What is the average pressure? Iii) What is the change in volume?
iv) What is the work done? Ans. – 400 J
c) Calculate the change in the helium’s energy content during the process.
i) Is helium diatomic or monatomic? Ii) Degrees of freedom per helium atom = ?
iii) What is the internal energy formula in terms of f, N, k, and T
iv) What is internal energy formula in terms of PV?
v) What is the change in the product of PV i. e. PfVf – PiVi
vi) What is the change in the internal energy? Ans. 1200 J
c) Calculate the amount of heat added to or removed from the helium during the process?
i) Write down the first law of thermodynamics
ii) What is Q = ? in terms of ∆U and W.
iii) From (b) and (c) find out what is Q = ? Ans. 1600 J

P P B
C B
A

C
A D

V V1 V

V2
Fig. 1.10 (a) Fig. 1.10(b)
Q. 1.33. An ideal gas is made to undergo the cyclic process shown in Fig. 1.10 (a). For each of the steps 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, (c) the heat added to the gas. (d) then determine the sign of each of these three quantities for the whole cycle. (e)What does

this process accomplish?
Hints. a) i) In Fig. 1.10 (a), is the initial volume higher or lower than the final volume in path A?
ii) So, is work done positive or negative? iii) Is Vi higher or lower then Vf in path C?
iv) Is the work done positive or negative? v) What is the change in V in path B?
vi)So, what is the work done?
vii) Out of the work done on different paths, does the volume change remain the same?
viii) Does the pressure or average remain higher in any path? Ix) What is the sign of the net work done for

the whole cycle?
b) i) What is the expression for energy from equipartition theorem?
Ii) What is the energy expression in terms of P and V?
iii) So, does any increase in P or V indicate energy increase?
iv) Does P and/or V increase or decrease in path A? So, what happens to energy?
v) Does P and/or V increase or decrease in path C? So, what happens to energy?
vi) For the whole cycle does P and/V increase or decrease, or remain the same? Or are the final values the same or different from the

initial values at the end of the cycle? So, what happens to energy for the whole cycle?
c) i) What is Q = ? in terms of ∆U and W
ii) Tabulate the result for W and ∆U by + and – signs. Then determine the sign of Q.
W ∆U Q
Path A
Path B
Path C
Whole cycle
e) Apparently, the net result of the cycle is to absorb energy as work and emit energy as heat.
Q. 1.34. An ideal diatomic gas in a cylinder with a movable piston undergoes the rectangular cyclic process shown in Fig 1.10 (b).

Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are frozen out. Assume

that the only type of work done on the gas is quasistatic compression-expansion work.
(a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy

content of the gas. Express all answers in terms of P1, P2, V1, and V2
Hint. i) What is the work done formula?
ii) What is the change in V in path A and what is the work done?
iii) What is the change in V and work done in path C?
iv) What is the change in V work done in path D?
v) What are the degrees of freedom for each molecule?
vi) What is the energy formula from equipartition theorem?
vii) What is the energy formula in terms of P and V?
viii) What is ∆U for path A in terms of P’s and V’s?
viii) What is ∆U for path B in terms of P’s and V’s?
viii) What is ∆U for path C in terms of P’s and V’s?
viii) What is ∆U for path D in terms of P’s and V’s?
ix)What is the formula for Q in terms of ∆U and W?
x) From ∆U and W, what is Q for path A in terms of P’s and V’s?
xi) From ∆U and W, what is Q for path B in terms of P’s and V’s?
xii) From ∆U and W, what is Q for path C in terms of P’s and V’s?
xiii) From ∆U and W, what is Q for path D in terms of P’s and V’s?
xiv) Tabulate the result below
W ∆U Q
Path A
Path B
Path C
Path D
Whole cycle

Compression of an Ideal Gas.
Isothermal compression. Very slow compression so that the temperature of the gas does not rise al all. Adiabatic compression. Very fast

so that no heat escapes from the gas during the process. Most real compressions are between these two extremes.
Ideal gas has P = NkT/VUse of Eqn. (1-29) yields
W = – NkTʃ(1/V)dV between the limits of Vi and Vf yields
= -NkT. ln V| = -NkT(ln Vf – ln Vi) = NkT ln (Vi/Vf) (1-30)
Work done is positive if Vi > Vf that is if the gas is compressed. If the gas expands isothermally, Vi < Vf, that is the work done is

negative.
If the gas is compressed isothermally, heat must be flowing out into the environment. This heat equals
Q = ∆U – W = ∆(1/2. N f k T) – W = 0 – W = NkTf ln (Vf/Vi) (1-31)
Which is negative of the work done in isothermal compression = > heat input = – work done. For compression Q is negative because heat

leaves the gas; for isothermal expansion, heat must enter the gas so Q is positive.

P
Isotherm

Vf Vi V
Fig. 1.11.
Adiabatic compression. No heat escapes. That is no change of heat. So, Q remains the same which causes the internal energy to rise.

Internal energy increase causes the temperature to increase for ideal gas.
= > ∆Q = 0; ∆U = Q + W = W
From equipartition theorem: U = ½ N f k T (1.33)
Whence the energy change along any infinitesimal segment of the curve is
dU = ½ N k f dT (1-34)
Work done during a quasistatic compression is – PdV
= > ½ N k f dT = – P dV (1-35)
For ideal gas, P = N k T/V

P
Adiabatic
Tf

V
Fig. 1.12. Adiabatic curve connects to isothermals
Substitute in (1-35), simplify to get f/2. dT/T = – dV/V (1-36)
Integrate both sides between Ti, Tf and Vi , Vf to get
f/2 ln (Tf/Ti) = – ln Vf/Vi (1-37)
= > ln (Tf/Ti)f/2 = – ln Vf/Vi
(Tf)f/2/(Ti)f/2 = (Vi/Vf)
After cross-multiplication VfTf f/2 = ViTif/2 (1-38)
Or VTf/2 = constant (1-39)
PV = NkT = > T = PV/Nk = > V(PV / Nk)f/2 = constant
= > Pf/2(V)V)f/2 = constant. x Nk)f/2
Raise both sides of the equation to the power 2/f, = > [Pf/2(V)V)f/2 ]2/f= [constant. x Nk)f/2 ]2/f
= > P V2/f V = another constant, = > PV(2 + f)/f = constant
The final result is VγP = constant (1.40)
where γ is called the adiabatic exponent= (f +2)/f
Q. 1.36. In course of pumping up a bicycle tire, a liter of air (which has γ = 7/5) at atmospheric pressure is compressed adiabatically

to a pressure of 7 atm
(a) What is the final volume of this air after compression?
(b) How much work is done in compressing the air?
© If the temperature of the air is initially 300 K, What is the temperature after compression?
Hint. (a) i) Write down the relation for adiabatic compression.
ii) What is Vf = ? in terms of Vi, Pf, and Pi
iii) Vi= ? iv) Pf = ? v) Pi = ? vi) Vf = ?

Ans. 0.25 L
(b) i) What is P = ? in terms of V and γ from the adiabatic compression eqn.
ii) Write down the work done formula.
iii) Integrate between the limits of Vi and Vf to get W= [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1]
iv) Substitute for P’s and V’s to get W = 1.86 L.atm = 188 J
© i) Write down the adiabatic relation in terms of T’s and V’s.
ii) Substitute for Ti, Vi, and f to get Tf = 520 K (approximate)
Q. 1. 37. In a Diesel engine atmospheric air is quickly compressed to about 1/20 of its original volume. (a) Estimate the temperature

of the air after compression. Assume γ = 5 for air.
(b) Explain why a Diesel engine does not require spark plug.
Hint. (a) i) Write down the eqn for adiabatic compression in terms of V’s and T’s.
ii) Tf = ? in terms of Ti, Vi, Vf and f
iii) Substitute the values to get Tf = 1000 K (approximate)
(b) The temperature is presumably hot enough to ignite the fuel as soon as it is injected, without the aid of a spark plug.
Thermodynamics Chapter Test # 1 Name Date

10 atm

B
P C
1 atm A

100 600 V (m3)
1. An ideal monatomic gas is made to undergo the cyclic process as shown in the Fig. above. For paths A, B, C, answer the

following questions:
i) WA = ? W = -PdV = – (1E5)(600-100) = -500E5 = -5E7 J

ii) ∆UA = ? 3/2∆(PV) =3/2(PfVf – PiVi) =3/2[(1E5)(600) – (1E5)(100)]=(1.5E5)(500) =7.5E7 J

iii) QA = ? ∆U –W = 7.5E7 – (-5E7) = 12.5E7 = 1.25E8 J

iv) WB = ? No change in volume, so W = 0

v) ∆UB = ? (3/2)∆PV = 1.5V∆P = (1.5)(600)(10E5 – 1E5) = 1.5 x 600 x 9E5 = 8.1E8 J

vi) QB = ? ∆U –W = 8.1E8 – 0 = 8.1E8 J

vii) WC = ? –PdV = -{(1E5 + 10E5)/2}x (100-600) = -5.5E5 x -500 = 2.75E8 J

viii) ∆UC = ? 3/2∆PV = 1.5(PfVf – PiVi) = 1.5x [1E5 x 100 – 10E5 x 600] = 1.5E7(1 – 60) = 8.85E8

ix) QC = ? ∆U –W = 8.85E86-2.75E8= 6.1E8

2. In course of pumping up a car tire, 3 L of air (which has γ = 7/5) at 27 oC and at 2 atmospheric pressure is compressed

adiabatically to a pressure of 10 atm.

i) What is the final volume of this air after compression?
PfVfγ = PiViγ = Vf = Vi. (Pi/Pf)1/γ = (3 L)(2 atm /10 atm)5/7= 0.95 L

ii) How much work is done in compressing the air? W = [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1]
[(3E-3 x 2E5)/0.4][(3/0.95)0.4 – 1] = 876 J

iii) What is the temperature after compression? 300 (3/0.95)0.4 = 475.20 K

3. What is the Kelvin temperature corresponding to night’s low of 60 oF?
273 + (5/9)(60 – 32) = 288.5 K

4. What is the volume of 10 moles of air at 27 oC and 0.5 atm pressure?

V= (10 moles)(8.31 J/mol.K)(300 K)/(0.5xE5 N/m2)=49,860 J.N/m2=49,860 N.m.m2/N= 49,860 m3
5. Estimate the number of molecule in your classroom of size 4 m x 4 m x 3 m at 1 atm pressure and at 23 oC.

N= PV/kT = (1E5 N/m2)(48m3)/[(1.383E-23 J/K). (300K)] = 1.16E27

6. Calculate rms speed of oxygen molecule at 27 oC.
vrms = (3RT/M)1/2 =[3 x (8.31 J/mol. K)(300 K)/(0.032 kg/mol)]1/2= 483 (N. m/kg)1/2 = [kg. m/s2. m/kg]1/2 = 483 [m2/s2]1/2= 483 m/s

7. Suppose a container has hydrogen and oxygen molecules in thermal equilibrium. Which molecule moves faster, on average? By what

factor?
vrms H / vrms O == (3RT/MH)1/2/ (3RT/ MO)1/2 = [MO/MH]1/2 = (0.032/0.002)1/2 = 4
Hydrogen, 4 times faster than oxygen

8. A monatomic gas in a cylinder with an initial volume of 4 L and an initial pressure of 2 atm is made to expand to a final

volume of 12 L and final pressure 6 atm.
i) What is the work done using the average pressure?
W= -PdV = – (4 atm)( 8 L) = – (4E5 N/m2)(8E-3m3) = -3,200 J

ii) What is the change in the internal energy? , U = 1.5 PV, ∆U = 1.5(PfVf – PiVi)
= 1.5[6E5 . 12E-3 – 2E5 . 4E-3] = 1.5[7,200 – 800] = 9,600 J

iii) Calculate the amount of heat added to the gas. Q = ∆U – W = 9,600+3,200 = 12,800 J
vrms = (3RT/M)1/2 ,Mnitrogen = 0.028 kg, Mhydrogen = 0.002 kg, MOxygen = 0.032 kg,
W = -P ∆V, U = 1.5 PV, Q = ∆U – W, PfVfγ = PiViγ ; W = [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1]; VfTff/2 = ViTif/2 , f = 5.
N= PV/kT, 1 atm =1E5 N/m2 , k = 1.38E-23 J/K, R= 8.315 J/mol.K, NA = 6.022E23; PV = nRT;
ToF = (9/5) ToC + 32, T K = ToC + 273, ToC = (5/9)(ToF-32) ; 1 L = 1E-3 m3

Heat Capacities. An object’s heat capacity C is the amount of heat Q needed to raise its temperature, per degree temperature increase

∆T. = > C ≡Q/∆T. (1-41)
Specific heat capacity, c, is the heat capacity per unit mass. = > c ≡ C/m = Q/(m ∆T) (1-42)
Heat gained for an increase in temperature ∆T or heat loss for a drop in temperature ∆T is given by
Q = mc∆T
For water, heat capacity per unit mass is cw = 4.186 J/g.oC
C = (∆U – W)/∆T (1-43)
Specific heat at constant volume, CV: Since Q =∆U – W, and no change in volume means no work done, W = 0; so, at constant volume Q =

∆U, and
CV= (∆U/∆T)V= (∂U/∂T)V (1-44)
Specific heat at constant pressure, CP: = [(∆U – (-P∆V))/∆T]P = (∂U/∂T)P + P(∂V/∂T)P (1-45)
Q. 1. If U = ½ NfkT, CV = ? (1-46)
Q. 2. What is f = ? for monatomic gas? What is CV = ?
Q. 3. What is CV per mole for a monatomic gas?
Q. 4. Solids have six degrees of freedom per atom. What is CV for solids?
Q. 5. What is the ideal gas equation in terms of P, V, N, k, and T? What is V= ?
Q.6. (∂U/∂T)P = (∂U/∂T)V = ? (1-56)
Q. 7. Prove that CP = V + Nk = CV + nR (1-48)
Q. 8. To measure the heat capacity of an object, all you usually have to do is put it in thermal contact with another object whose heat

capacity you know. Suppose that a chunk of 100-g metal is immersed in boiling water at 100 deg C, then quickly transferred into a

Styrofoam cup containing 250 g of water at 20 deg C. After a minute or so, the temperature of the contents of the cup is 24 deg C.

Assume that during this time no significant energy is transferred between the contents of the cup and the surroundings? The heat

capacity of the cup itself is negligible.
i) What is the mass of water?
ii) What is the heat capacity per unit mass of water?
iii) What is the increase in water temperature?
iv) What is the heat gained by water? Ans. 4186 J
v) What is the heat lost by the metal?
vi) Write down the formula for the heat capacity of the metal.
vii) What is the change in temperature of the metal?
viii) Calculate the heat capacity for the metal. Ans. 55 J/oC
ix) What is the specific heat capacity of the chunk of metal? 0.55 J/g.oC
Q. 9. The heat capacity of Brookshire’s Rotini Tricolore ia approximately 1.8 J/g.oC. Suppose you toss 340 g of pasta at 25 oC into 1.5

L of boiling water. What effect does it have on the temperature of the water (before there is time for the stove to provide more heat)?
i) If Tw is the water’s initial temperature and Tf is the final temperature, how do you write in symbols the drop of water temperature?
ii) If mw is the water mass and cw is the water’s heat capacity per unit mass, what is the expression for the heat lost by water?
iii) How do you write in symbols heat gained by the pasta if Tf is the final temperature and Tp is the initial temperature?
9. If mp is the water mass and cp is the water’s heat capacity per unit mass, what is the expression for the heat gain by pasta?
10. Equate heat loss expression= heat gain expression
11. Solve for Tf
12. Evaluate Tf Ans. 93.3 oC
Latent heat, L. Amount of heat absorbed during a phase transformation divided by mass.
= > L = Q/m (1-50)
The latent heat for melting ice is 333 J/g, or 80 cal/g, and that for boiling water is 2260 J/g or 540 cal/g.
Q. 10. Your 200-g cup of tea is boiling hot. About how much ice should you add to bring it down to a comfortable sipping temperature of

65 oC? Assume that ice is initially at -15 oC. The specific heat capacity of ice is 0.5 cal/g.oC. The heat capacity per unit mass of

tea is 1 cal/g.oC.
i) What is the tea’s initial temperature? ii) What is the tea’s final temperature?
iii) What is the temperature loss of tea? Iv) What is the heat loss formula?
v) How much heat is lost by tea? Ans. 700 vi) What is the temperature change

in ice to rise to the melting point?
vii) What is the heat required for this job if m is the mass of ice? Viii) What is heat required to melt the ice?

Ix) What is the temperature rise in water from 0 oC to 65oC? x) What is heat required

in the ice melting process?
xi) From heat loss = heat gain, find the mass of ice. Ans. 46 g.
Mathematics for Thermodynamics
Variables. The symbols u, w, x, y, z, t, P, Q, T, V, S, etc. symbols are used for variables i. e. physical quantities that take

different values. Out of these, independent variables are the ones that assume their values independently of any other. Time indicated

by the symbol t is independent variable. Our growth, if we represent it by the symbol g, will be a dependent variable.
Q. 1. Say you are driving. Your distance covered is represent by the variable x, and time by t. Out of these two, which one is

independent and which one dependent?
Q. 2. Say you are heating a kettle of water. The more you rise the temperature T, the more heat water accumulates. Find out dependent

and independent variables.
Q. 3. Say you cooked your food and then turn off the oven. As the time t passes, the oven becomes cooler by losing its heat Q. Which

ones are independent and dependent variables? Do both of them have increasing tendency at the same time in the cooling phase?
Function. A function is a relationship between two variables such that to each value of the independent variable there corresponds

exactly one value of the independent variable. Functional notation is y = f(x) where f is the name of the function, y is the dependent

variable, x is the independent variable, and f(x) is the value of the function at x.
Exponential Function. The exponential function f with base a is denoted by
f(x) = ax, where a > 0 but not = 1, and x is any real number (the variable symbol x can take any real number value). If a = 2.134, a is

replaced by the symbol e. And in that case f(x) = ex.
Change. A change in a variable’s value is represented by the Greek letter ∆x (delta). Thus, ∆t means a change in time t which is =

final time – initial time. Or, ∆t = tf – ti ; or , ∆t = t2 – t1; or, ∆t= tf – to.The suffixes f, i, or 2, 1, f and o are used to

indicate the final and the initial values of a variable. If a variable does not change, ∆ of that variable is zero. Binning an object,

say to measure its length or any other physical quantity, we use ∆.
Q.4. If your height is represented by the variable h, how do you write change in your height by using delta as well as the different

suffixes?
Q. 5. If ∆x is 1 inch, how many ∆x’s are there in 18 inches?
Rate of Change. Rate of change is defined as the ratio of the rates of change of the dependent variable and the independent variable.
Slope of a Line. A line is drawn by plotting a number of paired values of x and y. Here the ratio of the rate of change of y and x is

called the slope of the line. Also, you are aware that slope = rise/run. Rise is the change in y value or ∆y and the change in x value

or ∆x. => slope = ∆y/∆x. Keep in mind that the forward slash “/” stands for the mathematical operation of division. Rates of change or

slopes of the paired variables are found in the same way.
Derivative or the differentiation. When the change in the independent variable is minutely small i. e. almost zero, the rate of change

is then called the derivative of the dependent variable with respect to the independent variable. Derivative of y with respect x is

represented by dy/dx and is defined as
dy/dx = in the limit of ∆x →0, the value of ∆y/∆x.
Keep in mind that bits of changes in x and y i. e. ∆x and ∆y in other paired variables are countable or are discrete. In the limit of

incountable or nondiscrete deltas, dy and dx are used. If the ∆x bits are of the sizes of 1 cm, you can count them for your writing

pencil. However, if ∆x is of the size of an atom, dx has to be used.
Standard Formulas for Finding Derivatives:
There are some standard formulas for finding derivatives of functions.
d/dx(a) = 0, where a is a constant whose value does not change. => derivative of a constant number is zero
Q. 6. What is d/dx(3.45) = ?
d/dx(axn)= naxn-1, => derivative of variable raised to power n is n times the variable exponent reduced by 1. If the variable with the

exponent is multiplied by a constant a, the constant also multiplies the derivative.
Q. 7. What is d/dx(7×4)= ?
d/dx(eax) = a eax, => derivative of e raised to an exponent of a constant times a variable equals the constant times the same

expression
Q. 8. What is d/dx(e8x) =?
Second Derivative or Second Differentiation: Taking derivative twice of a function. Second time differentiation follows the same rule

as the first time differentiation. The second derivative symbol is
d2/dx2. Find the second derivative of the function f(x) = 3×3 + 2×2 + x + 1. First time differentiation of the function f(x) following

the rule d/dx(axn)= naxn-1, yields 9×2 + 4x + 1 + 0 = 9×2 + 4x + 1. Second time differentiation of the first time derivative gives 18x

+ 4 + 0 =18x + 4. Check this result. So,
(d2/dx2)[f(x)] = (d2/dx2)[3×3 + 2×2 + x + 1]=18x + 4.
Partial Differentiation. A function can be of more than one variable. For example, area of a rectangle is a function of both length and

width, and the volume of a cube is a function is a function of length, width, and height. Inflation of a balloon, or a tire is a

function of both pressure and temperature. Differentiation in this case can be done with respect one variable at a time keeping the

rest constant. We use del symbol ∂ in the form ∂/∂x instead of d/dx for partial differentiation. Let us consider a function of two

variables f(x,y). The partial derivative of f(x,y) could be written as
(∂f/∂x)y and (∂f/∂y)x
The first of these is read as “the partial derivative of f(x,y) with respect to x at constant y”.
Q. 9. How to read the second one?
Q. 10. (i) How to read β1= (∂V/∂T)P (ii) How to read β= (1/V)(∂V/∂T)P
Q. 11. Consider f = x2y3 + 5x and g(x,y,z) = x2 + y2 + z2 – r2 =0
(i) What is x2 + y2 + z2 = ? from x2 + y2 + z2 – r2 =0 (ii) What is y2=?
(iii) What is y = ?
(iv) Substitute y-value in f = x2y3 + 5x and find f(x,z) = ?
(v) (∂f/∂x)y = ? (vi) (∂f/∂y)x = ?
(vii) (∂f/∂x)z = ? (viii) (∂f/∂z)x = ?
(ix) What is z= ? from x2 + y2 + z2 – r2 =0
(x) Substitute the z-value in the expression for (∂f/∂z)x , simplify
(xi) Substitute the z-value in the expression for (∂f/∂x)z , simplify
(xii) Is (∂f/∂x)z = (∂f/∂x)y ?
Q. 12. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P
(i) Evaluate β= ? Is it 1/T or something else?
(ii) Evaluate κ = ? Is it 1/P or something else?

Exact Differential. Let u(x,y) be an arbitrary function of x and y. Then the differential of u(x,y) is written as
du = (∂u/∂x)ydx + (∂u/∂y)xdy = Mdx + Ndy
Q. 13. (i) What is M = ? (ii) What is N= ?
The condition for exact differential is (∂N/∂x)y = (∂M/∂y)x
Let du = (xy2 + 2x)dx + x2y dy
(iii) N = ? (iv) (∂N/∂x)y
(v) M = ? (vi) (∂M/∂y)x
(vii) Is du an exact differential?
Inexact Differentials. Inexact differentials is denoted by the symbol δ (lower case delta).
δw= y dx + x dy is inexact and there exists no function w(x,y) which satisfies this equation.
Useful Differential Relations for Systems with Two Independent Variables. Three useful mathematical relations can be derived for a

system with three coordinates x, y, and z, which are related by an equation of state
g(x,y,z) =0
Here x, y, and z can be P, V, and T, respectively, in thermodynamics relations.
Any pair of these variables may be chosen as independent and the equation of state may be expressed in three different forms:
x, y independent: z = z(x,y) => z can be written as an explicit function of only x and y
x, z independent: y = y(x,z) => y can be written as an explicit function of only x and z
y, z independent: x = x(y,z) => x can be written as an explicit function of only y and z
Q. 14. (i) What is the differential dx = ?
(ii) What is the differential dy = ?
(iii) What is the differential dz = ?
(iv) Substitute for dy in the expression for dx
(v) Simplify, collect like terms with dx and dz
(vi) Make sure you get something like[(∂x/∂y)z(∂y/∂x)z – 1]dx + [(∂x/∂y)z(∂y/∂z)x + (∂x/∂z)y]dz= 0
(vii) In (vi), the sum of some coefficient times the differentials of independent variables x and z is zero. This implies that the

coefficients of dx and dz must vanish independently. Set the coefficient of dx = 0,
Solve for (∂x/∂y)z to get (∂x/∂y)z=1/(∂y/∂x)z
(viii) Set the coefficient of dz = 0 => (∂x/∂y)z(∂y/∂z)x + (∂x/∂z)y=0
Prove that => (∂x/∂y)z(∂y/∂z)x(∂z/∂x)y= -1
Q. 15. Let x, y, z be P, V, and T respectively.
(i) Show that => (∂P/∂V)T(∂V/∂T)P = – (∂P/∂T)V
(ii) Show that the isothermal pressure coefficient α = (∂P/∂T)V = – (∂V/∂T)P/(∂V/∂P)T
Enthalpy of a system = internal energy of the system + work needed to create the system. For example, a magician makes a rabbit out of

nothing and place it on the table. The magician must summon up not only the energy U of the rabbit, but also some additional energy,

equal to PV at constant pressure P to push the atmosphere out of the way to make the room of volume V. The total energy required is the

enthalpy,
H = U + PV (1.51)
Change in enthalpy during a constant pressure is
∆H = ∆U + P∆V (1.53)
According to the 1st law of thermodynamics,
Change in energy = heat added to the system, plus expansion-compression work done on it, plus any other work done (e. g. (for example)

electrical) on it
= > ∆U = Q + (-P∆V) + Wother (1.54)
Substituting for ∆U in (1.53) = > ∆H = Q + (-P∆V) + Wother + P∆V
= >∆H = Q + Wother (1.55)
= > Enthalpy change can be caused by heat and other forms of work and not by compression-expansion work (during constant-pressure

processes). That is to say, you can forget about expansion-compression work if you deal with enthalpy instead of energy. In no “other”

types of work are being done, the change in enthalpy tells you directly how much heat has been added to the system. And that’s why the

symbol H is used.
When an object’s temperature is raised, the change in enthalpy per degree, at constant pressure, is the same as the heat capacity at

constant pressure, CP:
CP = (∂H/∂T)P (1-56)
CP should really be called “enthalpy capacity” as should CV be called “energy capacity”
Chemistry deals with ∆H in phase transformations, chemical reactions, ionization, dissolution in solvents, and so on.
Q. 1. When you boil 1 mole of water at 1 atm., ∆H = 40,660. What is the change in enthalpy when you boil one gram of water? (One mole

of water is about 18 grams)
Q. 2. What is the work needed to push the atmosphere away to make room for one mole of water vapor?
i) What is the ideal gas equation for n moles?
ii) What is the ideal gas equation for 1 mole?
iii) What is the Kelvin temperature for water boiling?
iv) What is the work done formula for constant pressure P and volume V?
v) What is the work done? Ans. 3100 J
Q. 3. Consider the combustion of one mole of H2 with ½ mole of O2 under standard conditions. How much of the heat energy comes from a

decrease in the internal energy of the system, and how much comes from work done by the collapsing atmosphere. ∆H for this reaction is

286 kJ, called the enthalpy of formation of water. (Treat the volume of liquid water as negligible and take the temperature as 25 oC).
i) What is the combined mole number of the reactants?
ii) What is the Kelvin temperature in question?
iii) What is the work done formula for the constant pressure P and the volume V?
iv) What is the value of the work done by the atmosphere? Ans. 3717 J = 4 kJ.
v) What is the total heat produced in the combustion reaction?
vi) How much comes from the internal energy (e. g. chemical energy in the molecular bonds) of the system
Q. 4. Consider the combustion of one mole of methane gas
CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)
The system is at standard temperature (298 K) and pressure (105 Pa) both before and after the reaction.
(a) Find ∆H in the process of converting a mole of methane into its elemental constituents ( carbon and hydrogen gas) given that ∆H

upon forming one mole of methane from elemental carbon and hydrogen is -74.81 kJ
(b) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H to

form one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.
(c) What is ∆H for the actual reaction in which methane and oxygen form carbon dioxide and water vapor directly?
i) What is ∆H in converting the mole of methane into elemental carbon and hydrogen?
ii) Add to (i) ∆H to form one mole of CO2
iii) Add to (ii) ∆H to form two moles of water Ans. -802.34 kJ
(d) How much heat is given off during the reaction, assuming that no “other” forms of work are done?
(e) The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy was

produced by methane combustion given that one mole methane combustion gives off 800,000 J?
CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)

i) What is one mole of methane mass + two moles of oxygen mass? 0.080 kg
ii) How many moles of methane would sun contain? 2.5E31
iii) How much energy could this mass of the sun give off? 2E37 J
iv) How many years could the fuel last if the sun was giving energy at this rate? 1600 yrs

Heat Conduction. The amount of heat Q conducted through a thickness ∆x and area A of a material of thermal conductivity kt and having

a temperature difference of ∆T on either side in time ∆t is
Q/∆t = -ktA(dT/dx) (1-60)
which is Fourier heat conduction law. The unit of Q is Joule, t is second, kt is watts per meter per Kelvin, T is Kelvin, x is in

meter. Some values of kt are air, 0.026; wood, 0.08; water 0.6; glass 0.8; iron, 80; copper, 400.
Q.1. A room window has an area of one square meter and a thickness of 3.2 mm, the temperature inside and outside the room are 20oC and

0o C, respectively. If the thermal conductivity of the material of the window is 0.8 W/m.K, find the rate of heat flow through it.

Ans. 5000 Watts
Q.2. Compute the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 meter square , for a

temperature difference of 20oC.
Conductivity of an ideal gas.

2r 2r
2r
A molecule of radius r can make a collision with another molecule of radius r if the distance between their centers are 2r. The

collision can happen at that distance between a molecule of radius 2r and one like a point. When a sphere of radius 2r moves in a

straight line of length l, it sweeps out a cylinder of volume 4πr2l. When the volume of this cylinder equals the average volume per

molecule in a gas, we are likely to get a collision. The average distance between two collisions is called the mean free path.
= > Volume of cylinder = average volume per molecule
= > π(2r)2l ≈ V/N
= > l ≈ (1/4πr2) (V/N) (1.62)
Average time between collisions is ∆t = mean free path, l/vrms (1.63)
Q. 3. Find the average time between collisions of nitrogen molecules in the air, given that r = 1.5E-10 m for nitrogen. Treat air as an

ideal gas.
i) What is V/N = ? in terms of k, T, and P
ii) What is V/N= ? at room temperature and at atmospheric pressure. Ans. 4E-26 cubic meter
iii) What is the mean free path, l = ? Ans. 150E-9 m
iv) Find vrms for nitrogen molecule = ?
v) Average time between collisions is l/vrms = ? Ans. 3E-10 s
Quantity of heat conducted in a gas medium through an imaginary barrier having difference of temperatures d = T2 – T1 is
Q =1/2 (U1 – U2) = -1/2(U2 – U1) = -1/2 Cv(T2- T1) = -1/2 CvldT/dx (1.64
Thermal conductivity is given by
kt = ½ CVl/(A∆t) = ½ (CV/Al) (l2/∆t) = ½( CV/V) lvavr (1-65)
where vavr is the average speed of the molecules.
Heat capacity of the gas per unit volume = CV/= 1/2Nk/V = f/2 P/T (1-66)
Q. 4. Air at room temperature and atmospheric pressure has f = 5. (a) Find CV/V = ? Ans. 800 J/m3.K
(b) Find the thermal conductivity for air. Ans. 0.031 W/m.K
Thermal conductivity of a given gas depends only on its temperature through vavr∞√T and possibly through f, and kt is proportional to

the square root of the absolute temperature.
Viscosity. Fluids have internal friction called viscosity. F = Aη(dux/dz) (1.69)
Where η is called the coefficient of viscosity = F(dz/dux)A . The unit of η is N.s/m2 = Pa.s
Q. 5. Find η = ? if F is 3 N, A = 10 cm2, and (dz/dux) = 0.1 s

∆z Fluid
Area A
Diffusion. Fick’s law Jx = – Ddn/dx (1-70)
Where Jx is the particle flux, D is the coefficient of friction. D for CO at room temperature and pressure is 2E-5 m2/s.
Q. 6. Find Jx for Co at room temperature and atmospheric pressure is dn/dx = 100/cm
Thermodynamics Mid-term Test Name—————————————————–Date
No going out during testing. Finish your outside jobs before the test starts. Empty your desk before sitting for the test. All cell

phones must be on the instructor’s desk. No goggles. No hats. No communication devices.
1. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P
(i) Evaluate β= ? Is it 1/T or something else?
(ii) Evaluate κ = ? Is it 1/P or something else?
2. Fractional increase in ocean volume due to global warming can be taken to be equal to the fractional rise of ocean water. Given

that the coefficient of volume expansion β for ocean water is 87.5E-6/oC, what will be the sea level rise of the original depth of 1000

m for 1oC rise of global temperature? Ans. 9 cm
3. You have 2.00 kg of water at a temperature of 20.0 oC. How much energy is required to raise the temperature of water to 95 oC

. If you pay just 10.0 cents for heating each 3.60E6 J, what is the cost of warming the water, given that the specific heat of cwater

is 4.19 kJ/kg. K.

Ans. Heat required is 629,000 J, and the cost is 1.75 cents
4. Suppose 10.0 g of water at a 100.0 oC is in an insulated cylinder equipped with a piston to maintain a constant pressure of

101.3 kPa. Enough heat is added to vaporize water to steam at 100.0oC. Water volume is 10.0 cc and steam volume 16,900 cc. (a) What is

the work done by the water as it vaporizes? (b) How much heat water needs to vaporize? (c) What is the change in internal energy of

water? Ans. 1710 J Q= 22,600 J ∆E = 20,900 J
5. Suppose you insulate above the ceiling of a room with an insulation material having a thermal resistance R = 5.2836 m2. K/W.

The room is 5.00 m by 5.00 m. The temperature inside the room is 21.0 oC and the temperature above the insulation is 40.0 oC. How much

heat enters the room through the ceiling in a day if the room temperature is maintained at 21.0 oC? Ans. 7.77E6 J
6. Suppose your classroom has a temperature of 22oC. What is the average kinetic energy of the molecules (and atoms) in the air?

Ans. 6.11E-21 J
7. What is the mean free path l of nitrogen molecules in the air at normal atmospheric pressure of 101.3kPa and a 293.15 K, given

that the radius of nitrogen molecule is 0.150 nm = 0.150E-9 m? Ans. 9.99E-8 m
8. The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy was

produced by methane combustion given that one mole methane combustion gives off 800,000 J?
CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)
i) What is one mole of methane mass + two moles of oxygen mass? 0.080 kg
ii) How many moles of methane would sun contain? 2.5E31
iii) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H to

form one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.
iv) How much energy could this mass of the sun give off? 2E37 J
v) How many years could the fuel last if the sun was giving energy at this rate? 1600 yrs

∆V = βV∆T; ∆V/V = ∆d/d, ∆d/d = β∆T; Q = cwatermwater∆T; 1 kW h =3.60E6 J; W = PʃdV integrated between Vi and Vf; Q = mLvaporization,

Lvaporization = 2.260E6 J/kg; ∆U = Q – W;
Q = ∆t. A. (Thigh – Tcool)/R; KEave = 1.5kT; k = 1.38E-23 J/K; l = kT/(17.77r2P); one mole of methane has 0.016kg and one mole of

oxygen has 0.064 kg ; ∆H = Q + Wother.
Thermodynamics Mid-term Test Name—————————————————–Date
No going out during testing. Finish your outside jobs before the test starts. Empty your desk before sitting for the test. All cell

phones must be on the instructor’s desk. No goggles. No hats. No communication devices.
1. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P
(i) Evaluate β= ?
(ii) Evaluate κ = ?
2. Fractional increase in ocean volume ∆V/V due to global warming can be taken to be equal to the fractional rise of ocean water

∆d/d. Given that the coefficient of volume expansion β for ocean water is 87.5E-6/oC, what will be the sea level rise of the original

depth of 1000 m for 1oC rise of global temperature?
3. You have 2.00 kg of water at a temperature of 20.0 oC. How much energy is required to raise the temperature of water to 95 oC .

If you pay just 10.0 cents for heating each 3.60E6 J, what is the cost of warming the water, given that the specific heat of cwater is

4.19 kJ/kg. K.
4. Suppose 10.0 g of water at a 100.0 oC is in an insulated cylinder equipped with a piston to maintain a constant pressure of

101.3 kPa. Enough heat is added to vaporize water to steam at 100.0oC. Water volume is 10.0 cc and steam volume 16,900 cc. (a) What is

the work done by the water as it vaporizes? (b) How much heat water needs to vaporize? (c) What is the change in internal energy of

water?
5. Suppose you insulate above the ceiling of a room with an insulation material having a thermal resistance R = 5.2836 m2. K/W.

The room is 5.00 m by 5.00 m. The temperature inside the room is 21.0 oC and the temperature above the insulation is 40.0 oC. How much

heat enters the room through the ceiling in a day if the room temperature is maintained at 21.0 oC?
6. Suppose your classroom has a temperature of 22oC. What is the average kinetic energy of the molecules (and atoms) in the air?
7. What is the mean free path l of nitrogen molecules in the air at normal atmospheric pressure of 101.3kPa and a 293.15 K, given

that the radius of nitrogen molecule is 0.150 nm = 0.150E-9 m?
8. The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy was

produced by methane combustion given that one mole methane combustion gives off 800,000 J?
CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)
vi) What is one mole of methane mass + two moles of oxygen mass?
vii) How many moles of methane would sun contain?
viii) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H to

form one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.
ix) How much energy could this mass of the sun give off?
x) How many years could the fuel last if the sun was giving energy at this rate?

∆V = βV∆T; ∆V/V = ∆d/d, ∆d/d = β∆T; Q = cwatermwater∆T; 1 kW h =3.60E6 J; W = P∫dV integrated between Vi and Vf; Q = mLvaporization,

Lvaporization = 2.260E6 J/kg; ∆U = Q – W;
Q = ∆t. A. (Thigh – Tcool)/R; KEave = 1.5kT; k = 1.38E-23 J/K; l = kT/(17.77r2P); one mole of methane has 0.016kg and one mole of

oxygen has 0.064 kg ; ∆H = Q + Wother.

The Second Law
Irreversible processes are not inevitable but are overwhelmingly probable. Temperature is a way of quantifying the tendency of energy

to enter or leave an object during the course of random rearrangements.
Q. 1. Take three coins – a penny, a nickel, and a dime. Flip them. (a) How many possible outcomes are there? Represent head by H and

tail by T. [ Probability of any event = how many of that event/out of how many total number] (b) What is the probability of getting

three heads? (c) What is the probability of getting three tails? (d) What is the probability of getting two heads and a tail? (e) What

is the probability of getting exactly one head and two tails?
Penny Nickel Dime
H H H
Each of the different outcomes in the flipping of the coins is called a microstate. In general, the microstate of a system refers to

the state of each individual particles. In this case, it refers to the state of each coin. For example, HHT or TTH, etc. Macrostate

refers to how many heads or tail. In HHT, there are two heads. Multiplicity refers to the number of microstates corresponding to a

macrostate. It is denoted by Greek letter capital omega Ω.
Q. 2. In the flipping of the coins in #1, fill out the chart.
Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities
3H Ω.(3)=?
2H Ω.(2) =?
1H Ω.(1) = ?
0H Ω.(0) = ?
Q. 3. In the above example, number of coins was 3 with each having 2 possibilities giving a total of 8 microstate possibilities. (a)

Write a mathematical formula to get 8 out of 2 and 3. (b) If you flip 100 coins, what is going to be the total number of microstates

written in the form as in (a)? (c) What is the total number of macrostates of 0 through 100 heads?
Ω(0) = ? when every coin shows tails-up
Ω( 1) = ? When only one coin shows heads-up. Any one can show heads-up – first one, second one, third one, etc. up to the 100th

one.
Ω(2)= ? When two coins, a pair, show heads-up. 100 choices for the first one, and for each of these choices, there are 99 remaining

choices for the second coin. Order of the pair is immaterial – #1 in the first and # 2 in the second as well as # 2 in the first and #

1 in the second. The total number of ways of selecting will be 100 . 99. The number of distinct pairs will be (100 . 90)/2.
Ω(3) = ? When three coins, a triplet, show heads-up. 100 choices for the first one, 99 choices for the second one , and 98 for

the third. And any triplet could be chosen in several ways – 3 choices for which one to flip first, and for each of these, 2 choices

for which to flip second. Thus the number of distinct triplets is (100 . 99 . 98)/(3. 2)
Ω (n) = ? [100. 99. 98. …….. (100- (-1))]/[n. (n-1). (n-2) … … 3. 2. 1]
= 100!/[n!. (100-n)!
If there are N coins, the multiplicity of the macrostate with n heads is
Ω (n) = N!/n! (N-n)!
is the of ways of choosing n objects out N.
Q. 4. Suppose you flip four fair coins.
(a) Make a list of all possible outcomes as in the table below.
Coin #1 Coin #2 Coin #3 Coin #4
H H H H
(b) Make a list of all the different “macrostates” of 0 heads to 4 heads and their probabilities

© Compute the multiplicity of each macrostate using the combinational formula Ω (n) = N!/n! (N-n)!
Two-State Paramagnet.
The two outcomes of a coin-flipping example can be applied to two-state paramagnets. A paramagnet is a material in which the

constituent particles act like tiny compass needles that tend to align parallel to any externally applied magnetic field. If the

constituent particles interact strongly enough with each other, the material can magnetize even without any externally applied field,

we call it a ferromagnet. Paramagnetism is a magnetic alignment that lasts only as long as an external field is applied. Individual

magnetic particles can be called dipoles because each has its own magnetic dipole moment vector. Each dipole could be an electron, a

group of electrons in an atom, or an atomic nucleus. For any such microscopic dipole, quantum mechanics allows the component of the

dipole moment vector along any given axes to take on only certain discrete values – intermediate values are not allowed. In the

simplest case only two values are allowed – one positive and the other one negative. In a two-state paramagnet, each elementary

compass needle can have only two possible orientations – either parallel or anti-parallel to the applied field.
We define N↑= # of elementary dipoles that point up at some particular time
N↓ = # of dipoles that point down at some particular time
The total number of dipoles is then N = N↑ + N↓ which will be taken to be fixed.

B ↑↓↓↓↓↑↑↑↓↑↓↑↓↑↓
Fig. 2.1. A symbolic representation of a two-state paramagnet, in which each elementary dipole can point either parallel or

antiparallel to the externally applied magnetic field.
This system has one macrostate for each possible value of N↑ from 0 to N. The multiplicity of any macrostate is given by the same

formula of coin-tossing:
Ω(N↑) = N!/(N↑! N↓!) (2-7)
The external magnetic field exerts a torque on each little dipole, trying to twist it to point parallel to the field. If the external

field points up, then an up-dipole has less energy than a down-dipole, since it needs additional energy to twist from up to down. The

total energy of the system is determined by the total number of up- and down-dipoles. So specifying which macrostate this system is in

is the same as specifying its total energy. In fact, in nearly all physical examples, the macrostate of a system is characterized, at

least in part, by its total energy.
Q. 1. How do you justify the example of coin-tossing for paramagnetism?

Einstein Model of a Solid.
Consider a collection of microscopic systems that can each store any number of energy “units” all of the same size. Equal-size energy

units occur for any quantum-mechanical harmonic oscillator whose potential energy function has the form 1/ksx2. The size of the energy

unit is hf where h is Planck’s constant (6.63E-34 J.s) and f is the natural frequency of the oscillator (1/2π√(ks/m). Examples of

quantum mechanical oscillators include the vibrational motions of diatomic and polyatomic molecules. In 3-D, each atom can oscillate in

3 independent directions. So, if there are N oscillators, there are only N/3 atoms. The model of a solid as a collection of identical

oscillators with quantized energy units was proposed by Einstein in 1907.
Let us consider a solid with N= 3 oscillators.
Oscillator #1 #2 #3 Oscillator #1 #2 #3
Energy 0 0 0 Energy 3 0 0
1 0 0 0 3 0
0 1 0 0 0 3
0 0 1 2 1 0
2 0 0 2 0 1
0 2 0 1 2 0
0 0 2 1 0 2
1 1 0 0 1 2
1 0 1 1 1 1
0 1 1
Each row in the table corresponds to a different microstate. There are 1 microstate with energy 0, 3 with 1, six with two, and ten with

three. The general formula for the multiplicity of an Einstein solid with N oscillators and q energy units is
Ω(N, q) = (q + N – 1)!/(q!(N-1)!))
The following is a graphical representation of the microstate of an Einstein solid. Each energy unit is represented by a dot. A

vertical line represents a partition between one oscillator and the next. A solid with 4 oscillators is represented by this sequence.

The sequence represents the microstate in which the first oscvillator has one unit of energy, the second oscillator has three, the

third oscillator has none, and the fourth oscillator has four. Any microstate can be represented uniquely in this way, and every

possible sequence of dots and lines corresponds to a microstate. There are always q dots and N -1 lines, for a total of q + N – 1

symbols. Given q and N, the number of possible arrangements is just the number of ways of choosing q of the symbols to be dots, that is

(q + N – 1)!/(q!(N-1)!))
Q. 2. For an Einstein solid with each of the following values of N and q, list all of the possible microstate, count them, and verify

the formula Ω(N, q) = (q + N – 1)!/(q!(N-1)!))
(a) N=3, q = 4; (b) N = 3, q = 5; (c) N = 3, q = 6; (d) N = 4, q = 2; (e) N=4, q =3; (f) N= 1, q = anything; (g) N= anything, q = 1
Soln. (a) 400 310 031 220 211
040 301 103 202 121
? ? ? ? ?
How many does the formula predict? How many did you get here?
(b) 500 410 041 320 032 311 221
? ? ? ? ? ? ?
? ? ? ? ? ? ?
How many does the formula predict? How many did you get here?

© 600 501 015 ? ? ? ?
060 150 420 ? ? ? ?
006 051 402 ? ? ? ?
510 105 240 ? ? ? ?
How many does the formula predict? How many did you get here?
(d) 2000 0020 1100 ? ?
0200 002 ? ? ?
How many does the formula predict? How many did you get here?
e) 3000 2100 ? ? ?
0300 2010 ? ? ?
0030 2001 ? ? ?
? ? ? ? ?
How many does the formula predict? How many did you get here?
(f) If N = 1, then all the energy must belong to the one and only one oscillator, so there is only one microstate, which we would

denote by “q”. According to the formula, the multiplicity is
q!/q! =1
(g) If q = 1, then there is only one unit of energy to distribute among N oscillators, so the allowed states would be 1000…, 0100…,

0010…, , and so on up to …0001. There are N places to put the unit of energy , so the number of possible microstates is N. According to

the formula
Ω(N,1) = N!/(1! (N-1)! = N
Q. 3. Calculate the multiplicity of an Einstein solid with 30 oscillators and 30 units of energy.
Soln. For N = 30 and q= 30, the number of microstates should be Ω (30, 30) = 59!/30!.29! = 5.91E16
Q. 4. For an Einstein solid with four oscillators and two units of energy, represent each possible microstate as a series of dots and

vertical lines.
i) How many vertical lines are needed for four oscillators
ii) How many dots are needed for two energy?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
There are ten microstates.
Paramagnetic Salt.
The work done to increase the magnetization of a sample, not including the energy stored in the magnetic field itself, is given by
δW = μoH dM (2-8)
The total work done is obtained by integrating between the limits of Mi and Mf.
= > W = ʃ μoH dM (2-9)
where magnetization M has been defined as the magnetic moment (either pole strength times pole separation distance) per unit volume.

Magnetic flux density B =permeability of free space, μo times the magnetic field intensity, H.
Q. 5. If a uniformly magnetized rod of volume v has N’ elemental magnetic dipoles of moment ∆m, find the magnetization of the bar.
i) What is the total dipole moment in volume v?
ii) What is the dipole moment per unit volume?
Q. 6. Given H = MT/D, what is W = ʃ μoH dM ? integrated between the limits of of Mi and Mf.
Also, if the equation of state is M = DH/T, what is W = ? in terms of μoH M. Ans. μoHfMf – μoHiMi
Paramagnetic Solid.
Isothermal Processes
Q. 7. From the first law of thermodynamics,
dU = δQ + δW (2-10)
i) Substitute (2-8) for δW in (2-10) and solve for δQ = ?
Take the internal energy function as a function of temperature T and magnetism M.
= > U = U(T, M) (2-11)
ii) Write down the differential for dU = ? (2-12)
(Check with your handout on “Mathematics for Thermodynamics”)
i) What is δQ = ? Ans. (∂U/∂T)MdT + [(∂U/∂M)T – μoH]dM (2-13)

ii) Imagine the process to take place at constant magnetization, (δQ/dT) = ? (2-14)

iii) Let CM = (∂U/∂T)M, (2-15)

substitute this in your answer to ii). δQ = ? (2-16)

Ans. δQ = CMdT + [(∂U/∂T)T – μoH]dM
iv) Consider a process at constant H. Differentiate your answer to v) with respect to T:

What is CH = (δQ/dT)H = ?
Ans, CH = CM + [(∂U∂M)T – μoH](∂M/∂T)H (2-17)
Many paramagnetic substances obey Curie law
M= DH/T = > Induced magnetic moment = (Curie constant x magnetic intensity)/Temperature (2-18)
and have an internal energy which is a function of temperature only. For such a material, called a perfect paramagnetic substance,

answer to vi) simplifies considerably.
CH = CM-μoH(∂M/∂T)H (2-19)
CH = CM + μoDH2/T2 (2-20)
CH = CM μo M2/D (2-21)
The internal energy for a perfect paramagnetic solid
U = ʃCMdT (2-22)
Adiabatic Processes for Perfect Paramagnetic Solid
Start with the answer to iii) : δQ = (∂U/∂T)MdT +[(∂U/∂M)T – μoH]dM (2-16)
In an adiabatic process, there is no change in the quantity of heat, = > δQ =0; however temperature varies = > (∂U/∂M)T vanishes. Use

the definition of CM, to find 0 = CMdT – μoHdM
Q. 8.
i) Solve for CMdT = ? (2-17)

ii) What is H = ? from Curie law (2-18)

iii) Substitute for H on the right hand side of i), CMdT = ? (2-19)

iv) Rewrite iii) to find (CM/T)dT = ? (2-20)

v) Integrate left hand side from To to T and and right hand side from Mo to M

Did you get CMln(T/To) = μo(M2 – Mo2)/2D ? (2-21)

Solve for T = ? Ans. T = Toexp[μo(M2 – Mo2)/(2DCM)] (2-22)
This relation shows how a paramagnetic material may be used to cool a sample by adiabatic demagnetization. A large magnetic field is

impressed on the sample as it is cooled to as low a temperature as is possible. The external magnetic field is then turned off and

temperature drops exponentially.
Simple Solid
A simple solid being a hydrostatic system, work is done at constant pressure called isobaric work.
W = P(Vi-Vf) (2-23)
For a solid well below the melting point and for moderate pressures up to several hundred atmospheres, the equation of state is given

by
V = Vo(1 + βT – κP) (2-24)
where β called the thermal volume expansibility is a measure of how much the material expands as its temperature is raised at a

constant pressure:
β≡ 1/V . (∂V/∂T)P (2-25)
and κ the isothermal compressibility is a measure of how much the solid changes volume as the pressure is changed
κ= -1/V . (∂V/∂P)T (2-26)
A better approximation for the equation of state for a solid, we write
V = Vo[1 + β(T –To)– κ(P-Po)} (2-27)
Where Vo, β, and κ refer to temperature To and pressure Po.
Q. 9.
i) Following Eqn (2-27), write down Vi = ? Vf = ?
ii) Substitute Vi and Vf in (2-23), to get W = ? Ans. W = PVoβ (Ti – Tf) (2-28)
iii) From (2-24) , dV = ?

iv) For an isothermal process (dT = 0) , dV = ?

v) Substitute in the work equation W = – ʃPdV between the limits Pi and Pf

vi) Integrate v) Ans. W = Voκ/2(Pf2 – Pi2) (2-29)

Hydrostatic equations are CV = (∂U/∂T)V (2-30)
CP = CV + [(∂U/∂V)T + P] (∂V/∂T)P (2-31)
vii) Solve (2-31) for [(∂U/∂V)T + P] ?

viii) Substitute in δQ = CVdT + [(∂U/∂V)T + P]dV and simplify.

ix) Did you get δQ = CVdT + [CP – CV)/βVodV (2-32)

x) For an adiabatic process, what is dT = ? from ix)

xi) Integrate dT to get T≡[(1- γ)/βVo]V + c (2-33)

xii) Combine xi) with (2-24) to find P = (-γ/κVo)V + c (2-34)

which is the relation between pressure and volume for an adiabatic process.
xiii) Work done during an adiabatic compression is W = ʃ – PdV between the limits of Vi and Vf.
Take differential of (2-34) to get dP = (-γ/κVo)dV. Solve for dV to substitute in the work integral. Did you get W = (Voκ/2γ) (Pf2 –

Pi2) (2-35)
Q. 10. The pressure on a 1 cm cube of aluminum of mass 2.7 g was increased from 1 to 1000 atmospheres at a constant temperature of

300K. Use the simple solid model to calculate (a) the amount of work done by using Eqn (2-29), (b) the amount of heat transferred ,

(iii) the change in internal energy of the sample. Given that the compressibility of Al is 1.4E-11 m2/N and expansivity = 69E-5 /K.
a) i) What is the volume Vo of Al sample? Ans. 1E-6 m3
ii) W = ? Ans. 0.07 J
b) i) From Eqn (2-32) with dT =0 for isothermal case, δQ = ?
ii) use dV = -VoκdP and CP – CV = mass of Al x38 J/kg.K in i) to find Q. Ans. -2.08 J
c) Use the first law of thermodynamics ∆U = Q + W to find ∆U = ?

Interacting Systems
To understand heat flow and irreversible processes, consider a system of two solids A and B. They are very weakly coupled meaning

exchange of energy between them is much slower than the exchange of energy among the atoms within each. Their energies are UA and UB

which can change on longer time scales. The total multiplicity for all allowed values UA and UB counting all possible microstates with

only the sum Utotal = UA + UB held fixed. qtotal = qA + qB
There are seven possible macrostates with qA = 0, 1, 2, 3, 4, 5, 6. The total multiplicity of any macrostate

Solid A of energy UA Solid B of energy UB
# of HOs NA=3, Units of # of HOs NB = 3, units of
energy qA = 3 energy qB = 3

Ωtotal is just the product of the individual multiplicities, since the systems are independent of each other. For each of the ΩA

microstates available to solid A, there are ΩB microstates available to solid B.
Q. 1. a) Fill out the table. b)What is the total number of microstates accessible to the system over long time scales (sum the last

column)? c) Check the number by applying the standard formula to the entire system of six oscillators and six energy units – ΩB (6,6)

= (6 + 6 – 1)!/(6!(6-1)!))
qA ΩA= (qA + NA – 1)!/(qA!(NA-1)!)) qB ΩB= (qA + NA – 1)!/(qA!(NA-1)!)) Ωtotal =ΩAΩB Ωtotal/sum
0 1 6 28 28 28/462
1 3 5 21 63 63/462
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
Individual numbers in column 5 represent macrostates, to be read as 1st macrostate has 0 energy in solid A and all the 6 energy units

in solid B, 2nd macrostate has 1 energy unit in solid A and 5 energy units in solid B, 3rd macrostate has energy 2 units in solid A

and 4 energy units in solid B, and so on. The sum of column 5 represents all microstates. Column 6 represents the chances of finding

the system in the 1st, 2nd, 3rd, etc. microstate.
Q. 2. Which macrostate is the most probable?
Q. 3. Plot the total multiplicity in a bar graph – qA along x-axis and Ωtotal along y-axis.
_______________________________________________________________________________________________________________________________________

_______________________________________________________________________________________________________________________________________

______________________________________________________________________
Fundamental Assumption of Statistical Mechanics. In an isolated system in thermal equilibrium, all accessible microstates are equally

probable.
Over long time scales, the energy gets passed around randomly in such a way that all 462 microstates are equally probable.
Q. 4. Produce a computer-generated table and graph for a system of two Einstein solids with NA = 300, NB = 200, qtotal =100. You can

use a computer spreadsheet program, or comparable software, or perhaps even a graphing calculator.
qA ΩA= (qA + NA – 1)!/(qA!(NA-1)!)) qB ΩB= (qA + NA – 1)!/(qA!(NA-1)!)) Ωtotal =ΩAΩB Ωtotal/sum
0 1 100 2.8E81 2.8E81 2.8E81/9.3E115
1 300 99 9.3E80 2.8E83 2.8E83/9.3E115
? ? ? ? ? ?
? ? ? ? ? ?
. . . . . .
. . . . . .
. . . . . .
100 1.7E96 0 1 1.7E96 1.7E96/9.3E115
The spontaneous flow of energy when a system is at , or very near, its most likely macrostate, that is, the macrostate with the

greatest multiplicity. The “law of increase of multiplicity” is one version of the famous second law of thermodynamics.
Q. 5. Following the above statement of the second law of thermodynamics, name the macrostate in Q. 1. for the spontaneous flow of

energy.
Q. 6. Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharing a total 20 units of energy. Assume the

solids are weakly coupled, and that the total energy is fixed.
a) How many different macrostates are available to this system?
b) How many different microstates are available to the system? Ω(20,20) = ? Ans. 6.89E10
c) Assume that this system is in thermal equilibrium, what is the probability of finding all the energy in solid A? Ω(10,20)/Ω(20,20)

= ? Ans. 1.00E7/6.89E10 = 1.45E-4
d) What is the probability of finding exactly half of the energy in solid A? Ω(10,10)Ω(10,10)/Ω(20,20) = 8.53E9/6.89E10
e) Under what circumstances would this system exhibit irreversible behavior?
The probability of energy being evenly distributed is greater than that of all the energy being in A by a factor of 1000. If the system

started with all the energy in one solid or the other, we would be pretty sure that it would evolve toward a state with energy more

evenly distributed. And if it started out with the energy evenly distributed, we could be pretty sure that at some later time we

wouldn’t find all the energy on one side or the other- this would happen less than one time in a thousand.
Large Systems.
In the previous section, it was shown that the multiplicity function becomes very sharp with the increase of the number of oscillators.

Here we look at what happens when the system is much larger, so that each oscillator contains, say 1020 or more oscillators.
Very Large Numbers
We encounter three kinds of numbers in statistical mechanics – small numbers, large numbers, and very large numbers. Examples of small

numbers are numbers less than one hundred.
Large numbers are written in powers of ten. The most important large number is Avogadro’s number. Large numbers do not change when

small numbers are added to them. For example 10E23 + 23 = 10E23.
Very large numbers can be made by exponentiating large numbers. Very large numbers can be multiplied by large numbers without changing

them. For instance 1010exp23 x 1023 = 10(10exp23 + 23) = 1010exp23
Q. 7. The natural logarithm function, ln, is defined so that elnx = x for any positive number x
(a) Sketch a graph of the natural logarithm function: y = ln x
(b) Prove the identities ln ab = ln a + ln b (start with e ln ab = ? = e lna e lnb = ?
© Prove that ln ab = b ln a (e ln a exp b ) = ab =

(d) Prove that (d/dx)(ln x) = 1/x

(e) Derive the approximation ln(1 + x) ≈ x which is valid for |x|<< 1. Use a calculator to check the accuracy of this approximation

for x = 0.1 and x = 0.01
Q. 8. (a) Simplify the expression e a ln b (Write it in a way that does not involve logarithm)
(b) Assume b <<a, prove that ln(a + b) ≈ln a + b/a (factor out the a from the argument of the logarithm, so that you can apply the

approximation of part (e) in the previous problem)
Q. 9. Write e10exp23 in the form of 10x, for some x. (take natural log of both sides. Then left side becomes 1023 and the right side

becomes ln 10x = x ln 10. = > 1023 = x ln 10. Noe solve for x. Ans. 4.34E22
Thermodynamics Chapter Test #2
Name——————————————————— Date____________________________
Q. 1. The temperature of a block of copper in increased from 127 oC to 137oC. Find the change in pressure to keep the volume constant,

given that β≈5.2E-5 K-1, and κ ≈ 7.6E-12 Pa-1.
i) If volume is to remain constant, what is the relation between V and Vo? V = Vo
ii) Following i), what is the change in pressure expression in terms of β, κ, and temperature change?
∆P = (β/κ) ∆T
iii) Substitute the given values to find the change in pressure. Ans. 6.8E7Pa ≈680 atm

Q. 2. Take four coins – a penny, a nickel, a dime, and a quarter. Flip them.
i) How many possible outcomes are there? 16
ii) What is the probability of getting four heads? 1/16
iii) What is the probability of getting four tails? 1/16
iv) What is the probability of getting two heads and two tails? 6/16
v) What is the probability of getting exactly one head and three tails? 4/16
Penny Nickel Dime Quarter
` H H H H

H H H T
H H T H
H T H H
T H H H

H H T T
H T T H
T H T H
T T H H
H T H T
T H H T

H T T T
T T T H
T T H T
T H T T

T T T T

Q. 3. In the flipping of the coins in #1, fill out the chart.
Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities
4H Ω.(4)=? 1
3H Ω.(3) =? 4
2H Ω.(2) = ? 6
1H Ω.(1) = ? 4
0H Ω.(0) = ? 1

Ω(N,n) = (N!)/[n!. (N-n)!]; V = Vo[1 + β(T –To)– κ(P-Po)} where Vo, β, and κ refer to temperature To and pressure Po. 0! = 1
Thermodynamics Chapter Test #2
Name——————————————————— Date____________________________
Q. 1. The temperature of a block of copper in increased from 127 oC to 137oC. Find the change in pressure to keep the volume constant,

Q. 2. Take four coins – a penny, a nickel, a dime, and a quarter. Flip them.
i) How many possible outcomes are there?
ii) What is the probability of getting four heads?
iii) What is the probability of getting four tails?
iv) What is the probability of getting two heads and two tails?
v) What is the probability of getting exactly one head and three tails?
Penny Nickel Dime Quarter

Q. 3. In the flipping of the coins in #1, fill out the chart.
Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities
4H Ω.(4)=?

3H Ω.(3) =?

2H Ω.(2) = ?

1H Ω.(1) = ?

0H Ω.(0) = ?
Ω(N,n) = (N!)/[n!. (N-n)!]; V = Vo[1 + β(T –To)– κ(P-Po)} where Vo, β, and κ refer to temperature To and pressure Po. 0! = 1

A macrostate (the total number in the same state) of a system , or configuration is specified by the number of particles in each of the

energy levels of the system. Thus, let there be Nj the number of particles that occupying the jth energy levels of n energy levels.

That is if n = 3, we understand that N1 in the 1st level, N2 in the 2nd level, and N3 in the 3rd level. The total number of particles

is N1 + N2 + N3 = N
A microstate (which one is in which state) is specified by the number of particles in each energy state. In general, there will be more

than one energy state (i. e. quantum state) for each energy level, a situation called degeneracy. A microstate is the most specific

description one can get. In general, there will be many, many differenent microstates corresponding to a given macrostate. The basic

postulate of statistical thermodynamics is that all possible microstates of an isolated assembly are equally probable. We are

interested in the number of microstates but not in their detailed description.
The number of microstates leading to a given macrostate is called the thermodynamic probability. It is the number of ways in which a

given configuration can be achieved. This is an “unnormalized probability, an integer between zero and infinity. For the kth macrostate

, the thermodynamic probability is taken to be wk. A true probability Pk could be obtained by dividing wk by the total number of

microstate Ω available to the system.
In the coin-tossing experiment, each macrostate is defined by the number of heads and the number of tails. If we toss N= 4 coins,

macrostates are (i) zero head and four tail, (ii) one head and three tails, (iii) two heads and two tails, (iv) three heads and one

tail, and (v) four heads and zero tail. A microstate is specified by the state – head or tail of each coin. It is the most detailed

description possible. Our interest is to find the number of microstates for each macrostate, the thermodynamic probability. The true

probability is the number divided by the total number of microstates. Thus
Pk = wk/Ω
where 5
Ω = ∑ wk = 16
k = 1

We can calculate the average occupancy numbers i.e. the average number of heads and tails. Let j = 1 or 2, where N1 is the number of

heads and N2 is the number of tails. Let Nk be the occupation number for the kth macrostate. Then the average occupation number is

Njav={∑Njk wk}/{∑wk } = ∑ Njkwk/Ω = ∑Njk wk
k k k k
The average number of heads is therefore
Njav = (1/16){(4×1) + (3×4) + (2×6) + (1 x4) + (0 x1)} = 2
Similarly, average N2av = 2, as we would expect. Then N1av + N2av = 4 = N. In a plot of thermodynamic probability w vs the number of

heads N1, the curve becomes symmetric about N1 = 2, the most probable configuration.
In a large number N of distinguishable coins, the number of ways to select from N candidates N1 heads and N-N1tails is w = N!/{(N!)(N-

N1)!} which is the binomial coefficient.
Macrostate Label Macrostate Specifications Microstate Therm. Prob. True Prob.
k N1 N2 Coin 1 Coin2 Coin 3 Coin 4 wk Pk
____________________________________________________________________________________
1 4 0 H H H H 1 1/16

2 3 1 H H H T 4 4/16
H H T H
H T H H
T H H H

3 2 2 H H T T 6 6/16
T T H H
H T H T
T H T H
H T T H
T H H T

4 1 3 H T T T 4 4/16
T H T T
T T H T
T T T H

5 0 4 T T T T 1 1/16
4.1 Cycles
Thermodynamic cycle. Processes which can continue indefinitely without permanently changing the system.
A heat engine is any device that absorbs heat and converts part of that energy into work. The net work done by the engine in some given

time period is denoted by W. The benefit of operating a heat engine is the work Produced. The cost of operation is the heat absorbed,

Qh.
Hot reservoir. Heat absorbed by the engine comes from a place called the hot reservoir. Its temperature is denoted by Th. Heat

absorbed from a hot reservoir in some given time period is denoted by Qh.
Cold reservoir. The waste heat is dumped into a place called the cold reservoir. Its temperature is denoted by Tc. Heat expelled to the

cold reservoir in some given time period is denoted by Qc.
Reservoir. Anything that is so large that its temperature does not change noticeable when heat enters or leaves. In steam engines,

the hot reservoir is the place where fuel is burned and the cold reservoir is the surrounding environment.
As a system undergoes a cycle, it goes through a series of transformations, after which the system has returned to its initial state.

During this cycle, the system may do work and absorb or exhaust heat to one or many heat reservoirs. The net result is the conversion

of heat to work, if the system is a heat engine, or the use of work to remove heat, if the system is a refrigerator.
Efficiency. The efficiency of an engine, e, is the benefit/cost ratio. = > e ≡ benefit/cost = W/Qh (4.1)
Q. 1. What is emax for given values of Th and Tc?
According to the first law, energy absorbed = energy expelled + work done = > Qh = Qc + W (4.2)
i) W = ? from Eqn. (4.2)
ii) Substitute W from i) in Eqn (4.1)
iii) Do you get e = 1 – Qc/Qh ?
iv) What has to be Qc = ? if e = 1
Coefficient of performance (COP) in heating mode = |Qh|/W
Coefficient of performance (COP) in heating mode = |Qc|/W
A good refrigerator should have a COP of 5 or 6.

Q. 2. A refrigerator has a COP of 5. When the refrigerator is running, it power is 500 W. A sample of water of mass 500 g and at

temperature 20 deg Celsius is placed in the freezer compartment. How long does it take to freeze the water to ice at 0 deg Celsius?
Hint. i)_ How much heat is to be extracted to reduce the water temperature to 0 deg Celsius?
Q1 = mc∆T= mass x specific heat of water x change of temperature = ? cwater = 4186 J/kg. oC
ii) What is the latent heat of fusion ? Lf =3.33E5 J/kg
iii) How much heat is to be extracted to freeze water at 0 oC?, Q2 = m Lf = ?
iv) What is the total amount of heat that has to be subtracted, sum of i) and ii). = ?
Ans. 2.08E5 J/kg
(v) COP = |Qh|/W = ? = > W = ? Ans. 4.17E4 J
vi) Power = Work/Time = > Time = Work/Power = ? Ans. 83.3 s
Configuration space. The coordinates x, y, and z specify coordinate space. A point in configuration space corresponds to one position,

while a point in momentum space corresponds to one momentum.
Momentum space. Momentum axes px, py, pz define momentum space.
Phase space. A combination of configuration and momentum space into a six dimensional is called the phase space. A trajectory in phase

space represents a curve of constant total mechanical energy. Since classically, we can predict No true trajectories can intersect.
Classically, the momentum and position of a particle can be determined to as high a precision as desired; however, quantum

mechanically, the Heisenberg uncertainty principle limits the precision with which we can simultaneously measure y and py-
∆y.∆py≥ h (1)
In 6-dimentional phase space, we can determine the position of a particle only to within a hyper-volume of h3.
Phase space for a system of N particles is 6N-dimensional spanned by the momenta and positions of the particles. Each point in this

space represents what is called a microstate of the system. Generally, we cannot know the microstate of the system, only the

macrostate. If the system is in a given macrostate, it usually can be in any of a very large number of microstates.
Ensembes and the Distribution Function. If we know nothing about a system of N particles, the system may be in any microstate. The

point representing the system may be anywhere within the phase space. If, on the other hand, one knows some macroscopic

characteristics, such as the total energy and volume, the system may be in only those microstates which are consistent with the known

values. When considering a system with some macroscopic characteristics, we actually consider a set of possible systems, each in a

different microstate. This collection of systems form an ensemble. An ensemble is represented in phase by a subset of all the possible

points. In quantum mechanics, the subset may be a collection of discrete, quantized points. In classical mechanics, this subset is

hyper-volume or hyper-surface.
The system is in only one microstate, corresponding to one point in phase space, at any instant of time; however, it will be

continuously changing microstates within the allowed set. The probability of finding a system in a given microstate is called is called

the distribution function.
The notation f(xi,pi) represents the distribution which is a function of all positions, xi, and momenta pi, where I = 1 to 3N.

Alternatively, we may choose to number each of the microstates , j = 1 to Ω, where Ω is the number of possible microstates.
The distribution function is defined so that it describes the probability of finding the system in a particular microstate. If the

ensemble consists of a continuous region,
f(xi,pi).dVx. dVp (2)
is the probability of finding the system in the hyper-volume dVx.dVp about the point (xi,pi) in phase space.
In rectangular coordinates, dVx = dx1dx2dx3dx4…dx3N-1 dX3N (3)
And dVp = dp1dp2dp3dp4…dp3N-1 dp3N (4)
The measurement of the observable property , A, of system such as energy or magnetization, three types of averages are important – time

average, ensemble average, and the most probable value.
Time average is obtained by repeatedly measuring the system as it evolves with time.
Ensemble average is found by averaging over all the microstates in the ensemble
<A>≡ ∑Ajfj/∑fj (5)
or <A> ≡∑A(xi,pi)f(xi,pi)/∑f(xi, pi) (6)
If the ensemble consists of a continuous region of phase ∑ will represent an integral
<A> ≡∑A(xi,pi)f(xi,pi)/ ∑f(xi, pi) →ʃA(xi,pi)f(xi,pi)dVx dVp /∑f(xi, pi) (7)
Q. 1.Transform the sum into integration for continuous x: ∑mixi →
The Canonical Ensemble and the Partition Function
Since systems are not generally isolated but rather in thermal equilibrium, we are much more likely to know a system’s temperature that

its energy. Let us separate the systems and the reservoir (a large system with a known temperature but isolated from the environment)

by a diathermal wall so that the number of particles in the system , N, and the number particles of the reservoir, No, are fixed. The

total energy of the system plus the reservoir is Uo and the energy of the system alone is ε. Then the energy of the reservoir must be

Uo – ε. Since energy may flow between the system and the reservoir, ε is not fixed but fluctuates about average value.
Boltzmann factor. The probability of finding the system in a single state with energy ε must be proportional to the number

of states Ω(Uo-ε) of the reservoir with energy Uo-ε. That is
f(ε) ∞ Ω(N, Uo, ε) (8)
Q. 2.What is f(ε1) ∞ ?
Q. 3. What is f(ε2) ∞ ?
Q. 4. What is f(ε1)/f(ε2) = ?
The ratio of the probability of finding the system in a single microstate with one of two different energies ε1 and ε2 is then
f(ε1)/f(ε2) = Ω(N, Uo, -ε1)/ Ω(N, Uo, -ε2) (9)
The microscopic definition of entropy is
S(E, V, N)≡k ln (Ω(E)) (10)
where k is a constant.
Q. 5. What is Ω(E) = ? (11)
And f(ε1)/f(ε2) = Ω(N, Uo, -ε1)/ Ω(N, Uo, -ε2) (12)
= > f(ε1)/f(ε2) = exp[S(N, Uo,- ε1)/k]/exp[S(N, Uo,- ε2)/k] (13)
Q. 6. What is a Taylor series? When can you expand a function in a Taylor series?
Since the reservoir is much larger than system, ε <<Uo, we can expand S(N, Uo-ε) about Uo.
S(N, Uo-ε) = S(N,Uo) –ε(∂S/∂Uo)No + ….0…… (14)
The statistical definition of temperature is
1/T = (∂S/∂U)N,X (15)
Q. 7. With the substitution of 1/T, S(N, Uo-ε) = ? (16)
Q. 8. S(N, Uo-ε1) = ? (17
Q. 9. S(N, Uo-ε2) = ? (18)
Neglecting higher order terms in the series, the ratio of probabilities may be written as
= > f(ε1)/f(ε2) = exp[S(N, Uo)- ε1)/kT]/exp[S(N, Uo)- ε2)/kT] (19)
= eS(N, Uo). – eε1/kT/ eS(N, Uo). – eε2/kT) = – eε1/kT/ – eε2/kT (20)
= > f(ε1)/f(ε2) = = – eε1/kT/ – eε2/kT (21)
This exponential term is called the Boltzmann factor and the ratio of Boltzmann factors represents the ratio of the probability that

the system will be in a given microstate of energy ε1 to the probability that the system will be in a single microstate of energy ε2.

The distribution function for the system must be of the form
f(εj) = Aexp(-εj/kT) = Ae-εj/kT, j is a suffix of ε. (22)
The constant A is determined by requiring the distribution function to be normalized –
∑f(εj)= ∑Ae-εj/kT = 1 (23)
j j

Thus A = (∑e-εj/kT)-1 (24)
j
= > A = [∑exp(-εj/kT)]-1 (25)
j

The Partition Function for the Canonical Ensemble
The sum of Boltzmann terms, called the partition function, is represented by a Z (from German zustandsumme) –

Z = ∑exp(-εj/kT) = sum of all Boltzmann factors (26)
j
Q. 10. Write partition function for four states j = 1, 2, 3, 4
Q. 11. The two loest energy states of atomic hydrogen have energies of -13.6 eV and – 3.4 eV. Determine the ratio of the number of

hydrogen atoms in the first excited state to the number in the ground state at T = 300 K. Repeat for T = 6000 K and T = 15 M K. Given

that k = 8.62E-5 eV/K
Hints. i) Write down the ration of probabilities for single states with energy ε1 and ε2.
f(ε2)/f(ε1) =?
ii) What is the formula for the number of electrons in n = 1, 2, 3, states etc.
iii) How many electrons are in the n = 1 state with energy ε1 = -13.6 eV
iv) How many electrons are there in the n = 2 states with energy ε2 = -3.4 eV?
v) What is the ratio of atoms in state n =2 to the number in state n = 1 ?
vi) (ε1 – ε2)/kT = ? for T = 300 K
vii) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 5.0E-172
viii) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans. 2.0E-171
Since the ratio of n=2 state to n=1 state is so small, all the hydrogen is in its ground state (n = 1)
ix) (ε1 – ε2)/kT = ? for T = 6000 K
x) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 2.7E-9
xi) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans.1.08E-8
xii) (ε1 – ε2)/kT = ? for T = 15E6 K
xiii) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 0.992
xiv) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans. 3.968
xv)
Q. 12. Prove that the probability of finding an atom in any particular energy level is P(ε) = (1/Z)e-F/kT
Where F = ε-TS and the “entropy” of a level is k times the logarithm of the number of degenerate states for that level (which defines S

= k ln n)
Hint. n degeneracy means the probability of the system being in that level is n times the probability of being in any one of the state.
P(ε) = nf(ε) = n. 1/Z . e-ε/kT
Write n as eln n = eS/k
= > P(ε) = eS/k. (1/Z). e-ε/kT = 1/Z . e –(ε-TS)/kT = 1/Z. e-F/kT
Q. 13. Consider a hypothetical atom that has two states: a ground state with energy 0 and an excited state with energy 2 eV. Draw a

graph of the partition function (PF) for the system as a function of temperature, and evaluate the PF mathematically at T = 300 K, 3000

K, 30,000 K, and 300,000 K, given that k = 8.62E-5 eV/K
Hints. i) Write down the PF expression, Z = ?
ii) Expand it for two-state system
iii) Where does Z → as T → to 0
iv) Where does Z → as T → to ∞
v) Substitute t = kT/ε, then Z = ?
vi) Fill out the table
T t = kT/ε Z
300 K ? ?
3000 ? ?
30,000 ? ?
300,000 ? ?
vii) Plot kT/ε (along x) vs Z (along y)
We rewrite Eqn (22) f(εj) = Aexp(-εj/kT) = Ae-εj/kT
as Ƥ(s) = (1/Z) e-E(s)/kT

(27)
= > Probability Ƥ(s) of finding state s with energy E(s).
Z = ∑e-E(s)/kT = the sum of all Boltzmann factors.
s
Z does not depend on any particular state, but does depend on temperature.

To interpretation Eqn (27), let us suppose for a moment that the ground state energy of a one-atom system is zero i. e. Eo = 0, while

the excited states have positive energies. Then the probability of the ground state is 1/Z, and all other states have smaller

probabilities. States with energies much less than kT have probabilities only slightly less than 1/Z, while states with energies much

greater than kT have negligible probabilities, suppressed by the exponential Boltzmann factor. Q. 15 Fig. shows a bar graph of the

probabilities for the states of a hypothetical system.

Q. 14. If the ground state energy is zero, what is Boltzmann factor = ?

Q. 15. The figure below represents a hypothetical system. (a) Estimate partition function for this. (b) Estimate the probability of

this system being in its ground state.
Bars represent relative probabilities of a hypothetical

system.
Horizontal axis is the energy. The smooth curve

represents the
Boltzmann distribution for one particular temperature.

At lower
Temperatures it would fall off more suddenly, while at

higher
Temperatures it would fall off more gradually. Bars

represent
Probability Ƥ(s) Boltzmann factors, too

E(s)
kT 2kT 3kT

Hints. i) What is the ground state energy in the problem?
ii) What does it contribute to the partition function?
iii) What is the height of all the 9 bars in the figure? Measure and sum them.
iv) What is the height of the first bar?
v) Relative to the height of the first bar, the total of all of them = ? (sum of heights/first bar height)
vi) What is Z = ? (the quotient obtained in # v)
vii) The probability of the ground state = ? (reciprocal of Z)

Q. 16. Imagine a particle that can be in only three states with energies – 0.05 eV, 0, and 0.05 eV. This particle is in equilibrium

with a reservoir at 300 K.
(a) Calculate the PF for the particle
(b) Calculate the probability for this particle to be in each of the three states.
© Because the zero point for measuring energies is arbitrary, we would just as well say that the energies of the three states are 0,

0.05 eV, and 0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn’t.
(a) i) kT = (8.62E-5 eV/K) (300 K) = ?
ii) Z= Z1 + Z2 + Z3 = e-(-0.05/0.026) + eo + e-(0.05/0.026) = ?
(b) Ƥ1 = Z1/Z = ? Ƥ2 = Z2/Z = ? Ƥ3= Z3/Z = ?
© i) Z= Z1 + Z2 + Z3 = e0 + e-(0.05/0.026) + e-(0.10/0.026) = ?
ii) Ƥ1 = Z1/Z = ? Ƥ2 = Z2/Z = ? Ƥ3= Z3/Z = ?
Nature possibly can’t care what we use as our zero-point for measuring energy.
November 28, 2011
The probability that a system is in any particular one of its microstate s, given that it is in equilibrium with a reservoir at

temperature T:
Ƥ(s) = (1/Z)e-βE(s) (6-12)
where β= 1/kT called the Boltzmann factor
and Z = ∑ e-βE(s) (6-13)
s

where the sum is carried over all the possible states.
Q. 1. In a system of 5 atoms, 2 of them in the ground state, 2 in the 4-eV state, and 1 in the 7-eV state. What is the average energy

of all the atoms? Ans. 3 eV

For a large sample of N atoms, and N(s) the number of atoms in any particular state s, the average value of energy is
<E>= ∑[E(s).N(s)]/N = ∑ [E(s) N(s)]/N = ∑E(s)Ƥ(s) (6-16)
Where Ƥ(s) is the probability of finding an atom in state s.
= > Average energy = sum of all energies weighted by their probabilities.
In statistical mechanics, each probability is given by (6-12).
So, <E>= (1/Z)∑E(s) e-βE(s) (6-17)
The sum is similar to the partition function (6-13), but with an extra factor E(s) in each term.
The average value of any other variable of interest can be computed in exactly the same way. If a variable X has the values X(s) in

state s, then
<X> = ∑X(s)Ƥ(s) = (1/Z)∑X(s)e-βE(s) (6-18)
Average values are additive. That is the average values of two objects is the sum of their individual average values. In a collection

of many identical, independent particles, the total average energy is just the product of average energy of just the one and their

number.
Q. 2. Suppose you have 10 atoms: 4 with energy of 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and one with energy 6 eV.
(a) Compute the average energy of all the atoms by adding up all the energies and dividing by 10. Ans. 1.7 eV
(b) Compute the probability that one of the atoms chosen at random would have energy E, for each of the four values of E that occur.
© Compute the average energy using the formula <E>= ∑E(s)Ƥ(s)
Hint. (a) You can do it.
(b) i) What is the total number of atoms?
ii) How many atoms have energy of 0 eV?
iii) What is the probability of finding at atom with energy of 0 eV?
iv) How many atoms have energy of 1 eV?
v) What is the probability of finding at atom with energy of 1 eV?
vi) How many atoms have energy of 4 eV?
vii) What is the probability of finding at atom with energy of 4 eV?
viii) How many atoms have energy of 6 eV?
ix)What is the probability of finding at atom with energy of 6 eV?
© i) What is the product of 0 eV and iii) of (b)?
ii) What is the product of 1 eV and v) of (b)?
iii) What is the product of 4 eV and vii) of (b)?
iv) What is the product of 6 eV and ix) of (b)?
v) What is the sum of i) through iv)? Does it agree with (a)
Q. 3. Z = ∑ e-βE(s), i ) find ∂Z/∂β = ?
s
ii) Divide i) by –Z. Is it the same as (6-16)?
s
iii) What is (-∂/∂β)(ln Z) Is it the same as (6-16)?

Q. 4. Go to Q. 1. a) i) What is the average value of energy?
ii) Compute the deviations of each of the energies from the average energy for all the five atoms.
iii) Compute the squares of each of the deviations.
iv)Find the sum of (iii)
v) Find the average
vi) Find the square root of v). Ans. 2.7 eV
This is called the root-mean-square (rms) deviation or the standard deviation.
(b) Prove that σE2 =<E2> – <E>2.
i) σE2 = <(∆Ei)2>= (1/N) ∑ (∆Ei)2 = (1/N)∑(Ei – <E>)2 = ?
Paramagnetism Q. 5. Each elementary dipole in an ideal two-state paramagnet has just two possible states: an “up” state with energy –μB

and a “down” state with energy + μB, where B is the strength of an externally applied magnetic field while the component of the

dipole’s magnetic moment in the direction of the field is ±μ.
i) What is E(up) = ? ii) What is E(down) = ? iii) What is e-βE(up) = ?
iv) What is e-βE(down) = ? v) What is Z = ? Express in hyperbolic functions Ans. 2 cosh(βμB)
vi) Probability of finding dipole in “up” state = ?
vii) Probability of finding dipole in “down” state = ?
viii) What is the sum of these two probabilities?
ix) Find the average energy of the dipole: Sum the products of i) and vi) and ii) and vii)
x) Simplify ix) to find – μBtanh(βμB)
xi) For a collection of N such dipoles what is the total energy?
xii) Differentiate Z in v)
xiii) Multiply xii) by (-1/Z)
xiv) Does xiii) equal x)?
xv) Find the average value of a dipole’s magnetic moment along the direction of B
xvi) <μz> = ∑μz(s)Ρ(s) = (+μ)(Ρ↑) + (-μ)(Ρ↓) = ? Ans. μtanh(βμB)
xvii) What is the total magnetization for the sample of N dipoles?
Q. 6. Consider a system of N identical harmonic oscillator at temperature T, the allowed energies of each oscillator being 0, hf, 2hs,

3hf, and so on.
(a) Prove by long division that 1/(1 -x) = 1 + x + x2 + x3 + ……………….
For what value of x does this series have a finite sum?
(b) Evaluate the partition function for a single oscillator. Use the result in (a) to simplify as much as possible
© Use <E> = (-1/Z)(∂Z/∂β) to find the average energy of a single oscillator at temperature T
(d) What is the total energy of N oscillator at temperature T?
(e) Use heat capacity C = ∂U/∂T expression to find the heat capacity.
(f) Use Mathematica to plot the function C/Nk = e1/t//{t2(e1/t – 1)2
Hint. (a) Write down 1 = 1 + 0.x + 0. x2 + 0.×3 + ………………. And carry out the division.
(b) Take the ground state to have an energy of 0, then sum the states with energies ε (=hf), 2ε (=2hf), 3ε (=3hf), 4ε (=4hf), 5ε

(=5hf) and so on to find Z = ? Ans. 1/(1-e-βε) by the result of (a)
© ans. <E>= ε e-βε /(1-e-βε)
(d) U = N<E> Ans. U = Nε/(eβε- 1)
(e) C ={Nε2/kT2}{(eε/kT)/(eε/kT – 1)2
When kT>>ε, what is eε/kT= ?, keep the first term in the numerator and up to the second term in the denominator. Simplify further.

What do you get? Ans. Nk
(f) Plot it.
Carnot Engine establishes an upper limit on the efficiencies of all other engines. It is of great importance both from practical and

theoretical viewpoints.
Carnot cycle. An ideal reversible cycle operating between two energy reseroirs.
Carnot theorem. No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating

between the same two reservoirs.
Carnot cycle description. Let the Carnot cycle takes place between two temperatures Tc and Th. Let the working substance be an ideal

gas contained in a cylinder fitted with a movable piston at the end. The cylinder’s wall and piston are thermally nonconducting. The

cycle consists of two adiabatic processes and two isothermal processes, all reversible:
1. Process A→B is an isothermal expansion. The gas placed in thermal contact with an energy reservoir at temperature Th. During

the expansion, the gas absorbs energy │Qh│from the reservoir through the base of the cylinder and does work WAB in raising the piston
2. In process B →C, the base of the cylinder is replaced by a thermally nonconducting wall, and the gas expands adiabatically. –

that is no energy enters or leaves the system by heat. During the expansion, the temperature of the gas decreases from Th to Tc and

the gas does work WBC in raising the piston.
3. In process C→D the gas is placed in thermal contact with an energy reservoir at temperature Tc and is compressed isothermally

at temperature Tc. During this time, the gas expels energy │Qc│to the reservoir, and the work done by the piston on the gas is WCD
4. In process D→A, the base of the cylinder is replaced by a nonconducting wall, and the gas is conmpressed adiabatically. The

temperature of the gas increases to Th, and the work done by the piston on the gasis WDA

The net work done is given by the area enclosed by the path ABCDA

A
B

C
D
The thermal efficiency is given by 1 – │Qc│/│Qh│ 1- Tc/Th

Q.1. The highest theoretical efficiency of a Carnot engine is 30%. If this engine uses the atmosohere, which has a temperature of 300

K, as its cold reservoir, what is the temperature of its hot reservoir?
i) Write down the formula you will use.
ii) What is 30% in decimal form?
iii) Substitute the values and solve for Th Ans. 429 K
Q. 2. A steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water to steam, and this steam then

drives a piston. The cold reservoir’s temperature is that of outside air 300 . What is the maximum thermal efficiency of this engine?
i) What is Tc=?
ii) What is Th= ?
iii) What is efficiency = ? Ans. 40%

Quantum Statistics
The ratio of probabilities of two different microstates with energy ε1 and particles N1 and energy ε2 and particle N2
f(ε1, N1)/ f(ε2, N2) = e-[ε1 –μ1)]/kT / e-[ε2—μN2]/kT (1)
= > Ratio of probabilities or relative probabilities with energy ε and number N
F(ε, N) = (1/Ƶ) e-[ε –μN]/kT (3)
Where Ƶ is called the grand partition function. Ƶ = ∑∑ e-[ε –μN]/kT (4)
N S
Q. 1. Assume a single hydrogen atom can be found in 3 possible states. State is ionized hydrogen and has 0 energy. State 2 contains

one electron and has an energy ε1 = -13.56 eV. State 3 contains two electrons and has an energy of ε2 = -14.31 eV. The atom may

exchange electrons with surrounding atoms, bu the hydrogen as a whole is neutral. Determine the grand partition function for the system

, treating μ as given. What value must μ have so that the gas is neutral (N) = 1)?
Hint. i) Write down the expression for the grand partition function
ii) Ƶ =? If ε = 0 and number of electron N =0
iii) Ƶ = ? if ε = ε1 and N = 1
iv) Ƶ = ? if ε= ε2 and N = 2
v) What is the sum of ii), iii), and iv)
vi) What is 0 times ii) = ?
vii) What is 1 times iii) = ?
viii) What is 2 times iv) = ?
ix) Sum vi), vii), and viii)
x) Divide ix) by v).
xi) Set x) equal to 1
xii) What is exp-(ε2 – 2μ)/kT = ?
xiii) What is μ= ?
xiv) Is xiii) ε2/2 ? Ans. -7.16 eV
Statistical Physics
It considers how the overall behavior of a system of many particles is related to the properties of the particles themselves. It

determines the most probable way in which a certain total amount of energy E is distributed among the N particles of a system of

particles in thermal equilibrium at the absolute temperature T. That is how many of the N particles are likely to have energy ε1, how

many ε2, and so on.
The greater the number W of different ways in which the particles can be arranged among the available states to yield a particular

distribution of energies, the more probable is the distribution. The most probable distribution , which corresponds to the system’s

being in thermal equilibrium, is the one for which W is maximum. The condition is that the system consists of a fixed number N of

particles whose total energy is some fixed amount E.
n(ε) = number of particles with energy = g(ε) f(ε) (1)
where g(ε) = number of states with energy
= statistical weight corresponding to energy
f(ε) = distribution function
= average number of particles in each state of energy ε
= probability of occupancy of each state of energy ε
For a continuous distribution g(ε) is replaced by g(ε) dε, the number of states with energies between ε+ dε.
We consider systems of three different kinds of particles.
Maxwell –Boltzmann distribution function for identical particles that are sufficiently far apart to be distinguishable , for instance,

molecules of a gas. Quantum mechanically speaking, wave function of particles overlap to a negligible extent.
Bose-Einstein distribution function is applicable to particles of 0 or integral spin that cannot be distinguished from one another.

Quantum mechanically speaking, their wave functions overlap. Photons fall in this category. Particles described by this distribution

are called bosons.
Fermi-Dirac distribution function is applicable to particles with half-integral spin (1/2, 3/2, 5/2, …) that cannot be distinguished

from one another. Particles following this distribution are called fermions. Electrons fall in this category.
MAXWELL-BOLTZMANN STATISTICS
The average number of particles fMB(ε) in a state of energy ε in a system of particles at the absolute temperature T is given by
fMB( ε)= A e- ε/kT (2)
where A depends on the number of particles in the system and is analogous to the the normalization constant of a wave function, and k

is Boltzmann constant = 1.381×10-23 J/K = 8.617 x10-5 eV/K.
Substituting the value of fMB( ε) in Eqn. (1), we get
n( ε) = A g( ε) e- ε/kT (3)
which gives the number n( ε) of identical distinguishable particles in an assembly at the temperature T that have energy ε.
Law of Atmospheres
Let us consider an ideal atmosphere in which gas molecules are in thermal equilibrium, and that there is no turbulence present and no

winds are pushing the molecules around. In such an atmosphere, density of molecules becomes a simple function of height, z.
Let m = mass of each molecule
N/V = number per volume
The force/area caused by the weight of the molecules are
ΔP = mgNΔz/V
In the limit Δz → 0,
dP/dz = -mgN/V = -mgn (4)
Using the ideal gas equation PV = NkT
n = P/kT (5)
Differentiating wrt t, dn/dP = 1/kT
Whence dn/dz = (dn/dP)(dP/dz) = -mgn/kT (6)
The solution of this eqn is
N = n0e-mgz/kT

(7)
Q. 1. Estimate the thickness of the atmosphere
Ans. z = kT/mg, m = 28 x 1.66E-27kg, z = ?
Q. 2. A cubic meter of atomic hydrogen at 0 degree Celsius and at atmospheric pressutre contains about 2.7E25 atoms. Find the number

of these atoms in the first excited states (n=20 at zero deg Celsius and at 10,000 deg Celsius.
Soln. n(ε1) = A g(ε1) e-ε1/kt and n(ε2) = A g(ε2) e-ε2/kt
The ratio of the number of atoms in the n = 1 and n=2 states is
n (ε2)/n(ε1) = g(ε2)/g(ε1) e- (ε2 – ε1)/kt
Since number of possible states that corresponds to the quantum number n is 2n2. The number of states of energy ε1 corresponding to n =

1 is 2. This is the case of 1s electron. It has l=0, ml= 0, ms = -1/ or +1/2.
The number of states n = 2 is 8 i. e. the number of states of energy ε2 is g(ε2) = 8; a 2s electron can have
l=0, ml= 0, ms = -1/ or +1/2, and 2p electron can have l=1, ml= 0, ±1, in each case of ms = -1/ or +1/2..
Since the ground state energy ε1 = – 13.6 eV, and ε2 = ε1/n2 = -3.4 eV, ε1 – ε2 = 10.2eV, T = 0 deg Celsius = 273 K
 (ε2 – ε1)/kt = 10.2 eV/(8.617E-5 eV/kt)(273 K) = 434
 n (ε2)/n(ε1) = (8/2)e-434 = 1.3E-188
Thus about 1 atom in every 1xE188 is in the first excited states at 0 deg Celsius. With 2.7E25 atoms in our sample, we are confident

that all are in the ground state.
(b) T = 10,273 K. and (ε2 – ε1)/kT = 11.5
 n (ε2)/n(ε1) = (8/2)e-11.5 = 4E-5
Now the number of excited atoms is about 1E21, a substantial number even though only a small fraction.
Q. 3. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in n=2 energy level?
Soln. g(ε2) = 8, g(ε1) = 2, (ε2 – ε1)/kt = -3ε1/4kt, since ε2 = ε1/4;
n (ε2)/n(ε1) =1/1000 = (8/2)e3ε1/4kT ; solving fot T: T = 1/k(-3/4ε1)/ln4000) = 1.43E4 K
Molecular Energy Distribution in an Ideal Gas
We will find the vdistribution of energies among the molecules of an ideal gas. In translational motion, energy is not quantized. Also,

number of molecules is quite large. We consider continuous distribution of energies. Let n(ε) dε be the number of molecules whose

energy lies between ε and ε+dε
n(ε) d ε = [g(ε)dε] [f(ε)] = Ag(ε)e-ε/kTdε (8)
where g(ε0dε = # of states that have energies between ε and ε+dε
Let us find it.
A molecule of energy ε has a momentum p = √(2mε) (since ε= p2/2m) = √(px2 + py2 + pz2)
Where each set of momentum components px , py, pz specifies a different state of motion. Let us imagine a space with momentum

coordinates px , py, pz. The number of states g(p) dp with momenta whose coordinates lie between p and p + dp is proportional to the

volume of a spherical shell in momentum space p in radius and dp thick. The volume of a sphere of radius p is 4πr2. If we multiply it

by the thickness dp, we get the volume. That is the volume is 4π p2dp.
So, we can write down
g(p) dp = B p2dp (9)
where B is a constant and g(p) is not the same as g(ε)
Since each momentum magnitude corresponds to a single energy ε, the number of energy states g(ε) dε between ε and ε+dε is the same as

the number of momentum states g(p)dp between p and p + dp, and so g(ε) dε = Bp2dp

(10)
From p2= 2mε, 2pdp =2 mdε => dp = mdε/p = mdε/(√(2mε)
Now Eqn (10) can be written as
Number of energy states g(ε) dε = B . 2mε. mdε/(√(2mε) = 2m3/2B √ε dε CHECK THIS (11)
The number of molecules with energies between ε and ε + dε is
n(ε) dε = C√ε e-ε/kT dε (12)
where C = = 2m3/2AB
Normalization
The constant C can be found in this way called normalization:
∞ ∞ ∞
N = ∫n(ε) dε = C ∫√εe-ε/kT dε = (C/2)(√π (kT)3/2 ) DO IT Use integral table ∫√x( e-ax)dx = (1/2a)√π/a
0 0 0
C = ?
n( ε) dε = ?

AVERAGE MOLECULAR ENERGY

∞
E = ∫εn(ε) dε Substitute for n(ε) dε and integrate to get 3/2 (NkT)
0
E/N = ? = average energy of an ideal gas molecule
Q. 4. Show that
∞
vavrg = 4π(m/2πkT)3/2 ∫v3 exp(-mv2/2kT) dv = √(8kT/πm)
0
vrms = √(3kT/m)
Q.5. Find the speed of oxygen at zero degree Celsius
Q. 6. Diferentiate the following expression wrt v, set it equal to 0, and then solve for v which will be vp.
n(v)= 4πN(m/2πkT)3/2 ∫v2 exp(-mv2/2kT)
Q. 6. Draw a bell-shaped graph to locate the approximate positions of vrms, vavrg, and the most probable speed vp = √2kT/m
Fermi-Dirac Distribution
The probability of a state being occupied by nFD phalf-spin particles of energy ε and chemical potential μ is given by
nFD = 1/[e(ε-μ)/kT + 1]
Q. 1. What is nFD = ? if ε>> μ
Q. 2. What is nFD = ? if ε < < μ
Q. 3. For a system of Fermi-Dirac particles at room temperature, compute the probability of a single-particle state being occupied if

its energy is
(a) ε – μ = -1 eV Ans. ≈ 1
(b) ε – μ = 0.01 eV Ans. 0.59
© ε – μ = 0 Ans. 0.50
(d) ε – μ = 0.01 eV Ans. 0.41
(e) ε – μ = + 1 eV Ans. 2E-17
Bose-Einstein Statistics
The probability of a state being occupied by nFD integral-spin particles of energy ε and chemical potential μ is given by
nBE = 1/[e(ε-μ)/kT -1]
Q. 4. What is nFD = ? if ε>> μ
Q. 5. What is nFD = ? if ε < < μ
The average occupancy of a state is given by n = 1/(ex -1) = e-x/(1 – e-x) where x = (ε – μ)/kT
The probability of a state being occupied by exactly n particles is
P(n) = (e-x) n (1 – e-x)
Thus we can compute everything we need from the quantity e-x = e-(ε – μ)/kT, where in this case kT = 0.026 eV.
Q. 6. For a system of Bose-Einstein particles at room temperature, compute the average occupancy of a single particle state and the

probability of the state containing 0, 1, 2, or 3 bosons, if the energy of the state is
(a) ε – μ = 0.001 eV Ans. n = 25.5; P(0) = 0.38, P(1) = 0.036, P(2) = 0.035, P(3) = 0.054
(b) ε – μ = 0.01 eV Ans. n = 2.13, P(0) = 0.319; P(1) = 0.217; P(2) = 0.148; P(3) = 0.101

(d) ε – μ = 1.0 eV Ans. n = 2E-17, P(0) ≈ 1; P(1) =2E-17; P(2) =4E-34; P(3) = 8E-51
Where the sum runs over all possible states and possible values of N.
For two types of particles, Gibbs factor is e-[E(s) –μANA(s) – μBNB(s)]/kT (5)
Example: Carbon monoxide poisoning
Hemoglobin molecules carry oxygen in the blood. We will use Gibbs factors in an adsorption site on a hemoglobin molecule. A single

hemoglobin molecule has four adsorption sites, each containing an Fe2+ ion surrounded by various other atoms. Each site can carry one

O2 molecule. Let the system be just one of the four sites. We assume that this site is completely independent of other three sites. If

oxygen is the only molecule that can occupy the site, the system has two possible states: unoccupied and occupied. Let us take the

energies of these two states to be 0 and ε where ε = – 0.7 eV.

Thermodynamics
Final Exam

1. An engine takes in 36 J of energy from a hot reservoir and performs 25.0 J of work in each cycle. The efficiency of the

engine is ——————————————
2. The exhaust temperature of a Carnot engine is 300oC. It has an efficiency of 30%. The intake temperature of the engine is

——————————–

3. An ice tray contains 500 g of liquid water at 0oC. The change in entropy of the water as it freezes slowly and completely at 0oC is

———————————

4. Linear thermal expansion coefficient α= (∆L/L)/∆T. Find α for steel if a 1-km steel bridge undergoes a variation in length of 77

cm between a cold winter night 0oC and a hot summer day 90oC. ———————————-
5. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in n=2 energy level? = > 1/1000 =

(8/2)e3ε1/4kT ; ε1 = -13.6 eV; k = 8.617E-5 eV/K; T =?
6. Given that Z = ∑ e-βE(s)
s
what is Z =? Is E(s) =0, 1, 2, and 3. ———————————————-

7. In # 7 what is –(∂Z/∂β)/Z =

8. If you toss two dice, what is the total number of ways in which you can obtain (a) a 12 and (b) a 7 ———-

9. The Fermi-Dirac statistics is governed by nFD = 1/[e(ε-μ)/kT + 1]. For a system of Fermi-Dirac particles at room temperature,

compute the probability of a single-particle state being occupied if its energy is ε – μ = 0. —————-

10. The Bose-Einstein is given by nBE = 1/[e(ε-μ)/kT -1]. If ε – μ = 0.01 eV, n = ? ————–

eC = 1 – Th/Tc; dS= dQ/T; e = {│Qh│-│Qc│}/│Qh│, Lf = 333000 J/kg; Q = mLf ; kT = 0.026 eV
Midterm Test
Thermodynamics
Print your name
The main span of a bridge made of steel has a length of 1158 m. Assume the temperature swing between -50oC to 50o C. The linear

expansion coefficient for steel is 13E-6/oC.
i) Wghat is ΔT = ?
ii) What is ΔL = ?
iii) ΔL = αLΔT = ?” 1.5 m
If P1 = 24.9 kPa, V1=0.100 m3, and V2 = 0.900 m3, P2 = ?
2.77 kPa

A weightlifter raises a mass of 180.0 kg through a height h = 1.25 m. Think him to be a thermodynamic system. How much heat energy does

he have to give off if his internal energy decreases by 4000.0 J?

i) How much mechanical work (mgh) does he do ?

ii) Q = ΔE + W = ? -1790 J = -0.428 kcal

Suppose this roomful air is at 22.0oC. What is the average kinetic energy of the molecules 9and atoms) in the air?
KE avrg = 1.5kT , k = 1.38E-23J/K, T = ? K Ans. 6.11E-21 J
What is the energy in electron volt, 1eV = 1.602E-19 J

Midterm Test
Thermodynamics
Print your name
The main span of a bridge made of steel has a length of 1158 m. Assume the temperature swing between -50oC to 50o C. The linear

expansion coefficient for steel is 13E-6/oC.
iv) Wghat is ΔT = ?
v) What is ΔL = ?
vi) ΔL = αLΔT = ?”
vii)

If P1 = 24.9 kPa, V1=0.100 m3, and V2 = 0.900 m3, P2 = ?

A weightlifter raises a mass of 180.0 kg through a height h = 1.25 m. Think him to be a thermodynamic system. How much heat energy does

he have to give off if his internal energy decreases by 4000.0 J?

iii) How much mechanical work (mgh) does he do ?
iv) Q = ΔE + W = ?
v)

Suppose this roomful air is at 22.0oC. What is the average kinetic energy of the molecules 9and atoms) in the air?
KE avrg = 1.5kT , k = 1.38E-23J/K,
What is the energy in electron volt? 1eV = 1.602E-19 J
Thermodynamics: Lecture 1
Diathermal wall.Very thin wall (could be metallic such as copper) that prevents particles passing through but allows some energy to

pass
Thermal contact. In thermal contact, two objects are separated by a wall that is a good conductor of heat.
Thermal equilibrium. Two objects are in thermal equilibrium if they do not change temperature when separated by a diathermal wall.
Temperature: A measure of hotness or coldness of an object.
Temperature is that property of an object that determines whether it is in thermal equilibrium with other objects
Two objects in thermal equilibrium with a third object will be in thermal equilibrium with each other.

Thermometry.
Empirical temperature scale.A reproducible experimental definition of a way to measure temperature.
Thermometric properties.Property of a substance that changes with temperature. All other properties must be held as constants.
System. A small part of the universe
Surroundings. Remaining portion of the universe that can directly interact with the system is called the surroundings or the

environment.
HJeat.Trasnsfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings.
Internal energy. All the energy of a system that is associated with its microscopic components – atoms and molecules – when viewed from

a reference frame at rest with respect to the center of mass of the system.
State variables. The state of a system can be described by specifying pressure, volume, temperature, and internal energy. The

macroscopic state of an isolated system can be specified only if the system is in thermal equilibrium internally. In the case of a gas

in a container, internal thermal equilibrium requires that every part of the gas be at the same pressure and temperature.
Transfer variables. While state variables are characteristic of a system in thermal equilibrium, transfer variables are characteristics

of a process in which energy is transferred between a system and its surroundings or the environment. Transfer variables are zero

unless a process occurs in which energy is transferred across the boundary of the system. Because a transfer of energy across the

boundary represents a change in the system, transfer variables are not associated with a given state of the system, but with a change

in the state of the system. Heat is a transfer variable. For a given set of conditions of a system, there is no definite value for the

heat. We can only assign a value of the heat if energy crosses the boundary by heat, resulting in a change in the system.
Work is another transfer variable in thermodynamics. Work = force x displacement
Imagine air in a syringe. If the cross-sectional area of the piston is A, and a push by a pressure P on the piston moves it through dx,

the volume it moves through is
dV = A.dx and the force on the piston is force = pressure x area of cross section => F = PA
Work done then W = force, PA .distance, dx =>PAdx = PdV
In the syringe, if the gas is compressed, work done is negative. If the gas expands, work done is positive
Q. The piston of a syringe has a diameter of 5 cm. A pressure of 10 N/sq m is applied to the piton to squeeze the gas from 10 cm to 4

cm. Find out the work done.
Hints.Pressure, P = ?, area A = ?, force F = P.A = ? ; dx = distance moved through = ?; volume change, dV = ?; work done = PdV = ?
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