Lecture 9 Overview (Ch. 1-3)Format of the first midterm: four problems with multiple questions.

• The Ideal Gas Law, calculation of W, Q and dS for various ideal gas processes.

• Einstein solid and two-state paramagnet, multiplicity and entropy, the stat. phys. definition of T, how to get from the multiplicity to the equation of state.

The test is an open textbook exam (but no open HW solutions!). I recommend to list all essential equations – you won’t have time to read the textbook!

Problem 1One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. T1 and T2 are given.

(a) (10) Calculate the work done by the gas. Is it positive or negative?

(b) (20) Using two methods (Sackur-Tetrode eq. and dQ/T), calculate the entropy change for each stage and for the whole cycle, Stotal. Did you get the expected result for Stotal? Explain.

(c) (5) What is the heat capacity (in units R) for each stage?

T

V

V1

V2

T1 T2

1

23

1 – 2 V T P = const (isobaric process) 0121212 TTRVVPW

2 – 3 V = const (isochoric process) 023 W

3 – 1 T = const (isothermal process) 0lnln2

11

2

11131

1

2

1

2

T

TTR

V

VTR

V

dVTRPdVW

V

V

V

V

0ln1ln1

2

1

21

2

11123112

T

T

T

TTR

T

TTRTTRWWWtotal

(a)

Problem 1 (cont.)

T

V

V1

V2

T1 T2

1

23

1 – 2 V T P = const (isobaric process)

mNfkN

UR

N

VRNVUS B ,lnln

2

3ln,,

i

f

i

f

i

f

i

f

T

T

V

VR

T

TR

V

VRS ln

2

3lnln

2

3ln

1

212 ln

2

5

T

TRS

2 – 3 V = const (isochoric process)1

2

2

123 ln

2

3ln

2

3

T

TR

T

TRS

3 – 1 T = const (isothermal process)1

2

2

131 lnln

T

TR

V

VRS

0ln2

3lnln

2

5

1

2

1

2

1

2 T

TR

T

TR

T

TRScycle

as it should be for a quasistatic cyclic process(quasistatic – reversible),

because S is a state function.

(b) Sackur-Tetrode equation:

Problem 1 (cont.)

T

V

V1

V2

T1 T2

1

23

1 – 2 V T P = const (isobaric process)

T

QSd

2 – 3 V = const (isochoric process)

3 – 1 T = const (isothermal process)

1

2

2

1

1

1

131 lnln

10

1

2

1

2T

TR

V

VR

V

dV

T

RTPdV

TSWQUd

V

V

V

V

ON

0ln2

3lnln

2

5

1

2

1

2

1

2 T

TR

T

TR

T

TRScycle

(b)

1

2 1

223 ln

2

3T

T

VV T

TR

T

dTCSdTCQQUd

2

1 1

212 ln

2

5T

T

PP T

TR

T

dTCSdTCQ

- for quasi-static processes

Problem 1 (cont)

(c) TdCQ Let’s express both Q and dT in terms of dV :

T

V

V1

V2

T1 T2

1

23

1 – 2 V T P = const (isobaric process)

RRRRCCC VP 2

5

2

3

2 – 3 V = const (isochoric process)

RCC V 2

3

3 – 1 T = const (isothermal process), dT = 0 while Q 0 C

At home: recall how these results would be modified for diatomic and polyatomic gases.

Problem 2 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. Process 3-1 is adiabatic; P1 , V1 , and V2 are given.

(a) (10) For each stage and for the whole cycle, express the work W done on the gas in terms of P1, V1, and V2. Comment on the sign of W.

(b) (5) What is the heat capacity (in units R) for each stage?

(c) (15) Calculate Q transferred to the gas in the cycle; the same for the reverse cycle; what would be the result if Q were an exact differential?

(d) (15) Using the Sackur-Tetrode equation, calculate the entropy change for each stage and for the whole cycle, Stotal. Did you get the expected result for Stotal? Explain.

1 – 2 012112 VVPW

2 – 3 023 W

P 1 2

3

V V1 V2

P1

P = const (isobaric process)

V = const (isochoric process)

(a)

3 – 1

012

3

3

511

1

1)(

3/2

2

1111

21

1

1111

31

1

2

1

2

V

VVP

VVVPdV

V

VPdVVPW

V

V

V

V

adiabatic process

Problem 2 (cont.)

3 – 1 adiabatic process 031 Q

(c) 1 – 2

2 – 3

P = const (isobaric process)

V = const (isochoric process)

02

5

2

5121

11221212

VVP

R

VP

R

VPRTTCQ P

012

3

2

3

2

12123111322323

V

VVPVPVPPPVTTCQ V

1

2

3

2

5

2

1211212312

V

VVPVVPQQQ

For the reverse cycle: QQreverse

If Q were an exact differential, for a cycle Q should be zero.

P 1 2

3

V V1 V2

P1

Problem 2 (cont.)

1 – 2 V T P = const (isobaric process)

NfkURVRNVUS B lnln2

3ln,,

i

f

i

f

i

f

i

f

T

T

V

VR

T

TR

V

VRS ln

2

3lnln

2

3ln

1

212 ln

2

5

V

VRS

2 – 3 V = const (isochoric process)

2

1

2

1

2

113

1

3

2

3

2

323 ln

2

5ln

2

3ln

2

3

V

VR

V

VR

V

VPP

P

P

T

T

T

TRS

3 – 1 031 S

0ln2

5ln

2

5

1

2

1

2 V

VR

V

VRScycle

as it should be for a quasistatic cyclic process(quasistatic – reversible),

because S is a state function.

Sackur-Tetrode equation:P 1 2

3

V V1 V2

P1

Q = 0 (quasistatic adiabatic = isentropic process)

(d)

Problem 3 Calculate the heat capacity of one mole of an ideal monatomic gas C(V) in the quasistatic process shown in the Figure. P0 and V0 are given.

00

0

00 1

V

VPV

V

PPP

RTPV RT

V

VVP

00 1

dTV

V

R

dVP

V

dV

R

VP

V

V

R

dVP

0

0

0

0

0

0 211

dVVPRdTWdUQ 2

3

30

20 we need to find the equation of

this process V=V(T)

40

0

0

21

425

)(

VVVV

R

VC

Td

QC

P

V

P0

V00

10 Start with the definition:

dT

dVVPR

Td

QVC )(

2

3)(

0

0 21

1

V

VP

R

dT

dV

00

21/12

3)(

2

3)(

V

V

V

VRR

dT

dVVPRVC

Problem 3 (cont.)

0

0

21

425

)(

VVVV

R

VC

P

V

P0

V00

T=constisotherm

S=const adiabat

V0/2 5V0/8

C/R

V/ V0

2.5

10 1/2 5/8

1.5

2/at )( 0VVVC the line touches an isotherm

8/5at 0)( 0VVVC the line touches an adiabat

50 Does it make sense?

Problem 4You are in possession of an Einstein solid with three oscillators and a two-state paramagnet with four spins. The magnetic field in the region of the paramagnet points “up” and is carefully tuned so that µB = , where µB is the energy of a spin pointing “down”, -µB is the energy of a spin pointing “up”, and is the energy level separation of the oscillators. At the beginning of the experiment the energy in the Einstein solid US is 4 and the energy in the paramagnet UP is -4 .

(a) (4) Using a schematic drawing of the Einstein solid, give an example of a microstate which corresponds to the macrostate US = 4 .(b) (4) Using a schematic drawing of the paramagnet, give an example of a microstate which corresponds to the macrostate UP = -4 .(c) (8) Considering that the “system” comprises the solid and the paramagnet, calculate the multiplicity of the system assuming that the solid and paramagnet cannot exchange energy.(d) (14) Now let the solid and paramagnet exchange energy until they come to thermal equilibrium. Note that because this system is small, there will be large fluctuations around thermal equilibrium, but let’s assume that the system is not fluctuating at the moment.What is the value of US now? Draw an example of a microstate in which you might find the solid.What is the value of UP now? Draw an example of a microstate in which you might find the paramagnet.

Problem 4 (cont.)

2 E1= - BB

E2 = + BB

Einstein solidTwo-state paramagnet

(a) US = 4

(b) UP = -4 E2 = + BB

E1= - BB

(c) 151

!2!4

!24

PSSP

Most of the confusion came from the fact that we usually measure the energy of an oscillator in the Einstein solid from its ground state (which is 1/2 above the bottom of the potential well), whereas for the two-state paramagnet we’ve chosen the zero energy in the middle of the energy gap between “spin-up” and “spin-down” levels. The avoid confusion, consider the number of energy quanta available for the system.

Problem 4 (cont.)

2 E1= - BB

E2 = + BB

Example of one of the equilibrium microstates:

(d)

24!1!3

!4

!2!2

!4 PSSP

constU SP 4In equilibrium, the multiplicity is maximum. The two-state paramagnet can absorb only multiples of 2. Two options: 2 is transferred from S to P, and 4 is transferred from S to P.

2 transfer

S – N = 3, q = 2, P – N = 4, N = 1

6!2!2

!41 PSSP

4 transfer

S – N = 3, q = 0, P – N = 4, N = 2

Thus, the equilibrium situation corresponds to the transfer of 2 from the Einstein solid to the two-state paramagnet

Note that the equilibrium condition UA/NA= UB/NB holds if both systems have only “quadratic” degrees of freedom.

Problem 5

Consider a system whose multiplicity is described by the equation:

(a) (10) Find the system’s entropy and temperature as functions of U. Are these results in agreement with the equipartition theorem? Does the expression for the entropy makes sense when T 0?

(b) (5) Find the heat capacity of the system at fixed volume.

(c) (15) Assume that the system is divided into two sub-systems, A and B; sub-system A holds energy UA and volume VA, while the sub-system B holds UB=U-UA and VB=V-VA. Show that for an equilibrium macropartition, the energy per molecule is the same for both sub-systems.

UkfN

VNkNfkkS BBBB ln2

lnlnln U

fNk

U

S

T BVN 2

1

,

BkfN

UT

2 - in agreement with the equipartition theoremTk

fNU B2

When T 0, U 0, and S - - doesn’t make sense. This means that the expression for holds in the “classical” limit of high temperatures, it should be modified at low T.

2/,, NfNUVNfNVU where U is the internal energy, V is the volume, N is the number of particles in the system, Nf is the total number of degrees of freedom, f(N) is some function of N.

(a)

Problem 5 (cont.)

2/3,, NNUVNfNVU

2/32/3,, BABA NA

NA

NA

NA UUUVVVNVU

02

3

2

3 12/32/32/312/3 BABA N

ANA

ANA

NA

A

A

UUUN

UUUN

U

B

B

A

AA

BA

A

N

U

N

UU

NUU

N 0

2

3

2

3

(b) BBVNVNVN

V Nkf

T

fNkT

T

STPdVTdSdU

T

U

T

QC

2

1

2,,,

(c)

Problem 6(10) The ESR (electron spin resonance) set-up can detect the minimum difference in

the number of “spin-up” and “spin-down” electrons in a two-state paramagnet N-N =1010. The paramagnetic sample is placed at 300K in an external magnetic field B = 1T. The component of the electron’s magnetic moment along B is B = 9.3x10-24 J/T. Find the minimum total number of electrons in the sample that is required to make this detection possible.

BEETk

EE

N

NB

B

2exp 1212

9955.03001038.1

1103.92exp

2exp

23

24

Tk

B

B

B - the high-T limit

10102

exp1

Tk

BNNN

B

B

Tk

B

Tk

BTk

BNNN

B

B

B

BB

B

2exp1

2exp1

102exp1

10

1210

104,40045.0

102

NN