Physics 425, Spring 2007

Final Exam May 8 4:00-5:50, Tuesday

1. A tank has a volume of 0.1 m3 and is filled with a monatomic ideal gas at a pressure

of 6 x 106 Pa and temperature 200 K. A second tank has a volume of 0.2 m3 and is filled with the same gas at a pressure of 3 x 106 Pa and temperature 300 K. A rigid piston separating the two tanks allows thermal interaction but no particle transfer between the two tanks. Assuming that the heat transfer between the two tanks is quasi-static, the walls of the tanks are adiabatic and rigid, and the gas has a constant molar specific heat cV = 3/2 R: (a) Find the final temperature of the system in final equilibrium [Hint: the internal

energy of the combined system is constant.] (14 points) (b) Find the entropy change of the system. [Hint: the change in the total entropy is

the sum of entropy change in each sub system.] (10 points)

!

Key to solutions :

(a) In final equilibrium, the gases have the same temperature T f .

The constant internal energy is E =3

2N1kT1 +

3

2N 2kT2 =

3

2(N1 + N 2)kT f

Using the ideal gas law : PV = NkT , there are :

N1T1k = P1V1, N 2T2k = P2V2, N1k =P1V1

T1,N 2k =

P2V2

T2

Therefore, T f =N1T1 + N 2T2

N1 + N 2=P1V1 + P2V2P1V1

T1+P2V2

T2

= 240 K

(b) Using the second law dS = (dE + PdV ) /T =CV dT

T+PdV

T

Here the volumes of the two tanks are fixed, so dV = 0,

thus dS =CV dT

T=ncV dT

T, yielding "S = ncV ln(T f /Ti)

Using PV = nRT , there is n1 =P1V1

RT1,n2 =

P2V2

RT2. Therefore,

"S = "S1 + "S2 =3

2

P1V1

T1ln(T f /T1) +

3

2

P2V2

T2ln(T f /T2) = 151 J/K

2. An engine works in the following idealized cycle: A-B: adiabatic compression B-C: isobaric expansion C-D: adiabatic expansion D-A: isobaric compression Assuming this cycle to be carried out quasi-statically for 1 mole monatomic ideal gas

with constant specific heats and the specific heat ratio : (a) Sketch the cycle in the P-V diagram (6 points) (b) Calculate heat Q and total work W during one cycle; express your answers in

terms of the gas pressure PA and PB, the gas volume VB and VC, and . (15 points) (c) Calculate the engine efficiency in terms of PA, PB, and . [Note: the engine

efficiency is the ratio of work to the amount of heat intake.] (6 points) Key to solutions:

!

(a) Note that A - B is the adiabatic compression with decreasing volume

and increasing pressure because of the relation PV " = const, where " > 0.

(b) The simplest way to solve this problem is to calculate heat Q first :

A - B : adiabatic process, hence QAB = 0

B - C : isobaric process, PC = PB , QBC = #EBC + PB#VBC

As #EBC = CV (TC $TB ) =CV

R(PCVC $ PBVB ) =

CV

RPB (VC $VB )

There is : QBC = (CV

R+1)PB (VC $VB ) =

"

" $1PB (VC $VB )

Since VC > VB , QBC > 0, the engine intakes heat.

C - D : same as A - B, QCD = 0

D - A : same as B- C : QDA ="

" $1PA (VA $VD )

Using the adiabatic relation for A - B and C - D processes, there is :

PAVA"

= PBVB", or VA =

PB

PA

%

& '

(

) *

1/"

VB , samely, VD =PC

PD

%

& '

(

) *

1/"

VC =PB

PA

%

& '

(

) *

1/"

VC

Therefore, the total heat :

Q =QBC +QDA ="

" $1PB (VC $VB ) $

"

" $1PA

PB

PA

%

& '

(

) *

1/"

(VC $VB ) ="

" $1PB $ PA

PB

PA

%

& '

(

) *

1/"+

,

- -

.

/

0 0 (VC $VB )

During one cycle, the net change in the internal energy #E = 0, thus the total work

W =Q $ #E =Q ="

" $1PB $ PA

PB

PA

%

& '

(

) *

1/"+

,

- -

.

/

0 0 (VC $VB )

(c) The engine efficiency is 1 =W

Qin=W

QBC=

"

" $1PB $ PA

PB

PA

%

& '

(

) *

1/"+

,

- -

.

/

0 0 (VC $VB )

"

" $1PB (VC $VB )

= 1$PB

PA

%

& '

(

) *

1

"$1

= 1$PA

PB

%

& '

(

) *

1$1

"

3. A certain liquid is made of identical molecules, each of them having only three values for the component of its magnetic moment along the direction of an external magnetic field: z = (1, 0, 1) 0 , and 0 is the Bohr magneton (0 = 0.93 x 10

-23 J/Tesla).

(a) Find the mean magnetic moment of one molecule

!

"

when the liquid is kept at a constant temperature T with the external magnetic field B. (10 points)

(b) Derive the approximate expressions for

!

"

at very high temperatures (kT >>0) and very low temperatures (kT > 1,e0B / kT >> 1,e"0B / kT % 0, therefore,

"

%0e

0B / kT

e0B / kT

= 0

(c) "

decreases monotonically with increasing temperature. At very high

temperatures, it shows a 1/T dependence, and at very low temperatures,

it reaches the maximum value 0.

(d) When B = 1 Tesla, 0B

kT= 0.002

4. Answer the following questions: (a) A gas consists of identical Fermions with the Fermi energy . When temperature

T = 0, find the occupation number for the state with energy . [Hint: distinguish

!

" > and

!

" < cases] (7 points) (b) A gas consists of N (N>>1) identical Bosons. Show that at very low temperatures

when all particles are in the lowest energy state = 0, the chemical potential varies with the temperature T according to:

!

(T " 0) # $1

NkT [Hint: for small x,

!

ln(1+ x) " x ] (7 points)

(c) At room temperature (T = 300 K) and pressure, 1 gram of liquid water (H2O) occupies a volume of 1 cm3. Water molecule has atomic weight of 18. Can we treat liquid water molecules using classical statistics? (5 points)

Key to solutions:

!

(a) ns

"

=1

e(#" ) / kT +1

, when T = 0 :

# > , (# " ) /kT = $, e(#" ) / kT = $, so ns

"

= 0

# < , (# " ) /kT = "$, e(#" ) / kT = 0, so ns

"

= 1

(b) ns

"

=1

e(#" ) / kT "1

When # = 0, ns

"

=1

e" / kT "1

= N

So e" / kT "1 =1

N,e" / kT =

1

N+1, " /kT = ln(

1

N+1)

As N >> 1, 1

N