PHYSICS 140A : ASSIGNMENT #2 SOLUTIONS
Problem 1. Solution:
(a) As we know VB and TB , we can directly use:
"RTB
= 1.04 ! 106 Pa
VB
PB =
(b) We can compute the area under the curve to get the work in this process.
B
W = " PdV = " PdV ! PB (VB ! VC )
A
VB
=
V
#RT A
dV ! #RTB (1 ! C )
V
VB
VA
"
' .V +
V $
= /RTB &ln,, B )) ( 1 + C # = 2.64 ! 10 3 J
VB "
% - VA *
C
C
C
dQ
dE + PdV
(c) "S = ! dS = !
=!
T
T
B
B
B
where E =
PV
, take it into (1), we get:
" !1
C
C
PdV + VdP P
$
dV
%R dP
+ dV = !
%R
+
($ " 1)T
T
$ "1
V
$ "1 P
B
B
#S = !
=
'V
)
(R ln%% C
) !1
& VB
Problem 2. Solution:
$
"" = !40.3 J / K
#
(1)
First of all, let’s consider the work during this period, i.e. S ABCD
B
C
D
A
W = S ABCD = ! PdV + ! PdV + ! PdV + ! PdV
A
B
C
D
B
D
PAV A#
PAVD#
= ! # dV + ! # dV + PA (V A " VD )
A V
C V
=
!
1
PA (V A " VD ) "
PAVB1"! (V A! " VD! )
! "1
! "1
Secondly, in this period, the ideal gas absorb heat during D->A, therefore:
A
A
A
' PV $
)
"" + PA (V A ! VD ) =
Q = ( dQ = ( dE + PdV = ( d %%
PA (V A ! VD )
) !1#
) !1
D
D
D &
Finally, we get:
W
1 r 1!" ( s " ! 1)
#=
= 1!
, where s and r have been defined in the problem.
Q
"
s !1
'V
Otto cycle has the efficiency of ) = 1 ! %% min
& Vmax
$
""
#
( !1
If r=18, s=2, ! Diesel = 81% < ! Otta = 85% , however an optimized Otto Cycle has
the parameter of r=8, ! max ofOtta = 75% , which is less than the efficiency in Diesel
cycle.
Problem 3. Solution:
(a) The unit of S is J/K, V is m 3 and E is J, therefore, a should has the unit of:
m6 K 5 / J 4 .
aS 5
5aS 4
2aS 5
(b) E = 2 2 " dE = 2 2 dS ! 3 2 dV
V N
V N
V N
(1)
On the other hand, we know:
dE = TdS ! PdV
Compare (1) and (2), we can easily get the relation:
T=
(2)
5aS 4
2aS 5
P
=
,
V 2N 2
V 3N 2
Then we can get the equation of state:
&V #
P$ !
%N"
1/ 2
5 5 / 4 a 1 / 4 = 2T 5 / 4
(c) As a isotherm process, the equation of state should be:
PV 1 / 2 = C = const.
Therefore, the work in this isotherm process is:
C
W = ! PdV = ! 1 / 2 dV = 2C (V f1 / 2 " Vi1 / 2 )
V
(d) Always be aware of the difference between isotherm process and adiabatic
process!
First, let’s rewrite the equation of state as:
16
P 4V 2 a =
N 2T 5
3125
We can then get the differential relation:
16 2 4
4 P 3V 2 adP + 2 P 4VadV =
N T dT
(1)
625
As adiabatic process, dE = ! PdV
(2)
aS 5
(VN )1 / 2 5 / 4
Where E = 2 2 = 5 / 4 1 / 4 T , and P is determined by the equation of state
V N
5 a
as above.
The (2) can give us the relation of dV and dT, dT = !
Invoking this relation to (1), we can get:
(
1 & 'V #
1
=
$
!
V % 'P " dQ =0 3P
Problem 4. Solution:
(a) Similar as the problem 3, dE = TdS ! PdV
!"
( )E %
T =&
# = 0
kB
' )S $V
!
!S
(N%
)
& # exp(
Nk B
'V $
2T
dV
V
( )E %
(N%
P = *&
# = !" 0 & #
' )V $ S
'V $
1+!
exp(
!S
)
Nk B
Therefore, PV = NkT = !RT . ! This system obeys the ideal gas law.
(b) As for adiabatic process, dE = dQ ! PdV = ! PdV = !
In addiation, we can rewrite E =
Therefore, dE =
Nk B T
dV
V
Nk B T
by using the relation in (a).
!
Nk B
dT , invoking this expression into (1):
!
Nk B
Nk T
dT = ! B dV ! TV " = const.
"
V
(c) From the derivation above, we can get cV =
1 ( )E %
R
&
# =
" ' )T $V !
1+!
' (V $
c P = cV + p%
R
" =
!
& (T # p
Problem 5. Solution:
(a)
The volume relation should be the same as the normal Carot Cycle. i.e.
dQ
! T = 0 , where only A->B and C->D contribute dQ. Thus,
(1)
(b) Let’s analysis the work in details.
AA’: Q AA' = W AA' = 0
&V #
A’B: Q A'B = W A'B = nRT2 ln$$ B !!
% V A' "
BC: Q BC = 0 ; W BC =
nR(T2 ! T1 )
" !1
& VC
% VD
CD: QCD = WCD = ' nRT1 ln$$
DA: Q DA = 0 ; W DA = !
#
&V #
!! = 'nRT1 ln$$ B !!
"
% VA "
nR(T2 ! T1 )
" !1
The total work in this cycle is:
&V #
&V #
W = W AA' + W A'B + WBC + WCD + WDA = nRT2 ln$$ B !! ' nRT1 ln$$ B !!
% V A' "
% VA "
By using the defined parameter x and r, we can get:
W = nR(T2 ! T1 ) ln r ! nRT2 ln (1 + (r ! 1) x )
(c)
Let W = nR (T2 ! T1 ) ln r ! nRT2 ln (1 + ( r ! 1) x ) = 0 , we can solve this problem, and the
answer is x =
r 1"T1 / T2 " 1
! 0.26
r "1
&
#
&
#
VB
r
!! = nRT2 ln$$
!! , therefore
(d) Q A'B = nRT2 ln$$
V
+
x
(
V
'
V
)
1
+
x
(
r
'
1
)
%
"
B
A "
% A
"=
T
T
W
ln r
= 1! 1
< 1! 1
Q A' B
T2 ln r ! ln(1 + (r ! 1) x)
T2