Physics 157 Written Homework 6
In this homework set, you’ll get more practice analyzing thermodynamic processes, with a
focus on calculating the efficiency of heat engines. In addition to the goals from last week, we
want you to be able:
 To calculate the net work done in a cyclic process
 To identify which in which parts of a cyclic thermodynamic process heat is flowing in to the
system, and in which parts heat is flowing out.
 To calculate the efficiency of a heat engine, given a description of the thermodynamics
processes associated with it, or a depiction of the cycle on a PV-diagram
A summary of the formulae you may need:
Ideal gas law:
P V = n R T or P1 V1 / T1 = P2 V2 / T2 if n is constant.
Work:
W = P ΔV (constant pressure) or area under curve on P-V diagram. Positive for
expansion, negative for compression.
Internal energy:
ΔU = n Cv ΔT
Heat: Q = ΔU + W (First Law of Thermodynamics)
__________________________________________________________________________
Constant volume: T/P is constant, W = 0 so Q = ΔU
Constant pressure: T/V is constant, W = P ΔV , Q = n Cp ΔT where Cp = Cv + R
Constant temperature: PV is constant, ΔU = 0, so Q = W. W = n R T ln(Vf/Vi),
Adiabatic: Q = 0, P Vγ = constant, T Vγ-1 = constant where γ = Cp/CV, W = -ΔU
_________________________________________________________________________
Efficiency of an engine: e = Wnet / Qin where Qin is the sum of all positive contributions to
Q
__________________________________________________________________________
Cr
=
ER
/
Q b
Ts
=
Iso
choriz
/
T3 3
I
Goo#
Cr
8
Fir
=
=
To
the
#2
Tz
Ti
I
#
#
#2
#3
Cakutate
W
e
T4
4
2
-
Isobaric
Ts
Ti
QH
T2
&
3/SV
=
-
W
#1 T ,
W
for
each
ER
R
=
Cr
find
E
e
we
positive Qin
only
need
to
calculate
calculate
To
PV=
RT
n
T 300K
=
P
n !
=
=
120kPA
=
V
10L
=
1
Q
:
Q
=
n(p
P, V ,
=
Qin
=
for
(300
n RT
-
T
,
)
P,
n
,
120 kPa
=
=
T
=
0
=
.
,
.
.
99 K
.
.
for
=
30L
R= 8 314
4811
899
V
n(p(300-899 991k
Q =-84005
,
process
A
,
=
T
280314
=
for all
mole
nCpst
T
Q
.
processes
Calculating
#
0 48111
is
n
0 48111 mol
.
=
process
#1
Q2
V
=
=
n(vIT
Ps 200kPa V
Cr= ER
is
=
Y
=
Q2
&3
n(psT
=
2
=
=
=
=
300k)
-
10L
! 8314
=
n(r(508-300)
=
Ty
n(r(T
=
=
500K
20005
n(p(Ty 500)
-
1500k
=
Py: 200kPa
V
=
30L
Q3
Qin
W
2
n(p(1500
=
=
-
500)
=
14000+
16000T
200kPa 20
.
=
=
W
-
①
in
=
↳5
16065
-(120-20)
=
I
=
=
10 %
16005
③
T
,
T
for
#3
=
W
nRT
=
-
↓
900k
Q
(n(E)
Us
V2
30L
=
10L
#2
=
68085 39555
:
+
:
PAU
W,
=
=
120
.
/10-0)
5
0 156
.
1
Wy
0
Whet= 1955
=
99535
Wat
+
W2
3955
=
=
10005
=
-
work
=
,
ner(Tz x)
=
means
Total
W3
0
=
xU(W
=
That
Qin
2
=
.
1V
W
300K
=
3935 0J
=
for
Q
same
same
T
900K
=
also the
is
&
the
is
=
=
-24005
15 6 %
.
=
,
T3
2
For
900K
T
=
,
Tz
=
and
=
& Te
,
#2 Q
,
respectively
=
sU
n(rAT
.
R
-
(Ts
calculate is
=
=
,
=
same
300K
V 30L 7 ,
Vz
the
are
300K
=n
To
T
900k
For
Q
:
-
:
Tz)
T V,
=
,
Ts
VsU-
900K
10L
=T
3
T3
=
1396 66k
.
SO
Q for #2
=
nER (1396
.
66
-
300)
=
/0966
.
6)
Whet
W3
W
=
+We
,
Wi
U
=
n(-1T
=
+
=
15 R(900
:
.
-
-1396
66)
.
4966 61017
=
.
W
,
Whet
e
=
P1V
=
=
-
2400
3661
120(10-30)
=
+
4966 61
=
=
-
2400
2566 6
.
.
=0 234= 23 4
.
.
10946 4
%
,
Process
To
have
&
calculate
to
how
H
Total
=
much
how
calculate
10 Cycles/min
Q
is
=
600
most
gas
much
.
generated
is
Q
used ,
is
cycles/hr
10966 61 5/cycle
heat
efficient
is
QH
·
we
generated
# of
cycles
10966 61x680
.
Vgas
69799665
63799665x
33x1065
=
-
=
=
1 879N0
I
.
88x18'L
Vgas=
1
,
L