19.64 a cylinder with a piston contains 0.150 mol 105 pa ...cohn/classes/206/chapter20.pdf · 19.64...
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19.64 A cylinder with a piston contains 0.150 molof nitrogen at 1.8 × 105 Pa and 300K. The nitrogenmay be treated as an ideal gas. The gas is firstcompressed isobarically to ½ its original volume.It then expands adiabatically back to its originalvolume, and finally it is heated isochorically to its original pressure.
a) Draw the p-V diagram,
b) Compute the temperatures at the beginning and endof the adiabatic expansion,
c) Compute the minimum pressure.
p
V
p0
V0
Isobaric compression to V0/2
V0/2
p
V
p0
V0V0/2
adiabatic expansion to V0
p
V
p0
V0V0/2
isochoric heating to p0
b) Compute the temperatures at the beginning and endof the adiabatic expansion
KTPap
n
3001080.1
150.0
0
50
=×=
= Given
p
V
p0
V0V0/2
Ti
Tf
inRTVp
nRTVp
=
=
20
0
000
Get Ti from ideal gas law:
KTTi 15020 ==
KTPap
n
3001080.1
150.0
0
50
=×=
= Given
p
V
p0
V0V0/2
Ti
Tf
b) Compute the temperatures at the beginning and endof the adiabatic expansion
Get Tf from:
10
10011
22, −
−−− =
= γ
γγγ VTVTVTVT fffii
KKTTf 7.1132
3002 4.1
0 === γ
c) Compute the minimum pressure.p
V
p0
V0V0/2
Ti
Tfpmin
PapKKPa
TT
pp f
4min
5
00min
1082.6
3007.113)1080.1(
×=
×==
000
0min
nRTVpnRTVp f
=
=
Thermal systems spontaneously change only in certain ways:
Spontaneous heat flow always occurs from a hotter body to a colder body
Ch. 20
2nd Law of Thermodynamics
(even though to do the reverse does not violate the 1st Law of Thermodynamics)
Reversible vs Irreversible Processes
Reversible Equilibrium (quasi-equilib.)Irreversible Nonequilibrium
Heat Engine: a device that converts heat to work
Simple engines involve a cyclic process of aworking substance (usually an ideal gas)
WQU =⇒=∆ 0Net heat flow into working substance equalsnet work done by engine
Two thermal reservoirs [one hot (H), one cold (C)]:exchange heat with working substanceat constant T (QH and QC)
Heat Engine: a device that converts heat to work
H
C
H
C
H QQ
QW
−=+== 11εEfficiency of engine:
20.39 A heat engine takes 0.350 mol of a diatomic idealgas around the cycle shown.
What is the efficiency?
HQW
=ε
TRnTnCQU V ∆
=∆==∆
25
Compute the W, Q for each process:
1 2: isochoric W=0 (dV=0)1st Law
JKmolKJmol
31018.2)300)(/314.8)(2/5)(350.0(
×=
=
2 3: adiabatic Q=0
)0(786)108)(/314.8)(2/5)(350.0(
>−=−⋅=
∆=−=∆
WJKKmolJmol
TnCWU V
3 1: isobaric
JKKmolJmol
TnRVpW
559)192)(/314.8)(350.0(
−=−⋅=
∆=∆=
RRCCTnCQ Vpp )2/7(, =+=∆=
JKKmolJmolQ
1956)192)(/314.8)(2/7)(350.0(
−=−⋅=
Net Work done:
JJJWW 2275597861332 =−=+ →→
Heat Input (QH):
JQ 321 1018.2 ×=→
HQW
=η
%4.10104.01018.2
2273
==×
=J
Jε
Engine Statement of the 2nd Law
It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending up in the same state in which it began.
Or equivalently…
Heat cannot be converted completely into work with no other change taking place.
Or equivalently…
There are no engines with 100% efficiency.
Internal Combustion Engines
Otto Cycle
fuel injected
Diesel Cyclefuel injected
allows forgreater r,efficiency
Stirling Cycle
V rV
Qbc
Qcd
Qda
Qab
Qin=Qbc+Qcd
Qout=Qab+Qda
heat enters, leaves during more than one process
in
out
in QQ
QW
−== 1ε
Brayton Cycle (jet engines)p
V
rp
p Taab
c dab: isobaric compressionbc: adiabatic compressioncd: isobaric expansionda: adiabatic expansion
What is the efficiency?
ab:
)(
)(
bap
abpC
TTnCTTnCQ−−=
−=
H
C
H QQ
QW
−== 1ε
cd: )( cdpH TTnCQ −=
express Tb, Tc, Td in terms of Ta, r
cd
ba
TTTT
−−
−=1ε
QC
QH
p
V
rp
p Taab
c d
γγγγad TpTrp −− = 11)(
γγ /)1( −=r
TT ad
dbb
bbb
nRTVrpnRTVp=
=)(
γ/1rTT a
b =
rTT d
b =
)/1/1()/11(1 1/2/)1(
/1
−− −−
−= γγγ
γ
rrTrT
a
a
cd
ba
TTTT
−−
−=1ε
γγ /)1(1 −−= r
p
QC
QH
V
rp
p Taab
c d
γγ /)1( −=r
TT bc
1/2 −= γrTT a
c
r0 2 4 6 8 10 12
e
0.0
0.1
0.2
0.3
0.4
0.5
0.6
γγ /)1(1 −−= r
QC
QH
V
rp
p Taab
c d
Refrigerators: heat engine in reverse
CH
CC
CH
CH
QQQ
WQ
K
WQQWQQ
−==
+=
−=−
Coefficient of performance
Real refrigerators
2nd Law of Thermodynamics
It’s impossible to convert heatcompletely to work (ε =1), with no other change taking place.
It’s impossible for any process to haveas its sole result the transfer of heat from a colder body to a hotter body.
Engine statement:
Refrigerator statement:
Equivalence of Engine and Refrigerator Statements of 2nd Law
Carnot Cycle: maximum efficiency
H
C
H
C
TT
−=−= 11ε
No engine can be more efficient than a Carnot engineoperating between the same two temperatures.
Equivalent statement of the 2nd Law
Proof: Suppose there is a hypothetical engine more efficient than Carnot (designate with ′)
Take Carnot refrigerator with hypothetical super-efficient engine supplying the work and QC’s the same:
H
C
′−=′ 1ε
H
C
−=1ε
HH
H
C
H
C
QQQQ
>′⇒
<′
⇒>′ εε
∆+=′ HH QQLet
CH QQW −′=′1st Law: CH QQ −∆+= ∆+= W
Violation of 2nd LawNo engine can be more efficient than a Carnot engineoperating between the same two temperatures.
Carnot Cycle: maximum efficiency
H
C
H
C
TT
−=−= 11ε
0=∑i i
i
TQ
In limit of infinitesimal T differencesbetween isotherms ∫∑ →
TdQ
TQ
i i
i
“closed cycle”
Entropy Statement of 2nd Law
In any thermodynamic process that proceeds from one equilibrium state to another, entropy of the system+environment (i.e. the “universe”) either remains unchanged or increases.
Equivalent to Engine statement: there are no perfect engines
0≥∆S