19.64 a cylinder with a piston contains 0.150 mol 105 pa ...cohn/classes/206/chapter20.pdf · 19.64...

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

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

QQ

−=−= 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

QQ

′−=′ 1ε

H

C

QQ

−=1ε

HH

H

C

H

C

QQQQ

QQ

>′⇒

<′

⇒>′ 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

QQ

−=−= 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