A summary of thermodynamic processes - Web Physics

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Copyright, 2004, John R. Newport, Ph.D.

Thermodynamic P
rocesses


Calorimetry


Change of phase?


Type of heat


Equation


Temperature





Calculation





Change?


No




specific heat


dQ = mCdT [1]

Yes

Yes




latent heat


dQ = mL


No


[1] Caution! Be careful of molar versus mass based specific he
at constants.


Ideal Gas Law




PV = nRT,




P => pressure in Pascals (N/m
2
)




V => volume in m
3




n => number of moles (dimensionless)




R => gas constant




T => temperature in Kelvin (not Celsius!)


Other Key Equations



dU = dQ


dW



(first law of
thermodynamics)


dQ = nC
V

T



(ideal gas, specific heat at constant volume)


dQ = nC
P

T



(ideal gas, specific heat at constant pressure)


dU = nC
V

T



(ideal gas, derivation attached)


C
P



C
V

= R



(statistical mechanics)


Internal Energy of an Ideal Gas


The internal energy depends only on the endpoints. Pick a constant volume and constant
pressure line segments to connect the endpoints. Using the first law:



U = nC
V
(T’


T
0
)

+
nC
P
(T
f



T’)


0


P
f
(V
f
-
V
0
) = nC
V
(T
f



T
0
), since


P
f
V
0

= nRT’
Copyright, 2004, John R. Newport, Ph.D.

Law
s

of Ther
modynamics for Ideal Gases


process


meaning


work (

W)


heat (

Q)

entropy (

S)


isobaric

constant pressure

P
0
(V
F



V
0
)


n
C
P
(T
F



T
0
)

n
C
P
ln(T
F
/T
0
)

isochoric

constant volume

0



n
C
V
(T
F



T
0
)

n
C
V
ln(T
F
/T
0
)

isothermal

constant temperature

(nRT
0
)ln(V
F
/V
0
)

[1]


(nR)ln(V
F
/V
0
)

adiabatic [2]

no heat exchange

(P
F
V
F
-

P
0
V
0
)

/(1
-


)

0


0






[3]


[1]

From the first law of thermodynamics, dU = dQ


dW; dU=0 for an isothermal
process, so dQ = dW, or

Q =

W


[2]

From the first law of thermodynamics, dU = dQ


dW; dQ
=0 for an adiabatic
process, so dU =
-
dW =>

nC
V
dT =
-
PdV;



from the ideal gas law, dT = d(PV/nR), so


n(C
V
/nR) d(PV) =
-
PdV


n(C
V
/nR) [PdV + VdP] =
-
PdV,


VdP =
-
PdV [1 + (C
V
/R)] / (C
V
/R);

since R = C
P

-

C
V
,



C
P

/ C
V



VdP =
-

PdV , or



P/P
0

= (V/V
0
)
-

,

or

PV


= P
0
V
0

.


[3]

dW = PdV, so

W =


(P
0
V
0

)

V
-

dV = (P
0
V
0

)

V
1
-

/(1
-


), V


[V
0
,V
F
]





W = (P
0
V
0

)

V
1
-

/(1
-


) = (P
F
V
F
-

P
0
V
0
)

/(1
-


)



Examples follow


(1) a simple example


(2) Carnot cycle


(3)
Otto cycle


(4) Diesel cycle


(5) Stirling cyc
le
Copyright, 2004, John R. Newport, Ph.D.

Example 1
:
A S
imple Example



Heat calculations:



Work calculations:



Q
12

= 8(C
P
/R)P
0
V
0




W
12

= 8P
0
V
0



Q
2
3

=
-
9(C
V
/R)P
0
V
0



W
23

=

W
41

= 0



Q
34

=
-
2(C
P
/R)P
0
V
0




W
34

=
-
2P
0
V
0



Q
41

=


3
(C
V
/R)P
0
V
0


Entropy calcu
lations:



Sums:



S
12

= nC
P
ln(3)




Q =

W = 6P
0
V
0



S
2
3

=
-
n
C
V
ln(
4
)




U =

Q
-


W = 0 (expected, closed cycle)



S
34

=
-
nC
P
ln(3)




S = 0


(reversible process)



S
41

=

nC
V
ln(
4
)


efficiency:


Q
H

=

Q
12

+

Q
41

= (8C
P

+
3
C
V
)P
0
V
0
/R


(sum of positive

heat results)

Q
C

= |

Q
23

+

Q
34
| = (
2
C
P

+
9
C
V
)P
0
V
0
/R


(sum of negative heat results)

e

= 1
-

(
2
C
P

+
9
C
V
)/

(
8
C
P

+
3
C
V
)

= 1
-

(
2


+
9
)/

(
8


+
3
);

for a monatomic gas,


= 5/3 and e = 0.24


Carnot efficiency:


T
C

= T
4

= P
0
V
0
/(nR)


T
H

= T
2

= 12P
0
V
0
/(nR)


e =

1
-

T
C
/T
H

= 0.92

(notice that the actual efficiency is much lower)
P

V

4P
0

P
0

3V
0

V
0

1

2

4

3

Copyright, 2004, John R. Newport, Ph.D.

Example
2
: Carnot Cycle


STATE

T


a


T
H

b


T
H

c


T
C

d


T
C


________________________________________________________________

STEP

TYPE



Q




W


U




S


a
-
>b

isothermal

nRT
H
ln(V
b
/V
a
)


Q

0



nRln(V
b
/V
a
)

b
-
>c

adiabatic

0




U

nC
V
(T
C

-

T
H
)


0

c
-
>d

isothermal

nRT
C
ln(V
d
/V
c
)



Q

0



nRln(V
d
/V
c
)

d
-
>a

adiabatic

0




U

nC
V

(T
H

-

T
C
)


0


______________________________________________________
__________


efficiency:



Qab = nRT
H
ln(V
b
/V
a
) > 0


Qcd =
-
nRT
C
ln(V
d
/V
c
) < 0


|Q
C
| / |

Q
H

| = (T
C
/T
H
)[

| ln(V
d
/V
c
)
/

ln(V
b
/V
a
) |
]


T
b
V
b

-
1

= T
c
V
c

-
1


|




| (adiabatic)

=> V
b
/V
a

= V
c
/V
d


=> |Q
C
| / |

Q
H

| = (T
C
/T
H
)

T
d
V
d

-
1

= T
a
V
a

-
1

|


e = 1
-

(T
C
/T
H
)


entropy:



S = 0, see efficiency calculation.
Reversible process.



Copyright, 2004, John R. Newport, Ph.D.

Example
3
:
Otto Cycle


STATE

P


V



T


a

Pa



V
a
= rV
b


T
a

b

P
b

= P
a
r



V
b



T
b
= T
a
r

-
1

c

P
c

= P
b
(T
c
/T
b
)


V
b



T
c
= T
d
r

-
1

d

P
d

= P
c
(1/r)



V
a
= rV
b


T
d


________________________________________________________________

STEP

TYPE



Q




W



U


S


a
-
>b

adiabatic

0


nC
V
(T
b



T
a
)


-

W

0

b
-
>c

isochoric

nC
V
(T
c



T
b
)

0




Q

nC
V
ln(Tc/Tb)

c
-
>d

adiabatic

0


nC
V
(T
d



T
c
)


-

W

0

d
-
>a

isochoric

nC
V
(T
a



T
d
)

0




Q

nC
V
ln(Ta/Td)

________________________________________________________________


efficiency:



Qbc

=
n
C
V
(T
c



T
b
)
> 0


Qcd =
nC
V
(T
a



T
d
)

< 0



|Q
C
| / |

Q
H

| =
(T
d



T
a
) / (T
c



T
b
)

= (T
d



T
a
)/

[r

-
1
(T
d



T
a
)]



=
1/

r

-
1
, or

e = 1
-

1/

r

-
1

NOTE:


Tc > Tb > Ta (since Pc>Pb);

Td/Ta = Tc/Tb > 1

=> Td > Ta
;



so that Tc = T
H

and Ta = T
C
OLD
;


using these te
mperatures,



the Carnot efficiency is e = 1


(1/

r

-
1
)(

T
a
/T
d
) > Otto efficiency

entropy:



S = nC
V
ln(Tc/Tb)

+ nC
V
ln(Ta/Td)


= nC
V
ln[(Tc/Tb)(Ta/Td)]


= nC
V
ln[(Td/Ta)(Ta/Td)]


= nC
V
ln (1) = 0


S = 0.
Reversible process.


Copyright, 2004, John R. Newport, Ph.D.

Example
4
:
Diesel Cycle


STATE

P


V



T


a

P
a



V
a
= rV
b


T
a

b

P
b

= P
a
r



V
b



T
b
= T
a
r

-
1

c

P
b



V
c
= r
c
V
b


T
c
= (V
c
/V
b
)T
b

= (V
c
/V
b
)T
a
r

-
1

d

P
d

= P
a
(V
c
/V
b
)


V
a
= rV
b


T
d

= (V
c
/V
b
)


(1/r)

-
1
T
b

= (V
c
/V
b
)


T
a


________________________________________________________________

STEP

TYPE



Q




W



U


S


a
-
>b

adiabatic

0


nC
V
(T
b



T
a
)


-

W

0

b
-
>c

isobaric

nC
P
(T
c



T
b
)

0




Q

nC
P
ln(Tc/Tb)


c
-
>d

adiabatic

0


nC
V
(T
d



T
c
)


-

W

0

d
-
>a

isochoric

nC
V
(T
a



T
d
)

0




Q

nC
V
ln(Ta/Td)

________________________________________________________________


e
fficiency:



Qbc = nC
P
(T
c



T
b
) > 0


Qcd = nC
V
(T
a



T
d
) < 0



|Q
C
| / |

Q
H

| = (1/


)

(T
d



T
a
)

/ (T
c



T
b
)


= (1/


) [(V
c
/V
b
)




1)T
a
]

/ [ (V
c
/V
b
)


1]T
b


= (1/


) [(V
c
/V
b
)




1)T
a
] / [ (V
c
/V
b
)


1]T
b


= (1/


) [(V
c
/V
b
)




1)] / [ (V
c
/V
b
)


1](1/
r

-
1
)


e
= 1


{
[
r
c




1
] / [
r
c



1
]
}

(
1/



r

-
1
)

NOTE: this
is
indeterminate
for r
c

= 1;



the ef
ficiency at this point is 1


(

-
1)
/

(


r

-
1
)

=
> 1


2
/

(5

r
2/3
) for monatomic

NOTE: for r
c

>> 1, e
-
> 1


(1/


)(r
c
/r)


-
1

=> 1


3
(r/r
c
)
-
2
/3

for monatomic


entropy:



S = nC
P
ln(Tc/Tb) + nC
V
ln(Ta/Td)


= nC
V

[

ln(Tc/Tb) + ln(Ta/Td)]


= nC
V

ln(Tc/Tb)

(Ta/Td)


= nC
V

ln(Vc/V
b)


(Vb/Vc)



= nC
V

ln(1)

= 0



S = 0.


Reversible process.


Copyright, 2004, John R. Newport, Ph.D.

Example
5
:
Stirling Cycle


STATE

P

V



T


a


Pa

V
a
= rV
b


T
C

b


Pb

V
b



T
C

c


Pc

V
b



T
H

d


Pd

V
a
= rV
b


T
H


________________________________________________________________

STEP

TYPE



Q




W




U


S


a
-
>b

isothermal


W



-
n
R
T
C

ln(r)


0

nRln(
Vb/Va
)

b
-
>c

isochoric

nC
V
(T
H



T
C
)


0




Q

nC
V
ln(Tc/Tb)


c
-
>d

isothermal


W



nRT
H

ln(r)


0

nRl
n(Vd/Vc)

d
-
>a

isochoric

-
nC
V
(T
H



T
C
)


0




Q

nC
V
ln(Ta/Td)

________________________________________________________________


efficiency:



Q
c
d

=

n
R
T
H

ln(r)

+ nC
V
(T
H



T
C
)
> 0


Q
ab

=
-
nRT
C

ln(r)
-

nC
V
(T
H



T
C
)
< 0



|Q
C
| / |

Q
H

| = [T
H

ln(r) + (C
V
/R)(T
H



T
C
)] / [T
C

ln(r) + (C
V
/R)(T
H



T
C
)]



= [T
H

(ln(r) + C
V
/R)


(C
V
/R))T
C
)] / [T
C

(ln
(r)
-

C
V
/R) +

(C
V
/R)T
H
]


e = 1
-

1/

(


r

-
1
)


entropy:



S = nRln(Vb/Va) + nRln(Vd/Vc) + nC
V
ln(Tc/Tb) + nC
V
ln(Ta/Td)


= nRln[(Vb/Va)(Vd/Vc)] + nC
V
ln[(Tc/Tb)(Ta/Td)]


= n
Rln[(1/r)(r)] + nC
V
ln[(T
H
/T
C
)(T
C
/T
H
)]


= nRln(1) + nC
V
ln(1



S = 0.
Reversible process.