# A summary of thermodynamic processes - Web Physics

Mechanics

Oct 27, 2013 (4 years and 8 months ago)

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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
)

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
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

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

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

|

=> 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

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

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

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

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.