Isothermal process on p-V, T-V, and p-T diagrams

p
p 1
p 2
<strong>Isothermal</strong> <strong>process</strong> on p-V, T-V, and p-T diagrams
V 1
a
T 0
W
Q
V 2
b
isothermal ⇒ T = T 0 = constant
a = (p 1 , V 1 , T 0 ) b = (p 2 , V 2 , T 0 )
V
T
T 0
pV = nRT 0
a
V 1
nRT0 p(V) = T(V) = T
V
0 p(T) = multivalued
PHYS 1101, Winter 2009, Prof. Clarke 1
b
V 2
V
p
p 1
p 2
T 0
a
b
T

p
p 1
p 2
Isochoric <strong>process</strong> on p-V, T-V, and p-T diagrams
V 0
a
b
isochoric ⇒ V = V 0 = constant
a = (p 1 , V 0 , T 1 ) b = (p 2 , V 0 , T 2 )
V
pV 0 = nRT
p(V) = multivalued T(V) = multivalued
T
T 1
V 0
nRT
p(T) =
V0
PHYS 1101, Winter 2009, Prof. Clarke 2
a
T2 b
p2 b
V
p
p 1
T 2
T 1
a
T

p
p 0
Isobaric <strong>process</strong> on p-V, T-V, and p-T diagrams
a
V 1
Q
W
p(V) = p 0
b
V 2
isobaric ⇒ p = p 0 = constant
a = (p 0 , V 1 , T 1 ) b = (p 0 , V 2 , T 2 )
V
T
T 2
T 1
a
p 0 V = nRT
V 1
p0V T(V) =
nR
p(T) = p 0
PHYS 1101, Winter 2009, Prof. Clarke 3
b
V 2
V
p
p 0
a
T 1
b
T 2
T

p (Nm –2 )
10 5
a
1
T 0
V c
Clicker question 1
Consider the p-V diagram below in which the system evolves from a → b
→ c. If T0 ~ 240K (and thus RT0 = 2,000 J mol –1 ), how many moles of
gas, n, are in the system?
a) 5
p c
isobar
b
isotherm
c
isochor
V (m 3 )
b) 10 5
c) 50
d) 1,000
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 4

p (Nm –2 )
10 5
p c
a
1
T 0
V c
Clicker question 1
Consider the p-V diagram below in which the system evolves from a → b
→ c. If T0 ~ 240K (and thus RT0 = 2,000 J mol –1 ), how many moles of
gas, n, are in the system?
a) 5
isobar
b
isotherm
c
isochor
V (m 3 )
b) 10 5
c) 50
d) 1,000
e) Not enough information to tell
pV 100,000
n = = = 50
RT0 2,000
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 5

Clicker question 2
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is V c , the volume at state c?
p (Nm –2 )
10 5
p c
a
1
T 0
V c
b
c
V (m 3 )
a) 0.5 m 3
b) 2.0 m 3
c) 4.0 m 3
d) 8.0 m 3
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 6

Clicker question 2
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is V c , the volume at state c?
p (Nm –2 )
10 5
p c
a
1
T 0
V c
b
c
V (m 3 )
a) 0.5 m 3
b) 2.0 m 3
c) 4.0 m 3
d) 8.0 m 3
e) Not enough information to tell
need to know p c
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 7

Clicker question 3
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is V c , the volume at state c?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
V c
b
c
V (m 3 )
a) 0.5 m 3
b) 2.0 m 3
c) 4.0 m 3
d) 8.0 m 3
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 8

Clicker question 3
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is V c , the volume at state c?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
V c
b
c
V (m 3 )
a) 0.5 m 3
b) 2.0 m 3
c) 4.0 m 3
d) 8.0 m 3
e) Not enough information to tell
pcVc = paVa ⇒ Vc = Va = 2 m3 pa pc first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 9

Clicker question 4
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net change in internal energy, ΔE int ?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
2
b
c
V (m 3 )
a) 0 J
b) 5.0 ×10 4 J
c) about 7.0 ×10 4 J
d) 10 5 J
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 10

Clicker question 4
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net change in internal energy, ΔE int ?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
2
b
c
V (m 3 )
a) 0 J
b) 5.0 ×10 4 J
c) about 7.0 ×10 4 J
d) 10 5 J
e) Not enough information to tell
ΔE int = nC V ΔT
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 11

Clicker question 5
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net work done by the system on its environment, W?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
2
b
c
V (m 3 )
a) 0 J
b) 5.0 ×10 4 J
c) about 7.0 ×10 4 J
d) 10 5 J
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 12

Clicker question 5
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net work done by the system on its environment, W?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
W
2
b
c
V (m 3 )
a) 0 J
b) 5.0 ×10 4 J
c) about 7.0 ×10 4 J
d) 10 5 J
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 13

Clicker question 6
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net heat transferred into the system, Q?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
2
b
c
V (m 3 )
a) –5.0 ×10 4 J
b) 5.0 ×10 4 J
c) –10 5 J
d) 10 5 J
e) Not enough information to tell
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 14

Clicker question 6
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is the net heat transferred into the system, Q?
p (Nm –2 )
10 5
5 ×10 4
a
1
T 0
2
b
c
V (m 3 )
a) –5.0 ×10 4 J
b) 5.0 ×10 4 J
c) –10 5 J
d) 10 5 J
e) Not enough information to tell
Q = ΔE int + W = 0 + 10 5 J
first law of thermodynamics: ΔE int = Q – W ( = nC V ΔT )
ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 15

type of gas degrees of
freedom
( f )
Internal Energy (revisited)
f
2
E int = nC V T = nRT = NkT C p = C V + R
n = number of moles; 1 mole = 6.0221 × 10 22 particles (N A )
N = number of particles
R = gas constant = 8.3147 J mol –1 K –1
k = Boltzmann’s constant = 1.3807 × 10 –23 J K –1
specific heat at
constant
volume (C V )
internal
energy
(E int )
specific heat at
constant
pressure (C p )
3 3
5
monatomic 3 R nRT R
2 2
2
5 5
7
diatomic 5 R nRT R
2 2
2
polyatomic (≥3) ~6 3 R 3 nRT 4 R
f
2
γ
(C p /C V )
PHYS 1101, Winter 2009, Prof. Clarke 16
5
3
7
5
4
3

p
p 1
p 2
reversible
a = (p 1 , V 1 , T 1 )
b = (p 2 , V 2 , T 2 )
γ
pV = constant
T 2
V 1
a
T 1
V 2
Adiabatic <strong>process</strong>es
adiabat
b
isotherms
V
isotherm
PHYS 1101, Winter 2009, Prof. Clarke 17
p
p 1
p 2
V 1
irreversible
a = (p 1 , V 1 , T 0 )
b = (p 2 , V 2 , T 0 )
p 1 V 1 = p 2 V 2
a
T 0
V 2
b
V

γ
pV = constant
Clicker question 7
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. This gas is…
p (kNm –2 )
16
1
b
adiabat
1 8
a
V (m 3 )
a) monatomic (γ = 5/3)
b) diatomic (γ = 7/5)
c) polyatomic (γ = 4/3)
d) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke 18

γ
pV = constant
Clicker question 7
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. This gas is…
p (kNm –2 )
16
1
b
adiabat
1 8
a
V (m 3 )
a) monatomic (γ = 5/3)
b) diatomic (γ = 7/5)
c) polyatomic (γ = 4/3)
d) not enough information to tell
γ γ
paVa = 1(8) = 23 γ =
p b V b = 16(1) = 2 4
⇒ γ = 4/3 ⇒ polyatomic
PHYS 1101, Winter 2009, Prof. Clarke 19
γ
γ

γ
pV = constant
Clicker question 8
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. How much heat is transferred to
the system?
a) 0 J
p (kNm –2 )
16
1
b
adiabat
1 8
a
V (m 3 )
b) 8 kJ
c) 16 kJ
d) 128 kJ
e) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke 20

γ
pV = constant
Clicker question 8
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. How much heat, Q, is
transferred to the system?
a) 0 J
p (kNm –2 )
16
1
b
adiabat
1 8
a
V (m 3 )
b) 8 kJ
c) 16 kJ
d) 128 kJ
e) not enough information to tell
By definition, Q = 0 for all adiabatic
<strong>process</strong>es.
PHYS 1101, Winter 2009, Prof. Clarke 21

Clicker question 9
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. How much work, W, does the
system do on its environment?
a) 20 kJ b) –20 kJ
p (kNm –2 )
16
1
b
adiabat
1 8
a
V (m 3 )
c) 24 kJ d) –24 kJ
e) 32 kJ f) –32 kJ
g) not enough information to tell
E int = nC V T = n3RT (for a polyatomic gas)
first law: ΔE int = Q – W ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 22

Clicker question 9
Consider the p-V diagram below in which the system evolves reversibly
along the adiabat from state a to state b. How much work, W, does the
system do on its environment?
a) 20 kJ b) –20 kJ
p (kNm –2 )
16
1
b
W
adiabat
1 8
a
V (m 3 )
c) 24 kJ d) –24 kJ
e) 32 kJ f) –32 kJ
g) not enough information to tell
ΔE int = 0 – W = 3Δ(nRT) = 3Δ(pV)
= 3(16 – 8) = 24 ⇒ W = –24 kJ
E int = nC V T = n3RT (for a polyatomic gas)
first law: ΔE int = Q – W ideal gas law: pV = nRT
PHYS 1101, Winter 2009, Prof. Clarke 23

p
p 1
p 2
p 3
p 4
S 4
S 3
a
V 1
e
S 2
S 1
Summary of Processes
isentrop: ΔS = 0
(reversible adiabat: Q = 0)
V 2
d
isobar: Δp = 0
isochor: ΔV = 0
isotherm: ΔT = 0
T 2
V
T 1
T 3
free expansion (irreversible adiabat: Q = 0)
b
c
a = (p 1 , V 1 , T 2 , S 3 )
b = (p 1 , V 2 , T 1 , S 1 )
c = (p 2 , V 2 , T 2 , S 2 )
d = (p 4 , V 2 , T 3 , S 3 )
e = (p 3 , V 1 , T 3 , S 4 )
PHYS 1101, Winter 2009, Prof. Clarke 24

Summary of Processes
<strong>process</strong> W Q
ΔE int =
nC V ΔT
pCp pCV isobar p(V2 – V1 ) (V2 – V1 ) (V2 – V1 ) nCp ln ( )
R
V1 V2 isotherm nRT ln nRT ln 0 nR ln
p1V1 – p2V2 p2V2 – p1V1 isentrop 0 0
γ – 1
γ – 1
free
expansion
( V2 V1 )
R
isochor 0 (p2 – p1 ) (p2 – p1 ) nCV ln ( p VCV VCV 2
R
R
p1 ( V2 V1 0 0 0 nR ln
PHYS 1101, Winter 2009, Prof. Clarke 25
)
ΔS
( V 2
V 1
( V 2
V 1
)
)
)

p
a
V
All <strong>process</strong>es on p-V, T-V, and T-S diagrams
p
T
S
V
T
a
V
An isobar (p), isotherm (T), isentrop (S), and isochor (V)
emanating from the same initial state (a) as manifest on
a p-V, T-V, and a T-S diagram.
PHYS 1101, Winter 2009, Prof. Clarke 26
p
T
S
V
T
V
a
S
p
T
S

p
b
Clicker question 10
Consider the p-V diagram below in which n = 1 mole of gas evolves reversibly
from state a to state b along the path shown. What is the net change in
entropy? (Note, e = 2.71828 = Euler’s number, and thus ln(e) = 1.)
a
1 e
some formulae, in case they help…
V
a) C p b) C p e
c) C V d) C V e
e) R f) R e
g) no where near enough information!!
ΔS = nC p ln (isobar)
ΔS = nC V ln (isochor)
ΔS = nR ln (isotherm)
PHYS 1101, Winter 2009, Prof. Clarke 27
)
( V2 V1 ( )
p2 p1 ( )
V2 V1

Clicker question 10
Consider the p-V diagram below in which n = 1 mole of gas evolves revers
-ibly from state a to state b along the path shown. What is the net change in
entropy? (Note, e = 2.71828 = Euler’s number, and thus ln(e) = 1.)
p
a
1 e
Since S is a state variable, it doesn’t
matter which path from a to b you
choose. Thus, choose the isobar.
b
V
a) C p b) C p e
c) C V d) C V e
e) R f) R e
g) Yes there is!!
ΔS = nC p ln (isobar)
ΔS = nC V ln (isochor)
ΔS = nR ln (isotherm)
PHYS 1101, Winter 2009, Prof. Clarke 28
)
( V2 V1 ( )
p2 p1 ( )
V2 V1

Clicker question 11
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. What is the change in internal energy?
p (Nm –2 )
2,000
b
a
1 V (m3 2 )
T (K)
500
250
b
10 S (JK –1 30 )
a) 2000 J b) –2000 J c) 5000 J
d) –5000 J e) 7000 J f) –7000 J
PHYS 1101, Winter 2009, Prof. Clarke 29
a
ΔE int = Q – W
= nC V ΔT
pV = nRT

Clicker question 11
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. What is the change in internal energy?
p (Nm –2 )
2,000
b
a
W = –2000 J
1 V (m3 2 )
T (K)
500
250
b
Q –7500 J
>
~
10 S (JK –1 30 )
Q ~ –7000 J
W = –2000 J
ΔE int = Q – W
a) 2000 J b) –2000 J c) 5000 J
d) –5000 J e) 7000 J f) –7000 J
~ –5000 J
PHYS 1101, Winter 2009, Prof. Clarke 30
a

Clicker question 12
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. About how many moles of gas are in the
system? (Take R = 8.)
p (Nm –2 )
2,000
b
a
1 V (m3 2 )
T (K)
500
250
a) 0.5 b) 1 c) 1.5
d) 2 e) not enough information to tell
b
10 S (JK –1 30 )
PHYS 1101, Winter 2009, Prof. Clarke 31
a
ΔE int = Q – W
= nC V ΔT
pV = nRT

Clicker question 12
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. About how many moles of gas are in the
system? (Take R = 8.)
p (Nm –2 )
2,000
b
a
1 V (m3 2 )
T (K)
500
250
a) 0.5 b) 1 c) 1.5
d) 2 e) not enough information to tell
b
10 S (JK –1 30 )
PHYS 1101, Winter 2009, Prof. Clarke 32
a
pV = nRT
at state b:
pV = 2000
RT ~ 2000
⇒ n ~ 1

Clicker question 13
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. With n = 1 and ΔE int = –5000 J, this gas is:
p (Nm –2 )
2,000
b
a
1 V (m3 2 )
T (K)
500
250
a) monatomic b) diatomic c) polyatomic
d) not enough information to tell
b
10 S (JK –1 30 )
PHYS 1101, Winter 2009, Prof. Clarke 33
a
ΔE int = nC V ΔT
C V ~ 12 (monatomic)
C V ~ 21 (diatomic)
C V ~ 25 (polyatomic)

Clicker question 13
The evolution of a system from state a to state b is shown on both the p-V
and T-S diagrams below. With n = 1 and ΔE int = –5000 J, this gas is:
p (Nm –2 )
2,000
b
a
1 V (m3 2 )
T (K)
500
250
a) monatomic b) diatomic c) polyatomic
d) not enough information to tell
b
10 S (JK –1 30 )
PHYS 1101, Winter 2009, Prof. Clarke 34
a
ΔE int = nC V ΔT
C V ~ 12 (monatomic)
C V ~ 21 (diatomic)
C V ~ 25 (polyatomic)

p
a
Three pictorial representations of an engine
Q out
W > 0
Q in
In a complete thermodynamical
cycle, gas
expands at high pressure
and compresses at low
pressure allowing work,
W, to be extracted in
each cycle.
c
V
Q in
a→c
T H
expansion
stroke
c→a
T C
compression
stroke
PHYS 1101, Winter 2009, Prof. Clarke 35
Q out
Q H = Q in
Q C = Q out
W out = W

T
T H
T C
a
S 1
Maximum efficiency and the Carnot cycle
Q in
Q out
c
S 2
S
T
T H
T C
a
d
S 1
Q in
W > 0 W > 0
non-optimal thermo
-dynamical cycle for
an engine
Q out
PHYS 1101, Winter 2009, Prof. Clarke 36
b
c
S 2
S
optimal thermodyn
-amical cycle for an
engine (Carnot cycle)
p
a
S 1
T C
Q in
S 2
d
adiabats
b
isotherms
T H
c
Qout V
the Carnot engine cycle
on a p-V diagram
Note that for an engine, the thermodynamical cycle is always clockwise.

p
a
Q in
W < 0
Q out
For a refrigerator, the
cycle is always
counterclockwise.
Expansion happens at
low pressure,
compression at high
pressure and this takes
work. Heat is drawn in at
T C and expelled at T H .
c
V
Refrigerators (heat pumps)
Q C = Q in
Q H = Q out
W in = –W
PHYS 1101, Winter 2009, Prof. Clarke 37
p
a
S 1
T C
S2 Qout d
adiabats
b
isotherms
T H
Q in
c
V
As for an engine, the most
optimal thermodynamical
cycle for a refrigerator is the
Carnot cycle traversed in
the counterclockwise
direction (opposite to the
engine).