Isothermal process on p-V, T-V, and p-T diagrams
isothermal ⇒ T = T0 = constant
a = (p1, V1, T0)
b = (p2, V2, T0)
pV = nRT0
p
p1
T
p
a
T0
p2
W
V1
Q
T0
V
nRT0
p(V) =
V
PHYS 1101, Winter 2009, Prof. Clarke
a
p2
b
b
a
b
V2
p1
V1
V2
T(V) = T0
V
T0
T
p(T) = multivalued
1
Isochoric process on p-V, T-V, and p-T diagrams
isochoric ⇒ V = V0 = constant
a = (p1, V0, T1)
b = (p2, V0, T2)
pV0 = nRT
p
T
p
p1
a
T1
a
p1
p2
b
T2
b
p2
V0
V
p(V) = multivalued
PHYS 1101, Winter 2009, Prof. Clarke
V0
V
T(V) = multivalued
a
b
T2
T1
T
nRT
p(T) =
V0
2
Isobaric process on p-V, T-V, and p-T diagrams
isobaric ⇒ p = p0 = constant
a = (p0, V1, T1)
b = (p0, V2, T2)
p0V = nRT
p
p0
T
a
p
T2
Q
b
b
p0
W
V1
T1
V2
V
p(V) = p0
PHYS 1101, Winter 2009, Prof. Clarke
a
b
T1
T2
a
V1
V2
p0V
T(V) =
nR
V
T
p(T) = p0
3
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 (Nm–2)
105
isobar
a
b
isotherm
T0
isochor
c
pc
1
b)
c)
d)
e)
105
50
1,000
Not enough information to tell
Vc V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
4
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 (Nm–2)
105
isobar
a
b
isotherm
T0
isochor
c
pc
1
Vc V (m3)
b)
c)
d)
e)
105
50
1,000
Not enough information to tell
pV
100,000
n =
=
= 50
RT0
2,000
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
5
Clicker question 2
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is Vc, the volume at state c?
p
a) 0.5 m3
(Nm–2)
105
a
b
T0
c
pc
1
b)
c)
d)
e)
2.0 m3
4.0 m3
8.0 m3
Not enough information to tell
Vc V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
6
Clicker question 2
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is Vc, the volume at state c?
p
a) 0.5 m3
(Nm–2)
105
a
b
T0
c
pc
1
Vc V (m3)
b)
c)
d)
e)
2.0 m3
4.0 m3
8.0 m3
Not enough information to tell
need to know pc
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
7
Clicker question 3
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is Vc, the volume at state c?
p
a) 0.5 m3
(Nm–2)
105
a
b
T0
c
5 ×104
1
b)
c)
d)
e)
2.0 m3
4.0 m3
8.0 m3
Not enough information to tell
Vc V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
8
Clicker question 3
Consider the p-V diagram below in which the system evolves from a → b
→ c. What is Vc, the volume at state c?
p
a) 0.5 m3
(Nm–2)
105
a
b
T0
c
5 ×104
1
Vc V (m3)
b)
c)
d)
e)
2.0 m3
4.0 m3
8.0 m3
Not enough information to tell
pa
pcVc = paVa ⇒ Vc = p Va = 2 m3
c
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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, ΔEint?
a) 0 J
p (Nm–2)
105
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
5.0 ×104 J
about 7.0 ×104 J
105 J
Not enough information to tell
V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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, ΔEint?
a) 0 J
p (Nm–2)
105
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
V
(m3)
5.0 ×104 J
about 7.0 ×104 J
105 J
Not enough information to tell
ΔEint = nCVΔT
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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?
a) 0 J
p (Nm–2)
105
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
5.0 ×104 J
about 7.0 ×104 J
105 J
Not enough information to tell
V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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?
a) 0 J
p (Nm–2)
105
W
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
5.0 ×104 J
about 7.0 ×104 J
105 J
Not enough information to tell
V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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
a) –5.0 ×104 J
(Nm–2)
105
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
5.0 ×104 J
–105 J
105 J
Not enough information to tell
V (m3)
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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
a) –5.0 ×104 J
(Nm–2)
105
a
b)
c)
d)
e)
b
T0
c
5 ×104
1
2
V (m3)
5.0 ×104 J
–105 J
105 J
Not enough information to tell
Q = ΔEint + W = 0 + 105 J
first law of thermodynamics: ΔEint = Q – W ( = nCVΔT )
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
15
Internal Energy (revisited)
f
f
nRT = NkT
Cp = CV + R
2
2
n = number of moles; 1 mole = 6.0221 × 1022 particles (NA)
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
Eint = nCVT =
type of gas
degrees of specific heat at internal specific heat at
freedom
constant
energy
constant
(f)
volume (CV )
(Eint) pressure (Cp )
monatomic
3
diatomic
5
polyatomic (≥3)
~6
PHYS 1101, Winter 2009, Prof. Clarke
3
R
2
5
R
2
3
nRT
2
5
nRT
2
5
R
2
7
R
2
3R
3 nRT
4R
γ
(Cp/CV )
5
3
7
5
4
3
16
Adiabatic processes
reversible
irreversible
a = (p1, V1, T1)
b = (p2, V2, T2)
a = (p1, V1, T0)
b = (p2, V2, T0)
pV γ = constant
p1V1 = p2V2
p
p
a
p1
adiabat
b
p2
V1
a
isotherms
T1
T2
p1
V2
PHYS 1101, Winter 2009, Prof. Clarke
T0
b
p2
V
isotherm
V1
V2
V
17
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
a) monatomic (γ = 5/3)
(kNm–2)
b) diatomic (γ = 7/5)
c) polyatomic (γ = 4/3)
d) not enough information to tell
b
16
adiabat
a
1
1
8 V (m3)
pV γ = constant
PHYS 1101, Winter 2009, Prof. Clarke
18
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
a) monatomic (γ = 5/3)
(kNm–2)
b) diatomic (γ = 7/5)
c) polyatomic (γ = 4/3)
d) not enough information to tell
b
16
adiabat
γ
a
1
8 V (m3)
1
pV = constant
γ
PHYS 1101, Winter 2009, Prof. Clarke
γ
paVa = 1(8) = 23γ =
γ
γ
pbVb = 16(1) = 24
⇒ γ = 4/3 ⇒ polyatomic
19
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)
b)
c)
d)
e)
b
16
adiabat
8 kJ
16 kJ
128 kJ
not enough information to tell
a
1
1
8 V (m3)
pV γ = constant
PHYS 1101, Winter 2009, Prof. Clarke
20
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)
b)
c)
d)
e)
b
16
adiabat
a
1
1
8 V (m3)
8 kJ
16 kJ
128 kJ
not enough information to tell
By definition, Q = 0 for all adiabatic
processes.
pV γ = constant
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)
c) 24 kJ
d) –24 kJ
e) 32 kJ
f) –32 kJ
g) not enough information to tell
b
16
adiabat
a
1
1
8 V (m3)
Eint = nCVT = n3RT (for a polyatomic gas)
first law: ΔEint = Q – W
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
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)
c) 24 kJ
d) –24 kJ
e) 32 kJ
f) –32 kJ
g) not enough information to tell
b
16
adiabat
W
1
1
ΔEint = 0 – W = 3Δ(nRT) = 3Δ(pV)
a
8 V (m3)
= 3(16 – 8) = 24 ⇒ W = –24 kJ
Eint = nCVT = n3RT (for a polyatomic gas)
first law: ΔEint = Q – W
PHYS 1101, Winter 2009, Prof. Clarke
ideal gas law: pV = nRT
23
Summary of Processes
p
S1
S3
S2
S4
p1
p2
p3
p4
isentrop: ΔS = 0
(reversible adiabat: Q = 0)
isobar: Δp = 0
isochor: ΔV = 0
a
b
isotherm: ΔT = 0
V1
d
V2
b = (p1, V2, T1, S1)
c = (p2, V2, T2, S2)
d = (p4, V2, T3, S3)
e = (p3, V1, T3, S4)
T1
c
e
a = (p1, V1, T2, S3)
T2
T3
V
free expansion (irreversible adiabat: Q = 0)
PHYS 1101, Winter 2009, Prof. Clarke
24
Summary of Processes
process
W
Q
ΔEint =
nCVΔT
isobar
p(V2 – V1)
pCp
(V2 – V1)
R
pCV
(V2 – V1)
R
nCp ln
isochor
0
VCV
(p2 – p1)
R
VCV
(p – p )
R 2 1
nCV ln
isotherm
V2
nRT ln
V1
V2
nRT ln
V1
isentrop
p1V1 – p2V2
γ –1
free
expansion
( )
0
PHYS 1101, Winter 2009, Prof. Clarke
( )
0
0
0
p2V2 – p1V1
γ –1
0
ΔS
V2
V1
( )
p2
p1
( )
V2
nR ln
V1
( )
0
V2
nR ln
V1
( )
25
All processes on p-V, T-V, and T-S diagrams
p
T
a
V
T
p
p
p
a
a
T
V
S
T
S
S
V
V
T
V
S
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
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.)
p
a) Cp
c) CV
d) CV e
e) R
f) R e
g) no where near enough information!!
b
a
1
e
b) Cp e
V
some formulae, in case they help…
ΔS = nCp ln(V2) (isobar)
ΔS
ΔS
PHYS 1101, Winter 2009, Prof. Clarke
V1
p
= nCV ln 2
p1
V
= nR ln 2
V1
( )
( )
(isochor)
(isotherm)
27
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) Cp
c) CV
d) CV e
e) R
f) R e
g) Yes there is!!
b
a
1
e
b) Cp e
V
Since S is a state variable, it doesn’t
matter which path from a to b you
choose. Thus, choose the isobar.
PHYS 1101, Winter 2009, Prof. Clarke
ΔS = nCp ln(V2) (isobar)
ΔS
ΔS
V1
p
= nCV ln 2
p1
V
= nR ln 2
V1
( )
( )
(isochor)
(isotherm)
28
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
T (K)
b
a
250
1
a
500
2 V (m3)
= nCV ΔT
pV = nRT
b
10
ΔEint = Q – W
30
S (JK–1)
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
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
T (K)
a
b
250
W = –2000 J
1
a
500
2 V (m3)
b
Q >~ –7500 J
10
30
Q ~ –7000 J
W = –2000 J
ΔEint = Q – W
~ –5000 J
S (JK–1)
a) 2000 J
b) –2000 J
c) 5000 J
d) –5000 J
e) 7000 J
f) –7000 J
PHYS 1101, Winter 2009, Prof. Clarke
30
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
T (K)
b
a
250
1
a
500
2 V (m3)
= nCV ΔT
pV = nRT
b
10
ΔEint = Q – W
30
S (JK–1)
a) 0.5
b) 1
d) 2
e) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke
c) 1.5
31
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
T (K)
b
a
500
250
1
a
2 V (m3)
b
10
30
pV = nRT
at state b:
pV = 2000
RT ~ 2000 ⇒ n~1
S (JK–1)
a) 0.5
b) 1
d) 2
e) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke
c) 1.5
32
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 ΔEint = –5000 J, this gas is:
p (Nm–2)
2,000
T (K)
b
a
250
1
2 V (m3)
a) monatomic
a
500
CV ~ 12 (monatomic)
CV ~ 21 (diatomic)
CV ~ 25 (polyatomic)
b
10
b) diatomic
ΔEint = nCV ΔT
30
S (JK–1)
c) polyatomic
d) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke
33
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 ΔEint = –5000 J, this gas is:
p (Nm–2)
2,000
T (K)
b
a
250
1
2 V (m3)
a) monatomic
a
500
CV ~ 12 (monatomic)
CV ~ 21 (diatomic)
CV ~ 25 (polyatomic)
b
10
b) diatomic
ΔEint = nCV ΔT
30
S (JK–1)
c) polyatomic
d) not enough information to tell
PHYS 1101, Winter 2009, Prof. Clarke
34
Three pictorial representations of an engine
p
Qin
a
Qin
a→c
TH
W>0
Qout
c
V
In a complete thermodynamical cycle, gas
expands at high pressure
and compresses at low
pressure allowing work,
W, to be extracted in
each cycle.
PHYS 1101, Winter 2009, Prof. Clarke
expansion
stroke
c→a
TC
Qout
QH = Qin
QC = Qout
Wout = W
compression
stroke
35
Maximum efficiency and the Carnot cycle
p
T
TH
T
a
Qin
TH
a
W>0
TC
Qout
S1
Qin
b
W>0
c
TC
S2 S
non-optimal thermo
-dynamical cycle for
an engine
d
S1
Qout
S1
a Q S2 adiabats
in
b
TC
c
isotherms
TH
d
c
Qout
S2 S
optimal thermodyn
-amical cycle for an
engine (Carnot cycle)
V
the Carnot engine cycle
on a p-V diagram
Note that for an engine, the thermodynamical cycle is always clockwise.
PHYS 1101, Winter 2009, Prof. Clarke
36
Refrigerators (heat pumps)
p
a
p
Qout
a Q S2 adiabats
out
W<0
Qin
S1
c
b
TC
V
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
TC and expelled at TH.
PHYS 1101, Winter 2009, Prof. Clarke
isotherms
TH
d
Qin
c
V
QC = Qin
QH = Qout
Win = –W
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).
37