20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #3
page 1
EXPANSIONS, ENERGY, ENTHALPY
Isothermal Gas Expansion
( ∆T = 0)
gas (p1, V1, T) = gas (p2, V2, T)
Irreversibly (many ways possible)
(1)
Set pext = 0
p= 0
T
T
p= 0
p 1,V1
p 2,V2
v2
w(1) = − ∫ pext dV = 0
V1
(2)
Set pext = p2
p2
T
p2
T
p 2,V2
p 1,V1
v2
w(2) = −∫ p2dV = −p2 (V2 −V1 )
V1
p
p1
p2
V1
-w(2)
V2
Note, work is negative: system expands against surroundings
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
(3)
Lecture #3
page 2
Carry out change in two steps
gas (p1, V1, T) = gas (p3, V3, T) = gas (p2, V2, T)
p1 > p3 > p2
p2
p3
p3
T
T
T
p 2,V2
p 3,V3
p 1,V1
v3
v2
V1
V3
w(3) = − ∫ p3dV − ∫ p2dV = − p3 (V3 −V1 ) −p2 (V2 −V3 )
p
p1
More work delivered to
surroundings in this case.
p3
p2
V1 V3
V2
-w(3)
(4)
Reversible change
p = pext throughout
V
p
wrev = − ∫ 2 pdV
V1
p1
p2
V1
-
V2
rev
For ideal gas:
V
wrev = − ∫ 2
V1
Maximum work delivered to
surroundings for isothermal gas
expansion is obtained using a
reversible path
V
p
nRT
dV = −nRT ln 2 = nRT ln 2
V
V1
p1
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #3
page 3
The Internal Energy U
dU = d-q + d-w
(First Law)
dU = C pathdT − pext dV
And U (T ,V
)
⇒
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜
⎟ dT + ⎜
⎟ dV
∂
T
∂
V
⎝
⎠V
⎝
⎠T
Some frequent constraints:
dU = d-qrev + d-wrev = d-qrev – pdV
Reversible
⇒
•
Isolated
⇒
d-q = d-w = 0
•
Adiabatic
⇒
d-q = 0
•
Constant V
⇒
•
(p = pext )
⇒ dU = d-w
reversible
=
-pdV
w = 0 ⇒ dU = d-qV
Constant V
∂U ⎞
⎛ ∂U ⎞
dU = ⎛⎜
⎟ dT + ⎜
⎟ dV
⎝ ∂T ⎠V
⎝ ∂V ⎠T
⎛ ∂U ⎞
⇒
d- qV = ⎜
⎟ dT
⎝ ∂T ⎠V
But also
d-qV = CV dT
So
⇒
⎛ ∂U ⎞
⎟ = CV
⎜
⎝ ∂T ⎠V
very important result!!
⎛ ∂U ⎞
⎟
⎝ ∂V ⎠T
dU = CV dT + ⎜
dV
what is this?
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #3
⎛ ∂U ⎞
⎟ )
⎝ ∂V ⎠T
Joule Free Expansion of a Gas
gas
(to get ⎜
vac
gas (p1, T1, V1) = gas (p2, T2, V2)
Since q = w = 0
Recall
⇒
⎛ ∂U ⎞
⎜
⎟
⎝ ∂V ⎠T
⎛ ∂U ⎞
⎟
⎝ ∂V ⎠T
•
q=0
Expansion into Vac.
(pext=0)
w=0
Constant U
dV = 0
dVU = −CV dTU
⎛ ∂U ⎞
⎜
⎟ = −CV
⎝ ∂V ⎠T
Joule did this.
Adiabatic
dU or ∆U = 0
dU = CV dT + ⎜
page 4
⎛ ∂T ⎞
⎜
⎟
⎝ ∂V ⎠U
measure in Joule exp't!
⎛ ∆T ⎞
⎜
⎟
⎝ ∆V ⎠U
⎛ ∆T ⎞
⎛ ∂T ⎞
∴ dU = CV dT − CV ηJ dV
⎟ =⎜
⎟ ≡ ηJ
⎝ ∆V ⎠U ⎝ ∂V ⎠U
Joule coefficient
lim ⎜
∆V → 0
For Ideal gas
⇒
ηJ = 0
dU = CV dT
U(T)
exactly
Always for ideal gas
only depends on T
The internal energy of an ideal gas depends only on temperature
Consequences
⇒
⇒
∆U = 0
∆U = ∫ CV dT
For all isothermal expansions or
compressions of ideal gases
For any ideal gas change in state
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #3
H(T,p)
Enthalpy
page 5
H ≡ U + pV
Chemical reactions and biological processes usually take place under
constant pressure and with reversible pV work. Enthalpy turns out
to be an especially useful function of state under those conditions.
reversible
=
gas (p, T1, V1)
gas (p, T2, V2)
const . p
U1
U2
∆U = q + w = qp − p ∆V
∆U + p ∆V = qp
define as H
∆U + ∆ ( pV ) = qp
H ≡ U + pV
Choose
H (T , p )
∂H
•
∆H = q p
⇒
⎞
What are ⎛⎜
⎟
⎝ ∂T ⎠ p
⎛ ∂H ⎞
⎜
⎟
⎝ ∂T ⎠ p
∆ (U + pV ) = qp
⇒
for a reversible constant p process
⎛ ∂H ⎞
⎛ ∂H ⎞
dH = ⎜
⎟ dp
⎟ dT + ⎜
⎝ ∂T ⎠ p
⎝ ∂p ⎠T
⎛ ∂H ⎞
and ⎜
⎟ ?
⎝ ∂p ⎠T
⇒
for a reversible process at constant p (dp = 0)
⇒
dH = đq p
⇒
⎛ ∂H ⎞
and dH = ⎜
⎟ dT
⎝ ∂T ⎠ p
∂H ⎞
⎟ dT
⎝ ∂T ⎠ p
đq p = ⎛⎜
∴
but
⎛ ∂H ⎞
⎜
⎟ = Cp
⎝ ∂T ⎠ p
đqp = C p dT
also
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
•
⎛ ∂H ⎞
⎜
⎟
⎝ ∂p ⎠T
Lecture #3
⇒
page 6
Joule-Thomson expansion
adiabatic, q = 0
porous partition (throttle)
gas (p1, T1)
w = pV
1 1 − pV
2 2
⇒
∴
gas (p2, T2)
=
∆U = q + w = pV
1 1 − pV
2 2 = −∆ ( pV
∆U + ∆ ( pV ) = 0 ⇒
∴
)
∆ (U + pV ) = 0
∆H = 0
Joule-Thomson is a constant Enthalpy process.
⎛ ∂H ⎞
⎟ dp
⎝ ∂p ⎠T
dH = C p dT + ⎜
⇒
⎛ ∂H ⎞
⎛ ∂T ⎞
⎜
⎟ = −C p ⎜
⎟
⎝ ∂p ⎠T
⎝ ∂p ⎠H
Define
∴
⎛ ∂H
⎜⎜
⎝ ∂p
⎛ ∂H ⎞
⎟ dpH
⎝ ∂p ⎠T
C p dT = − ⎜
⇒
← can measure this
⎛ ∆T ⎞
⎜
⎟
⎝ ∆p ⎠ H
⎛ ∆T ⎞
⎛ ∂T ⎞
lim ⎜
=⎜
⎟
⎟ ≡ µJT ← Joule-Thomson Coefficient
∆p → 0
⎝ ∆p ⎠H ⎝ ∂p ⎠H
⎞
⎟⎟ = −C p µJT
⎠T
and
dH = C p dT − C p µJT dp
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #3
U(T),
For an ideal gas:
page 7
pV=nRT
H ≡ U (T ) + pV = U (T ) + nRT
H (T )
only
⎛ ∂H
⎜
⎝ ∂p
⇒
depends on T, no p dependence
⎞
⎟ = µJT = 0 for an ideal gas
⎠T
For an ideal gas C p = CV + R
⎛ ∂H ⎞
⎟ ,
⎝ ∂T ⎠ p
⎛ ∂U ⎞
⎟
⎝ ∂T ⎠V
Cp = ⎜
CV = ⎜
H = U + pV ,
↓
⎛ ∂H ⎞
⎛ ∂U ⎞
⎜
⎟ =⎜
⎟ +R
∂
∂
T
T
⎝
⎠p ⎝
⎠p
↑
↑
⎛ ∂U ⎞ ⎛ ∂V ⎞
C p = CV + ⎜
⎟ ⎜
⎟ +R
⎝ ∂V ⎠T ⎝ ∂T ⎠ p
= 0 for ideal gas
∴
C p = CV + R
pV = RT