EXPANSIONS, THERMODYNAMIC CYCLES V T

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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 #4
page 1
EXPANSIONS, THERMODYNAMIC CYCLES
•
Reversible Adiabatic Expansion (or compression) of an Ideal Gas
1 mole gas (V1,T1) = 1 mole gas (V2,T2)
đq = 0
Ideal gas ⇒
adiabatic ⇒
From 1st Law
∴
CV dT = − pdV
T2
CV ∫
T1
⇒
p =RT
V
V dV
dT
= −R ∫
V V
T
Define
2
dU = -pd V ⇒
CV
đw = -pd V
CvdT = -pdV along path
dT
dV
= −R
T
V
R CV
⇒
1
Cp
γ ≡
CV
Reversible ⇒
dU = Cv d T
⎛T2 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
Cp
C p −CV =R for i.g.
⎯⎯⎯⎯⎯⎯→
⎛T2 ⎞ ⎛V1 ⎞CV
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
γ −1
⇒
⎛T2 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
3 ⎫
For monatomic ideal gas:
CV = R ⎪
2 ⎪
5
( > 1 generally)
⎬ γ =
5 ⎪
3
Cp = R
2 ⎪⎭
[More generally the heat capacity for a molecular species has a
temperature dependence that can be approximated as
C p (T ) = a + bT + cT 2 with a, b, and c tabulated.]
In an adiabatic expansion (V2 > V1), the gas cools (T2 > T1).
And in an adiabatic compression (V2 < V1), the gas heats up.
−1
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 #4
pV
T =
R
For an ideal gas (one mole)
pV γ
page 2
γ
⇒
⎛ p2 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝ p1 ⎠ ⎝V2 ⎠
γ
γ
⇒ pV
1 1 = pV
2 2
is constant along a reversible adiabat
Recall, for an isothermal process T = constant ⇒
pV = constant
p
isotherm
pV=constant
p1
pVγ=const.
p2
adiabat
V2adiabat < V2isotherm
because the gas
cools during reversible
adiabatic expansion
V2ad V2iso
V1
• Irreversible Adiabatic Expansion of an ideal gas against a constant
external pressure
1 mol gas (p1,T1) = 1 mol gas (p2,T2)
adiabatic
Constant pext = p2
Ideal gas
1st Law
∴
Integrating:
Using pV = RT
⇒
⇒
⇒
⇒
(pext=p2)
đq = 0
đw = -p2d V
dU = Cv d T
dU = -p2 dV
Cv d T = - p 2 dV
Cv ( T2 - T 1 ) = - p 2 ( V2 - V1 )
⎛
T2 (CV + R ) =T1 ⎜ CV +
⎝
p2 ⎞
R⎟
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 #4
p2 < p1
Note
⇒
T2 < T1
(-wrev) > (-wirrev)
Note also
page 3
Again, expansion cools
as expected, less work is
recovered through an
irreversible process
Thermodynamic Cycles
•
Reversible Ideal Gas processes:
Find ∆U , ∆H , q , w
p
p1
(I)
(T1)
p1
A (isotherm)
p2
p3
B
(T1)
(rev.
adiabat)
(T2)
V1
(II)
p
C (const. V)
p2
V2
(T1)
D (const. p )
A
(isotherm)
V1
(T3)
E (const. V)
(T1)
V2
For (I)
1gas(p1,V1,T1)
A
1gas(p2,V2,T1)
B
C
1gas(p3,V2,T2)
There are two paths from initial to final states (A) and (B+C). As far
as functions of states (e.g. U, H) are concerned it doesn’t matter
which path is taken.
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
[A]
Lecture #4
1 mol gas (p1,V1,T1)
const. T
=
V
wA = −RT1 ln 2
V1
[B]
1 mol gas (p1,V1,T1)
Adiabat:
Ideal gas:
1st Law:
[C]
1 mol gas (p2,V2,T1)
∆UA = 0
Ideal gas isotherm:
page 4
∆HA = 0
V
qA = RT1 ln 2
V1
rev.adiabat
=
1 mol gas (p3,V3,T2)
qB = 0
∆UB = CV (T2 −T1 )
∆HB = C p (T2 −T1 )
w B = CV (T2 −T1 )
1 mol gas (p3,V2,T2)
reversible
=
const. V
1 mol gas (p2,V2,T1)
Constant V:
wC = 0
1st Law:
qC = CV (T1 −T2 )
Ideal gas:
∆UC = CV (T1 −T2 )
∆HC = C p (T1 −T2 )
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
[A]
Lecture #4
vs.
∆UA = 0
∆HA = 0
qB + qC = CV (T1 −T2 ) ≠ qA
wB + w C = CV (T2 −T1 ) ≠ w A
∆UD = CV (T3 −T1 )
∆HD = C p (T3 −T1 )
∆UE = CV (T1 −T3 )
∆HE = C p (T1 −T3 )
[E]
[A]
∆UA = 0
∆HA = 0
V
qA = RT1 ln 2
V1
V
w A = −RT1 ln 2
V1
[B] + [C]
∆UB + ∆UC = 0 = ∆UA
∆HB + ∆HC = 0 = ∆HA
V
qA = RT1 ln 2
V1
V
w A = −RT1 ln 2
V1
[D]
page 5
vs.
qD = C p (T3 −T1 )
wE = 0
wD = −R (T3 −T1 )
qE = CV (T1 −T3 )
[D] + [E]
∆UD + ∆UE = ∆UA
∆HD + ∆HE = ∆HA
qD + qE = R (T3 −T1 ) ≠ qA
wD + w E = −R (T3 −T1 ) ≠ w A
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