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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
•
Lecture #5
page 1
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
2
dU = -pd V ⇒
Define
CvdT = -pdV along path
dT
dV
= −R
T
V
CV
Cp
R CV
⇒
1
Cp
γ ≡
CV
đw = -pd V
Reversible ⇒
dU = Cv d T
⎛T2 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
C p −CV =R for i.g.
⎯⎯⎯⎯⎯⎯→
⎛T2 ⎞ ⎛V1 ⎞CV
⎜ ⎟=⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
−1
γ −1
⇒
⎛T2 ⎞ ⎛V1 ⎞
⎜ ⎟ =
⎜ ⎟
⎝T1 ⎠ ⎝V2 ⎠
3 ⎫
For monatomic ideal gas:
CV = R ⎪
2 ⎪
5
( > 1 generally)
⎬ γ =
5 ⎪
3
Cp = R
2 ⎪⎭
In an adiabatic expansion (V2 > V1), the gas cools (T2 > T1).
And in an adiabatic compression (V2 < V1), the gas heats up.
For an ideal gas (one mole)
pV γ
pV
T =
R
γ
⇒
⎛ p2 ⎞ ⎛V1 ⎞
⎜ ⎟=⎜ ⎟
⎝ p1 ⎠ ⎝V2 ⎠
γ
γ
⇒ pV
1 1 = pV
2 2
is constant along a reversible adiabat
For an isothermal process
T = constant
⇒
pV = constant
5.60 Spring 2008
Lecture #5
page 2
p
V2adiabat < V2isotherm
because the gas
cools during reversible
adiabatic expansion
p1
p2
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:
⇒
⇒
⇒
⇒
đq = 0 đw = -p2dV
dU = CvdT
dU = -p2dV
Cv d T = - p2 dV
Cv ( T2 - T 1 ) = - p2 ( V2 - V1 )
⎛
Using pV = RT
Note
(pext=p2)
p2 < p1
T2 (CV + R ) =T1 ⎜ CV +
⎝
⇒
T2 < T1
Note also (-wrev) > (-wirrev) p2 ⎞
R⎟
p1 ⎠
Again, expansion cools
Less work is recovered through
an irreversible process
5.60 Spring 2008
Lecture #5
page 3
Some Thermodynamic Cycles
•
Reversible Ideal Gas processes: Find ∆U , ∆H , q , w ,
p
p1
p2
p3
(I)
(T1)
B
(T1)
(rev.
adiabat)
(T2
)
[A]
C (const. V)
1 mol gas (p1,V1,T1)
V
wA = −RT1 ln 2
V1
1 mol gas (p1,V1,T1)
Adiabat:
Ideal gas:
1st Law:
D (const. p)
E (const. V)
(isotherm)
(T1)
V1
const. T
=
V
qA = RT1 ln 2
V1
=
V2
1 mol gas (p2,V2,T1)
∆UA = 0
rev.adiabat
(T3)
A
p2
V2
Ideal gas isotherm:
[B]
(T1)
p1
A (isotherm)
V1
(II)
p
∆H A = 0
đq
∫T
1 mol gas (p3,V3,T2)
qB = 0
∆UB = CV (T2 −T1 )
∆HB = C p (T2 −T1 )
w B = CV (T2 −T1 )
V2
V1
=R ln
đqB
∫T
=0
đq
∫T
5.60 Spring 2008
Lecture #5
1 mol gas (p3,V2,T2)
[C]
reversible
wC = 0
1st Law:
qC = CV (T1 −T2 )
đqC
∫T
vs.
∆UA = 0
∆HA = 0
∆UC = CV (T1 −T2 )
∆HC = C p (T1 −T2 )
[B] + [C]
∆UB + ∆UC = 0 = ∆UA
∆HB + ∆HC = 0 = ∆HA
V
qA = RT1 ln 2
V1
V
w A = −RT1 ln 2
V1
đqA
Ideal gas:
⎛T ⎞
= CV ln ⎜ 1 ⎟
⎝T2 ⎠
[A]
∫T
1 mol gas (p2,V2,T1)
=
const. V
Constant V:
page 4
V2
V1
= R ln
qB + qC = CV (T1 −T2 ) ≠ qA
wB + w C = CV (T2 −T1 ) ≠ w A
đqB
∫T
+∫
đqC
T
⎛V ⎞
đq
= R ln ⎜ 2 ⎟ = ∫ A
T
⎝V1 ⎠
⎛ đq ⎞
This result suggests that ⎜ ∫ rev ⎟ is a state function!
⎝ T ⎠
[D]
∆UD = CV (T3 −T1 )
∆HD = C p (T3 −T1 )
wD = −R (T3 −T1 )
qD = C p (T3 −T1 )
đqD
∫T
⎛T ⎞
= C p ln ⎜ 3 ⎟
⎝T1 ⎠
5.60 Spring 2008
Lecture #5
∆UE = CV (T1 −T3 )
∆HE = C p (T1 −T3 )
[E]
wE = 0
đqE
∫T
[A]
[D] + [E]
∆UD + ∆UE = ∆UA
∆HD + ∆HE = ∆HA
V
qA = RT1 ln 2
V1
V
w A = −RT1 ln 2
V1
đqA
∫T
qE = CV (T1 −T3 )
⎛T ⎞
= CV ln ⎜ 1 ⎟
⎝T3 ⎠
vs.
∆UA = 0
∆HA = 0
page 5
V2
V1
= R ln
qD + qE = R (T3 −T1 ) ≠ qA
wD + w E = −R (T3 −T1 ) ≠ w A
đqD
∫T
+∫
đqE
T
⎛V ⎞
đq
= R ln ⎜ 2 ⎟ = ∫ A
T
⎝V1 ⎠
⎛ đq ⎞
Here again ⎜ ∫ rev ⎟ looks like a state function!
⎝ T ⎠