5.60 Thermodynamics & Kinetics: Mit Opencourseware
5.60 Thermodynamics & Kinetics: Mit Opencourseware
5.60 Thermodynamics & Kinetics: Mit Opencourseware
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5.60 Spring 2008 Lecture #9 page 1
Entropy
• For a reversible ideal gas Carnot cycle:
−w q2rev T
Efficiency ε= = 1 + rev = 1 − 2
qrev q1 T1
q1 q2 đqrev
⇒ +
T1 T2
=0 ⇒ v∫ T
=0
T1 (hot) ( −w ) ( −w ′ )
ε= ε′ =
q1 q1 q1′
w w
Assume ε ′ > ε (left
Some
reversible q2 q2 Carnot engine less efficient
engine T2 (cold) cycle than Carnot cycle)
Since the engine is reversible, we can run it backwards. Use the work
(-w’) out of the Carnot engine as work input (w) to run the left engine
backwards.
∴ Total work out = 0 (-w’ = w > 0)
−w ′ −w w −w w
But ε ' > ε ⇒ > ⇒ > = ⇒ q1 < −q1′ since q1 < 0, q1′ > 0
q1′ q1 q1′ q1 −q1
⇒ − (q1′ + q1 ) > 0
This contradicts the 2nd law (Clausius). This says that we have a net
flow of heat into the hot reservoir, but no work is being done!
T2
∴ The efficiency of any reversible engine is ε = 1 −
T1
5.60 Spring 2008 Lecture #9 page 2
đqrev
∴ For any reversible cycle v∫ =0
T
đqrev 2 đqrev
dS = ⇒ ΔS = S2 − S1 = ∫
T 1 T
p
1→2
1 irreversible
isotherm with pext = p2
( −w )irrev < ( −w )rev ⇒ wirrev > wrev
adiabat 2 ΔU = qirrev + wirrev = qrev + w rev
adiabat
4
isotherm (rev.) 3 ∴ qirrev < qrev
q2rev q2rev
ε irrev = 1 + irrev < 1 + rev = ε rev (q2 < 0)
q1 q1
⎧ đqrev
đq ⎪⎪ v∫ T = 0
• Leads to Clausius inequality v∫ T ≤0 contains ⎨
⎪ đqirrev < 0
⎪⎩ v∫ T
(A) irreversible
(A): The system is isolated and
1 2 irreversibly (spontaneously) changes
(B) reversible from [1] to [2]
đq
=0 !
2 đqirrev 1 đqrev
Clausius v∫ T ≤0 ⇒ ∫ +∫ ≤0
1 T 2 T
1 đqrev
⇒ ∫ = S1 − S2 = −ΔS ≤ 0
2 T
∴ ΔS = S2 − S1 ≥ 0
1 2 ΔS = S2 − S1 independent of path
(b) Reversible:
Initially T1 ≠T2
dS = dS1 + dS2 =
đq1 đq2 (T2 −T1 )
T1
−
T2
= đq1
TT
( đq1 = −đq2 )
1 2
adiabatic
1 mol gas (V,T) =
1 mol gas (2V,T)
ΔU = 0 q=0 w=0
đqrev đw V RdV 1
ΔSbackwards = ∫ = −∫ = ∫V = R ln
T T 2 V 2