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Adiabatic Free-ExpansionA
B
vacuum
Q = 0 <— adiabatic
thermal energy transfer to gas?
work done by gas?
W = 0 <— free
1st Law: ΔU =Q −W = 0
eqn of state: U = 32nRT so ΔU = 3
2nRΔT
and TA = TB
Adiabatic Free-Expansion: Entropy Change
ΔSgas =dQrev
Ti
f
∫ = ?process is irreversible — the gas will not spontaneously occupy one half of container
—> entropy change must bepositive
P
V
A
B
to calculate change in entropy, connect A and B by a reversible, quasi-static isothermal “model” process
ΔSgas: adiabaticfree−expansion
= ΔSgas: quasi−staticisothermalexpansion
= nR lnVBVA
> 0
reservoir
gasQ
VA VB
quasi-static isothermal expansion
System Entropy ChangeA
B
vacuum ΔSsystem = ΔSgas = nR lnVBVA
> 0 irreversible
ΔSsystem = ΔSgas + ΔSreservoir
gas
adiabatic free-expansion
reversible
ΔSsystem = +QT
+ −QT
= 0
Work Done by Gas
ΔSsystem > 0
A
B
vacuum
irreversible adiabatic free-expansion
ΔSsystem = 0
reservoir
Q
VA VB
reversible quasi-static isothermal expansion
gas
<— entropy change —>
Wa. f .e = 0 Wq.s.i.e. = nRT lnVBVA
<— work done —>
Note that energy of each system did not change, yet in one case the gas did work while in the other it did not
System ISystem II
Availability of Energy, or “Lost Work”• work could have been done by System I
—> the quality of the energy was lowered byfree-expansion/irreversible process
• in an irreversible process energy equal to becomes unavailable to do work, where is thetemperature of the coldest available reservoir—> this is known as “lost work”
TCΔSUniverseTC
Wlost = TCΔSUniverse
Entropy as a Measure of Quality
• example – heat engine:—> extracts thermal energy at high temperature—> exhausts thermal energy at low temperature
this energy is of lower quality because we need an even lower temperature reservoir to convert
some of it to work• given an amount of energy, it is high quality & lowentropy if it is localized, coherent, or at a high temp
QHQL