任意可逆循环的热温比的和为零

∆S = ∆S o + ∆S i = 0.01( J / K )
Qi Qo + =0 Ti To
An arbitrary reversible cycle can be approximated by a series of Carnot cycles . For each of the small Carnot cycle , we know that :
∆Qi ∆Qo + =0 Ti To
Entropy
P 330 - 334 1. Energy and entropy In studying the efficiency of a Carnot engine we learned that if an ideal engine absorbs an amount of heat Qi at temperature Ti and then discards heat Qo at To , the quantities are related by
For an irreversible cycle To j Qij Qoj − Qoj or + <0 1− < 1− Tij Toj Qij Tij Summing over all such cycles , and passing to the limit we obtain ：
dQ ∫ T <0
Example 1 An ideal gas is allowed to expand freely and adiabatically from a volume V1 to a volume V2 . What is the change in entropy for this irreversible process ? ( A free expansion is one in which no work is done . ) P 335 We compute the change in entropy by considering a reversible process connecting the two states : a quasi - static isothermal expansion from V1 to V2 at constant temperature .
∆S = ∫
V2
V1
dQ V2 pdV V2 nR V2 =∫ =∫ dV = nR ln V1 V1 V V1 T T
Example 2 A copper wire connected between two large pieces of metal conducts 40J of heat from one piece at 400K to the other at 350K . What is the change in entropy for this process ? P 337
j j
By letting the heat become arbitrarily small , the sum approaches an integral
dQ ∫ T =0
For an reversible cycle , the line integral taken around a complete cycle is zero . In the other words , for an reversible cycle the line integral between two states depends only on those states not on the path connecting them . 可逆循环热温比的积分与路径无关
The change in entropy of the hot piece of metal is :
Q − 40 ∆S i = = = −0.1( J / K ) Ti 400
The change in entropy of the other metal piece is : Q 40 ∆S o = = = 0.11( J / K ) To 350 The overall change in entropy is :
P
This equation means that in any irreversible process the entropy of an isolated system always increase .
irreversible R1 B
A reversible R2
o
V
不可逆过程的熵总是增加的
Summing over all the isothermal paths along which heat ∆Qj is either absorbed or rejected at temperature Tj , we obtain : 任意可逆循 环的热温比 ∆Q j ∑ T =0 的和为零
SB − S A > 0
The second law of the thermodynamics can be cast into another form : The entropy of the universe always increases towards a maximum .
heat death theory （热寂说） 热寂说）
To understand why the line integral is independent of the path , consider the system moving reversibly along path R1 from A to B , then back to A along a different reversible path . We have
B
A
A reversible R2
o
B
V
B
or
∫
( R1 )
d QI d QR < ∫ so T A T ( R2 ) A
∫
( R1 )
d QI < SB − S A A T
证明不可逆过程的熵总是增加的
B
∫
d QI < SB − S A A T
For an isolated system , the entropy of the system increases after any irreversible process .
Consider a irreversible process shown in the diagram .
B
∫
( R1 )
d QI d QR + ∫ <0 T A T ( R2 ) B
B
A
P
irreversible R1 B
or∫Biblioteka ( R1 )d QI d QR <− ∫ T A T ( R2 ) B
B
dQ ∆S = S B − S A = ∫ A T
由于任意可逆循环的热温比的积分与路径无关， 由于任意可逆循环的热温比的积分与路径无关， 将这个积分叫做熵变。熵是状态参量， 将这个积分叫做熵变。熵是状态参量，熵与能 量同样重要。 量同样重要。
B
What is the change in entropy for an irreversible process ? An irreversible process can not be represented by a continuous path on a PV diagram because it does not move through a series of equilibrium states . However , we can join the P irreversible R 1 ends of an irreversible B process on a PV diagram by a reversible process , then approximate the entire cycle A reversible R2 by a series of small cycle . V o 不可逆过程的熵变？ 不可逆过程的熵变？
dQ dQ ∫ ) A T + ( R∫ ) B T = 0 ( R1 2
B A
P
R1
A
B
R2
Because each path is reversible
dQ dQ ∫ ) A T = −( R∫ ) B T ( R2 2
B A
o
V
dQ dQ = ∫ so that ∫ T T ( R1 ) A ( R2 ) A
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B
B
证明可逆循环 热温比的积分 与路径无关
dQ In a reversible process , the line integral ∫ A T is independent of the path , therefore , it has physical significance . The line integral is called the change in entropy between any two states A and B connected by a reversible process .