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# 提升宝宝视觉一空间智力的良方

PART ONE THERMODYNAMICS

QQ 空间

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Chapter I General Laws of thermodynamics 热力学的基本规律
?热力学研究的对象是由大量微观粒子组成的宏观系统， 它们与外界的相互作用表现为能量的交换和物质（粒子） 的交换。由此分为三个系统，如表所示：

1. Equilibrium States and Equilibrium Processes
We can use some parameters(参数) to describe the state(状态) of the thermodynamic system： a gas or a liquid: P(pressure压强), V(Volume),T(Temperature) a film of a liquid: ?(surface tension coefficient),?(film area), T

a rod(棒): l, ?(cross-sectional area), f(tensile force), E(Young’s modulus) But, all these parameters describing the systems are based on “Equilibrium States”
What is Equilibrium States?

In the figure, a vessel(容器) is divided into two halves by a valve(阀). Left half: containing gas; Right half: evacuated(抽 真空的). The question is: when the valve is opened, a stream of gas will flux(流动), what state we call this? What process the system passes? Can we describe this process? If we can, How we describe?

The system will finally approach a state in which all the external parameters remain constant. Such state is called equilibrium state If the system has gradients(梯度) of macroscopic parameters (P,V,T), such a state is referred to as a non-equilibrium state The process from a non-equilibrium state to a equilibrium state is called relaxation (弛豫)

Relaxation
? 选择物理量“密度”。考虑上例中两个部分的密度变

?当t = ? 时:

1 n n ? ( n n ) ?? ? 0? ? e

? 选择物理量“温度”。考虑一个长棒，两端初始时刻

? 当时间变化为t = ? 时，温

? 1）上述的两个例子为“理想的弛豫过程”；

?2）所选的变量n和T被物理学称之为“物理控制参量”；

?3）上述过程实际上为“非平衡过程”；
?4）非平衡过程原则上不能用物理量描述； ?5）热力学的主要目的是研究“热力学过程”；

?6）实际的热力学过程会是非平衡的，怎么办？

If we slow down the rate of relaxation process, and assume that there will be enough time for the parameters to equalize over the entire system. Therefore:

quasi-state

?由于分子处于不停的热运动之中，所谓的平衡只是

?By virtue of symmetry(由于对称性)，the process may

operate in a forward and reverse direction. For reverse process an equilibrium process can be reversed in time. During the reverse process the system will pass through the same succession of states as during the forward process, but in a reversed sequence. This equilibrium process is referred to as “reversible process”.
?在时间上相反的过程为“逆过程”， ?一个过程可正可反，则为“可逆过程”。

? 1.行列式的值为0； ? 2.两个独立变量； ? 3.第三个变量可以用以上两个变量线性表示。

?任意选取两个独立变量，这两个变量为“状态参

?三个变量之间的关系称之为状态方程：f (P,V,T ) = 0 .

?对于均匀的(homogeneous)系统，状态参量选 P、V
?The state are depicted(描绘) by points and processes by

lines on a P~V plane. (系统的状态用P~V平面的点及过 程用线描述)。

Thermodynamic Units 热力学单位
? 牛(N) =kg m s-2
? 压强：帕斯卡Pa (N m-2) ?

1大气压强 (pn)= 101325 Pa ? 能量： 焦耳(J) ? 1J = 1N m ? (见中文教材P6)

2. Temperature. The temperature principle 热平衡定律和温度
? Temperature is a rather subjective (主观的) term(术语)-degree of

? ?

?
? ?

?
? ?

?
?

hotness of body. But, the term can be given a more objective meaning through a easily measured physical parameters depending on the degree of hotness. Examples: (测量温度的物质) height of the column of liquid mercury in a glass tube (汞柱); the pressure of a gas in a constant volume vessel; the resistance of a conductor; the emissive power of a red-hot body, etc. The limit of applicability(应用的限制): Mercury solidification point(凝固点); gas condensation point(凝聚点)； fusion point of metal(金属熔点)。 The fields of application of these thermometers overlap(重叠)。

Thermostat, 恒温热源
? The concept of the thermostat: ? a thermostat has heat capacity(热容) greater

than that of any reference bodies. When contact, the temperature of thermostat does not change and , on the other hand, any reference body acquires its temperature after a short relaxation time.

? (恒温热源的温度不变，与其接触的物体瞬时达到恒温

?

f AC ( p A ,V A ; pC ,VC ) ? 0 pC ? FAC ( p A ,V A ;VC )

f BC ( p B ,VB ; pC ,VC ) ? 0 pC ? FBC ( p B ,VB ;VC )

f AB ( p A ,V A ; p B ,VB ) ? 0

? Placing a vessel containing a gas into a thermostat, and changing the volume of the vessel with the help of a piston (活塞), we measure simultaneously the pressure(压强). Then, plotting the results on a P-V

plane, a isotherm curve is obtained.

P

gas in vessel

?
piston
1 2 3

?The form of isotherm curve depends on the conditional
V

temperature,?, of the thermostat. Let the isotherm corresponding to the conditional temperature be described by the equation ?(? | P,V) = 0. Solving this equation with respect to ? , we obtain ? = ? (P,V) .

?Changing the ? in a infinitely small and repeating

the procedure(流程) above, different isotherm curves are obtained as the Fig.
P
20

10

?1

?2

?3

1

?It has been shown experimentally that the isotherms

V

2

of the same gas never intersect(相交) and, consequently, the conditional temperature is a single-valued function of state specified by parameters P and V. ?Experiments also shown that at levels far from the gas liquefaction points the isotherms can be approximately represented by equiangular hyperbolas ? PV = const (2.1)

Consequently, the equation ? = ? (P,V) will be referred to as
the equation of state (状态方程). If a gas satisfies Eq.(2.1) and equation of state, this gas is called thermally ideal gas or ideal gas. The real gases close to it, when the temperature is high and density is low.

Result:
A postulate(基本原理) is introduced ------

temperature principle:
There exists (more then one) function of states of a system that remains constant for any process operating in a thermostat, referred to as the conditional temperature. (在恒温热源所进行的过程，保持不变的状态函数被称之为：

3. Entropy. The Entropy Principle
piston Heat-insulated walls

? A vessel with ideally heat-insulated walls is called: ?

? An equilibrium process in conditions of he

at

?

Varying the volume of the gas by means of the piston and measuring the pressure, we can plot a curve on P-V plane, which is a adiabat curve. If we take the vessel out temporarily and contact with a body at higher or lower temperature, then restore the vessel and repeat the procedure described above, we can plot another adiabat curve on the P-V plane, then the other.

T> T0 T = T0

piston

T < T0

P

40

20

?1
0
0

?2

?3

1

V

2

The number of the nth adiabat can be replaced by a varying parameter ? ------ conditional entropy. We represent the procedure by a equation: ? = ? ( P,V). The quantity ? remains constant along each adiabat and instead of the number n, the adiabat can be defined by setting the values of conditional entropy ?1 ?2 … . Shown: 1. adiabats of the same gas never intersect; 2. conditional entropy is a single-valued function of state.

In adiabat, the ideal gases are described by the equation PV? = const

? The experimental findings indicate: each adiabat intersects each isotherm only at one point.
80

P
60

?3 ?2

?3 ?2

40

20

?1

?1

0 0 1 2

V

The relation of two pairs of variables(变量) P、V and ?、?
? 1. one-to-one correspondence(一一对应). ? 2. For each point on the P-V plane, a isotherm

curve ? = const and a adiabat curve ? = const pass through that point. ? 3. The state of the gas can be specified equally by taking a pair of number ?、? as a pair of numbers P、V . 气体状态可以在P-V和?-?平面等价表示.
?
?

? = ? ( P,V): caloric equation of the state of a substance

P
40

intersecting point
20

0 0 1

V

2

? ? = Cp/Cv > 1 ? Each adiabat intersects each isotherm at one and only

one point. At that point, PV = a, and PV ? = b. Solution:

Result:
A postulate(基本原理)--- the entropy principle:
There exists a single-valued function of state, remaining constant for any processes operating in a thermally-insulated vessel, called the conditional entropy. It depends only on the product PV ?. “绝热过程中存在着一个单值的、不变的状态函 数 ------ 条件熵,它仅仅依赖于乘积项PV ? ”

4. Absolute Temperature and Absolute Entropy
? define accurately the transformations

? ’ = f1(? ) and ? ’ = f2(?),
which is monotonically increasing function of their arguments.
?According to the one-to-one correspondence of

the pairs of variables P-V and ?-? on the P-V and ?-? planes. The difference of the Jacobian transformation is applicable
D? ? ( P, V ) ? (? , ? )

Preceding 前面的， monotonically 单调地， argument 自变量

The geometrical meaning of the Jacobian lies in that its module gives the variation factor from P-V plane to ?-? plane.
dPdV ? ?? | D | d?d?
? There is a complete

equivalence for the two planes,

and the | D | = 1. This normalization (归一化) leads to a defined selection of the temperature and entropy scale. ? PV均匀有确定值，根据| D | = 1，可以推断， ?-? 也应该有确定值，即绝对值:即PVD的确定 导致了?、?的确定值为ST。 absolute temperature and absolute entropy.

? ( P, V ) ?1 ? (T , S )

? (T , S ) ?1 ? ( P, V )

(4.1)可以确定T、S：

? 在等温方程和绝热方程已知的条件下，

? (T , S ) ? T ' S ' (? ? 1) y ? 1, ? ( P, V ) T' ? 1 S ' (? ? 1) y (4.2)

? (4.2)式的意义： ? T ’ 和S ’ 表示它们对自变量的偏导数: T ’=? T /? x, S ’=? S /? y.
? (4.2)式的右边方程中，the

right-hand side depends only on y and left-hand side only on x. Each side is equal to 1/R
1 T ( x) ? , R
'

R S ( y) ? , (? ? 1) y
'

(4.3)

?(4.3)式的求解：
PV T ? ?? , R R S ? ln( PV ? ) ? C ? ?1 ( 4.4)

?如果我们知道了任意一点的状态S0、P0、V0、T0，就

PV ? RT , S ? S0 R T ? P1?? ? ln ? 1?? ? ? 1 T0 P 0 ( 4. 6 )

? 由此确定了理想气体的绝对温度和绝对熵 ? Now, look at on P-V and T-S planes the isochore,

isobar, isotherm, isoentropic curves (adiabat) for a perfect gas.(V、P、T、S = const)

P

V = const

T

S = const P = const

P = const T = const S = const

T = const V = const

V

S

On the T-S plane, P and V are expressed by the equation:
PV ? RT , PV ? ? conste(? ?1) S / R

? Eliminating V or P from above equations, the

isochores and isobars on the T-S plane:

V ? const, T ? T1 (V )e (? ?1) S / R P ? const, T ? T2 ( P )e (? ?1) S / ?R
? Which is the final expression ? Homework : question one and two.

5. Work 功
? In mechanics(力学)， the work is determined by formula:

?A ? f dx

? In thermodynamics, a perfect gas undergoes expansion

or compression(压缩), the work is determined by the quantities P and V, formula:

?A ? PdV ,
P

A?

V2 ?V1

PdV

(5.2)

?此表达式为系统对外作功。

V1 V

? The pressure of a gas is not determined only by V, but

by the state of the gas(V and T) .
? So, the pressure is concerned with the process,

consequently, the work involved in the process is not determined by the initial and final states of the gas but depends on the entire course of the process.
? ?A denoting by the infinitely small amount of work. ?Results:

?T,S,P,V

are functions of the state of the gas.
?Work is a function of the process, expressing

the compression or expansion of the gas .

? 1)the isothermal process:
P V1 ? P2V2 ? PV , 1 i.e. P? P 1V 1 V
V2 V2 AT ? ?V PdV ?P V1 ?V 1 1 1

? ? ?

1 V dV ?RT ln 2 V V1
i.e. PV1 P? 1 ? V
?

P V1 ? P 1 2V2 ? PV ,
V2 AS ? P V1 ?V 1 1

?

? 3)the isobaric process: P 1 ? P 2 ? P

1 RT1 V1 ? ?1 dV ? [ 1 ? ( ) ] ? ?1 V2 V?

? 作业：
? 英Page 25

V2 AP ? P ?V dV ? P (V2 ? V1 ) 1

? 4)the isochoric process: ? Answer the Problem AV ? 0 V1 ? V2 ? V ? 中P66：1.6，1.7， 1.9

6. Adiabatic and Isothermal Potentials Mechanics:力学 , coordinate: 坐标, analogy:类比
? In one dimension case of the mechanics, the

force f is a function of coordinate x, the work , ?A = f d x, can be represented as the decrease of a function U(x), which is referred to the potential of force f or potential energy.
? ?A = f d x= - d U(x), f = -?U/? x ?U(x) is an opposite sign of work; 因为是对外作功

?In thermodynamics, analogy, f?P and x?V,

there is a thermodynamics potential:
?

P = -?U/? V

? 结论：系统对外作功会改变系统的势能!

? 焦耳从1840年开始，在20多年的时间内，反复进行了

? 实验1. 在绝热的水箱中，重物下降带动叶片转动，使

? 实验2. 在绝热的水箱中，电流通过电阻器使水温升高。

? 结果发现：使物体升高一定的温度，所需的功在实验误

U B ?U A ? W

? 此态函数被称之为“内能”

? Equate the work done in an adiabatic process to the

decrease in the adiabatic potential U, denoted by 绝热过程做的功＝绝热势的减少
? In an arbitrary non-adiabatic process, the general equation is:

? AS ? PdVS ? ?dU S

dU ? ? PdV ? ? dS

(6.2)

(在非绝热过程中，应该考虑熵S的变化量。)

? Considering a total differential expression(全微分表达式) ? ?? ? ? ?P ?
? ?? ? ? ? ? V ? S ? ?S ? ?V

? Translated to Jacobian form, and use its property, we find
? (? , S ) ? ( P, V ) ? , ? (V , S ) ? (V , S ) ? (? , S ) ?1 ? ( P, V )

? ?? ? ? ? ? 1, i.e. ? ?T ? S

? ? T ? ?( S )

? In (? ? / ? T)S, S is fixed. So, ? (S) is an arbitrary function

of entropy, and ? ? (S) dS is an arbitrary function too. (? (S) 是任意的， ? ? (S) dS 也是任意的)

?For normalizing U, assuming ? (S) = 0, ? = T, then

dU = T dS – P dV
derivatives with respect to two vari

ables are

(6.3)

?Adiabat potential,U, has its own variables S and V, the

T = (?U/ ?S)V , P = – (?U/ ?V)S

(6.4)

Isothermal process
? Equating the work done in an isothermal process to the

decrease in the isothermal potential F, denoted by 等温过程做的功＝ 等温势的减少

? AT ? PdVT ? ?dFT
? In an arbitrary non-isothermal process, the general equation is:

dF ? ? dT ? PdV
? Considering a total differential expression(全微分表达式) ? ?? ? ? ?P ?
? ?? ? ? ? ? V ? T ? ?T ? ?V

? Translated to Jacobian form, and use its property, we find
?( ? , T ) ? ( P, V ) ? , ? (V , T ) ? (V , T ) ?( ? , T ) ?1 ? ( P, V )

? ?? ? ?? ? ? 1, ? ?S ?T

i.e.

? ? ? S ? ? (T )

? In (? ? / ? S)T, T is fixed. So, there is an arbitrary function

of ?(T), ? and then isothermal potentials F.

(? (T) 是任意的， ?是任意的， F 也是任意的) ?For normalizing F, assuming ? (T) = 0, ? = - S. then

dF = -S dT – P dV
derivatives with respect to two variables are

(6.5)

?Isothermal potential,F, has its own variables T and V, the

S = - (?F/ ?T)V , P = – (?F/ ?V)T
?From (6.3) and (6.5), we find

d(U-F) = T dS +S dT = d(TS) F = U - TS

7.The Energy Principle. Supply and Removal of Heat

? Introducing a concept of the “internal energy”(内能),

defining it as the total energy of the thermodynamic system.
? 改变 内能的两种方法： ? (1)performing work ?A on a gas by applied forces; ? (2)contacting with a body heater or cooler, the

system will obtain or loss the amount of energy, the quantity of heat, denoted(定义为) by the symbol ? Q.

?This assertion, being in essence the law of conservation of

energy, will be referred to as the energy principle or the first law of thermodynamics. ?这个基于能量守恒原理的论断被称之为“能量原理”或 “热力学第一定律”。 ?值得注意的是，在绝热过程中，系统对外作功消耗了内 能。因系统对外作的功等于绝热势能的减少。故绝热势 能与内能是一致的。 ?The adiabatic potential U is identical to internal energy.
dU ? ?Q ? ?A ? ?Q ? PdV (7.1)

?Q ? TdS 热量与微“熵”的关系是：

dU ? TdS ? PdV

(7.2)
(7.3)

?A special form of the law of conservation. ?It is valid for any processes: equilibrium and non-

equilibrium ?方程(7.3)所涉及的内容比方程(7.1)更加广泛。 因为熵具有更深刻的含意。 ?(7.3) is often referred to as the basic thermodynamic identity for equilibrium processes. ?(7.3)式被称之为热力学恒等式。 ??(T,S)/?(P,V)=1 is a corollary(推论) of Eq.(7.3)。

? 比较两个公式：

? A ? PdV

?Q ? TdS

? Gibbs(吉布斯) suggested that V is a mechanical

coordinate(力学坐标) in

work, S and T is a thermodynamic coordinate and a thermodynamic force, respectively.
? 为系统加热相当于为系统增加热力学坐标---熵。
? 平衡过程的热力学相当于力学的扩展---被认为是

? 焦耳的实验：

? 1840年焦耳做了自由膨胀的实验：
? 两个容器均浸没在水中。实验的目的是要检测气体自由

?分析：打开活门，气体扩散。在扩散过程中，不受任何

?在理想气体的自由扩散过程中，内能与体积无关，U(T，V) = U(T)。 ?焦耳定律
? ?U ? ? ?V ? ? ?T ? ? ? ? ? ? ? ? ?1 ? ?V ?T ? ?T ?U ? ?U ?V
?T ? ? ? ?V ?U ? 0

?焦耳系数： ? ?

?导致

? ?U ? ? ? ? ?V ?T

? 0

? The isothermal potential F is also called the free energy(自由能), the

difference U – F = TS,the bound energy(束缚能). ? F 被看成是在等温过程中内能花在产生功上的一部分。

TdS

gas in vessel

PdV
piston

? Since temperature is a function not only of

entropy, but also of the state of a gas (for instance, S, V). The area (quantity of heat) depends not only on the initial and final states, but also on the entire course of the process.
? 温度 T 不是 S 的单值函数，
T

1 2

S S S ? 中间曲线 I的吸热不同于其它两条曲线，即吸热

? PdV and TdS are not total differentials(全微分), only t he

difference(差) TdS-PdV is a total differential of the internal energy, that a function of the state of a gas.

P

T

V ? 设两个图的初、终两点表示相同的两个状态，虽然三

V1

V2

S1

S

S2

? 热力学第一定律的物理含义

? P-V图上的曲线为过程量，因为隐含的S、T不确定。

? 一旦参量确定，则过程确定。但描述过程的是用热或功，
? ? ?

?
?

? Isothermal process T= constant
S2 QT ? ?S TdS ? T ( S 2 ? S1 ), 1

P V1 ? P law) 1 2V2 ( Boyle QT ? RT V RT ln( 2 ) ? ?1 V1

R PV ? S ? S0 ? ln( ), ? ? ?1 P V 0
0

?Isobaric process P = constant
dS ? R?dV , ? ?1 Qp ?

?R (T2 ? T1 ) ? ?1
R(T2 ? T1 ) ? ?1

?Isochoric process
dS ? RdP , (? ? 1) P QV ?

QS ? 0

?作业：证明上述结论

? 1. 玻意耳定律(1662年)：

? 等温条件下，PV为常数。
? 2. 阿伏加德罗定律(1811年)：

? 相同的T、P条件下，相等体积所含的摩

8. Heat Capacity of Gases(热容量)
? Define: the quantity of heat supplying to the

gas to raise its temperature by one degree.
C ?

?Q ?T

? Function: State or Process ? ? Process Function ! ? There are different heat capacities

depending on different process. ? Such as CP, CV, CS, CT 。。。Infinite set . ? But CS = 0 and CV = 1/0

Is negative heat capacity possible ?
? Yes, it is possible for two having opposite signs. ? For example, heat a gas, and do the work, but

? W>?Q, here ?Q >0, and ?T<0.

? We only focus on two heat capacities: CP,
? From definition, there are
CV ? T ( ?S )V , ?T CP ? T ( ?S )P ?T

CV.

? According to the calculation of expression of entropy
R CV ? , ? ?1 R? CP ? ? ?1
CP ??, CV C P ? CV ? R

? dF = -S dT – P dV 描述了等温过程的作功量。
? dU = T dS – P dV 描述了绝热过程的作功量。
? 反之，如何描述热量？ ? 过程的需要：the heat quantity in isobaric and isochoric?

? We need a function : dH = VdP+ T dS
? In isobaric process: dH = T dS = ? Q
? 因此，可以用状态函数H描述等压吸热过程 (忽视做功与内能)。

? 并且能够得到 CP = (? Q / ?T)P = (?H / ?T)P (enthalpy, 焓)

? In isochoric process: dU = T dS = ? Q
? 因此，可以用状态函数U描述等容吸热过程。 ? 并且能够得到 CV = (? Q / ?T)V = (?U / ?T)V ? 即：

U ? ? Cv dT
dU ? CV dT dH ? C P dT

?For a perfect gas:

? heat capacity is dependent of process. If we restrict heat

capacity as a constant, ?Q = C dT, what will take place? ? If variables are T and V, the differential of inte

rnal energy is

?Q ? CdT ? Cv dT ? PdV
(C ? Cv )dT ? ? PdV ? ? RT dV V

dU ? Cv dT

(C ? Cv )

dT dV ?R ?0 T V Cp ? C dT dV ? ( ? ? 1) ? 0, ? ? T V Cv ? C

Therefore:

PV ? const

?

Homework,P52:Problem(a) 中文：1.13, 1.16

9. Cyclic Processes, The Carnot Cycle 循环过程，卡诺循环

? P-V plane, reversible cyclic process, clockwise direction. ?Along the path abc, the gas expands, and positive work.

?Along the path cda, the gas compressed, negative work,

the area below the curve cda.. ?Along the path abcda, the work is the area of the cycle.

The same cyclic process represented on a T-S
? S-T plane, reversible cyclic process, clockwise

? a’b’c’d’a’ is the same cycle as abcda in PV plane?
?Along the path a’b’c’, the entropy increases, heat is supplied.
?Along the path c’d’a’, the entropy diminishes, heat is removed. ?Along the path a’b’c’d’a’, there is a excess quantity of heat.

? a and a’ in two planes are the same state. The

same cycle represented by two cycles on two planes. According to the energy principle:

? ?Q ? ? dU ? ? ?A
? The internal energy is a function of state, so

? dU ? 0
? And consequently, 其结果为

Q? A

? Areas of two cycles must be equal to each other . ? 结论：经过了一个循环，内能不变，气体

Represent the cycle schematically(图示)
? 热循环过程用过程图表示为

? a’b’c’过程吸收的热量Q1与c’d’a’过程放出的热量之差

Q1 ? Q2 A ?? ? Q1 Q1

? A cycle proceeding in an anticlockwise direction. ? (逆时针方向)，热和功的符号相反。什么过程？ ? 外界对系统作功，放出热量，内能不变。机？

? The cooler transfers heat from a body at a lower

temperature to a body at a higher temperature . ? 将热量从冷物体传到热物体。

Carnot cycle(使用绝热、等温构成循环)
? Considering a cycle consisting of two

Fig.11

Fig,12

? The Carnot cycle on P-V and T-S planes
? T1过程: ? 等温条件下熵增大的过程。 ? 吸热为：Q1=T1(S2-S1) ? T2过程, 放热为: Q2=T2(S1-S2) ? S2过程: ? 绝热条件下温度降低的过程。

? S1过程: ? 绝热条件下温度升高的过程。

Carnot cycle 的过程
? 循环过程类似于活

What is the efficiency of a reversible Carnot cycle ?

??
Q1 ? Q2 Q1 ? T1 ? T2 T1 (Q1 ? 0, Q2 ? 0)

? 过程的效率为：绿 / (绿+黄)

? 在上两图中，T2降低，效率增加，

? 可以理解为：高温热源与低温热源温度

? Perpetual motion machine 永动机？

Discussion:
? 1. 效率与工作介质无关。 ? 2 .效率与冷热物体的温度比有关，

T2/T1越小，卡诺循环的效率越高。 ? 3 .效率总是小于1(unity)，当冷物体的温度趋近 于0时，效率趋近于1。 ? 4 .卡诺循环是热机可能的工作循环，它将一个 恒温吸热和一个恒温制冷结合(incorporate)，中 间过程为绝热。 ? 5 .定理(theorem)：在最高温度吸热及在最低温 度放热的卡诺循环效率最高。(设其他循环在中 间温度有吸热和放热发生。) Home Work, P37： problem (1)

? 对卡诺循环的分析，导致了一系列热力

10.Axiomatics(公理) of thermodynamics. Generalization of the Concept of Entropy to Arbitrary thermodynamic systems Nernst’s(能斯特) Heat Theorem The second law of thermodynamics (中文P39) ? Clausius(克劳修斯) and Kelvin(开尔文) Expressions: ? a) heat can never pass from a colder to a warmer body without some other changes, connected therewith, occurring at the same time;(非自发性) ? b) it is impossible, by means of inanimate (无生命的) material agency, to derive mechanical effect from any portion of matter by cooling it below the T of the coldest of the surrounding objects.

? 不可能将物体降低到比其周围最冷的物体还要冷而作功。

? 1）如果克氏表述不成立，则开氏表述也不成立。

? 2）如果开氏表述不成立，则克氏表述也不成立

? A是可逆机；B是不可逆机。
?若两者吸热相同：Q1＝ Q1’， ?做功不同，设W’(B)＞W(A)

?则ηB＞ηA， Q2> Q2’
?结果：1）功W’-W产生于低温热源。

2） ηB≤ηA

(因实际热机存在非有效功，使效率降低。)

? 开尔文发现热机的效率与吸热放热比有关，且与工作

Q2 ?2 ? Q1 ?1

T2 Q2 ? Q1 T1

?Nernst’s law(the third law): at absolute zero the entropy

of any equilibrium system is constant, independent of any varying parameters. (绝对零度不可能达到，只能无限接近。)

? 由卡诺定理：任何热机的效率均低于可逆热机的效率：
? ? 1?
T2 Q2 ? 1? Q1 T1
? Q2 T2 ? 0

?则有 ?取吸热为正：

Q1 T1

Q1 T1

?

Q2 T2

? 0

?此式为“克劳修斯等式和不等式”
Qi ?对于一个普遍的热循环，将会有： ? T ? 0 i

dQ ?0 其积分表达式为： ? T

? 两个可逆过程：
B dQ dQ R ? T ? 0 ?A T

??

? 定义态函数： S B ? S A

dQR ' ?0 B T B dQ r ?? A T
A

? 对于任意可逆过程，态函数相同。 ? 若有一个过程是不可逆的（设为R）：先经过不

? 在绝热条件下:

dS ? 0

?熵增加原理

?T ?p ( ) ? ( ) s? v ?V ?S ?T ?V ( ) ( ) s? p ?p ?S

?S ?p ( ?S ) ? ? ( ?V ) ( ) ? ( ) T p T v ? p ? T ?V ?T
? 说明：每个偏微分代表一个

“物理效应”

The van der Waals Gas
? For the equation: PV=RT

if T? 0K, then P or V ? 0.
Problem? the molecule of perfect gas has not volume. The Clausius equation: P (V-b) = RT Van der Waals: there are attration and repulsion forces between molecules, at low temperatures the repulsion force is large, which will increase pressure and is proportional to NA/V. So, van der Waal’s equation is ( P ? a )(V ? b) ? RT V2 作业：2.10

? 根据物态方程：
a )(V ? b) ? RT 2 V ?P ( )V (V ? b) ? R ?T CV ?S ?P dT ? ( )T dV ? dT ? ( )V dV ?V T ?T CV R dS ?

dT ? dV T V ?b (P ?

dS ? (

?S )V ?T

? 微商： ? 内能：

dU ? TdS ? PdV ? T ( ? CV dT ? a dV 2 V

CV R dT ? dV ) ? PdV T V ?b

? 自由能，利用：? ln

T V ?b ) ? R ln( )]dT ? PdV T0 V0 ? b 1 R ln(V ? b)dT ? PdV ? d [ RT ln(V ? b)] ? ad V

xdx ? ? yde y ? ( y ? 1)e y ? (ln( x) ? 1) x

Onnes(昂尼斯) equation
? 有许多实际气体的近似表达式，其中之一为
PV ? RT (1 ? B(T ) C (T ) ? ? ...) 2 V V

? 此方程将影响内能的温度关系、熵、热容量等。 ? 右图为B(T)~T的关系图。 ? 其特点是： ? 1）低温时为负，高温为正。 ? 2）B(T)的一阶导数为正。 ? 3）高温时B(T)趋向恒定值。 ?低温下，分子的动能小，分子间的吸引力(a/V2)使压强降低，B(T)<0； ?高温下，分子的斥力(b)显著，使压强增加， B(T)>0。

Homework, 英P60: 1

? 体胀系数α是： ? 压强系数β是：
1?V ? ?( ) p V ?T
? ?
1 ?p ( )V p ?T

? 等温压缩系数κT是：

1?V ? ?( ) T? T V ?p

?V?p ?T ( ) ( ) ( ) ? 1 T V p? ?p ?T ?V

? ? ? ?p T

13. Gas Cooling Methods, Gay-Lussac and JouleThomoson Processes
?A reversible adiabatic expansion of a

gas can be used to cool it. ?Gay-Lussac first carried out this experiment. ?Thermal equilibrium sets in in the vessel, and the change in temperature ,T2-T1,can be measured. The experiment show that T2=T1, i.e. U is independent of volume.

It should be noted that this process is an irreversible adiabatic process, and we cannot claim that the entropy remains constant. Later, we will know that the entropy increases in irreversible adiabatic process.

? The meaning of the derivative(导数) (?T/?V)U。
? (T , U ) ? (T , U ) ? (V , S ) ? ?T ? ? ? ? ? ? (V , U ) ? (V , S ) ? (V , U ) ? ?V ?U ?T ?U ?T ?U ?S ? [( )S ( )V ? ( )V ( ) S ]( )V ?V ?S ?S ?V ?U ?T P ?T ?( )S ? ( )V ?V T ?S ?U ?S ? (U , S ) ? ( S , V ) ?S ( )S ( )V ? ? ?( )U ?V ?U ? (V , S ) ? (U , V ) ?V ?S dU ? TdS ? PdV ? 0, ( )U ? P / T ?V

For the real gas, and van der Waals’ equation
( ?T T P ?P a )U ? [ ? ( )V ] ? ? 2 ?V CV T ?T V CV

Joule-Thomson Process 焦耳-汤姆逊过程
? The gas filling a heat-insulated tube is forced through a ? ? ?

?

porous(多孔的) plug with the aid of two pistons. Constant pressures P1 > P2 are maintained by two pistons. A steady-state(稳态的) gas flow through the porous partition。 Now, analyze the process: abrupt volume change from V1 to V2. The process is not a quasi-static state process. The Joule-Thomson process is adiabatic.

dH ? TdS ? VdP,

H ? U ?

PV

U 2 ? U1 ? P2V2 ? P 1V1 H 2 ? H1

U 2 ? P2V2 ? U1 ? P 1V1

So, the Joule-Thomson process is an isoenthalpic process.(等焓过程)
?总之，过程为非准静态的突变、是绝热的、等焓的。 ?然而，为了理论分析的方便，必须对过程进行假设。

?Let us replace the Joule-Thomson by an imaginary

isoenthalpic process. The purpose of analyzing is to determining the temperature change. That is to say, in a isoenthalpic process, the question of temperature~pressure. Joule and Thomson defined a Joule-Thomson coefficient:

? ?? ?

?T ? ? ? P ? ?H

? ??
? 其中

? (T , H ) ? (T , H ) ? ( P, T ) ?H ? ?T ? ? ? ? ? ( )T / C P ? ? ( P, H ) ? ( P, T ) ? ( P , H ) ?P ? ?P ? H

1 ?V ? ?H ? CP ? ? , ? ? ( )P ? V ?T ? ?T ? P ?H T?S ? V?P ?S ( )T ? ( )T ? T ( )T ? V ?P ?P ?P ?V ? ?T ( )P ?V ?T
? 因此：

? ??

1 T ?V V ? ?T ? [ ( ) P ? 1] ? [T? ? 1] ? ? C P V ?T CP ? ?P ? H

? 内容参考：英文P64：(13-7)

( ? 对于理想气体：

?V V )P ? ? 0, ?T T

??

1 , T

? ?0

? ??

?T ?? 0 ? ? ?T ? ? ? 0? ? ? ?P ? 0 ? ?P ? H

? 在节流过程中，气体体积类似于自由膨胀，但因活塞作用于两

? 即： ? ?

? ? ?

? 0 ? 0 ? 0

(P ? a )V ? RT V2

（1）

P(V ? b) ? RT （2）

? 将方程（1）两边对T求偏导，可以得到
a ?V T ?V 2a / V 2 ( P ? 2 )( ) P ? R 以及 ( )P ?1 ? ?0 2 ?T V V ?T P ? a /V

P( ?V )P ? R ?T

T ?V b ( )P ?1 ? ? ? 0 V ?T V

? 将节流过程分为两个部分讨论，如图所示

(P 1 ? a V1
2

)V1 ? RT1

( P2 ?

a )V2 ? RT2 2 V2

P1 (V1 ? b) ? RT1 和 P2 (V2 ? b) ? RT2
? 与理想气体的物态方程相比，可以认为，真实气体的

? 对于实际气体, B(T)可为正，也可为负。

RT ? PV PV a ?( )V ~ ( P ? 2 )V 1 ? B /V V ?B V

? 当B(T)> 0时， B(T) = B ：
PV V2 RT ? ? P( ) ~ P(V ? b) 1 ? B /V V ?B

??

1 dB(T ) 1 2a [T ? B(T )] ? [ ? b] CP T C P RT

?真实气体分子之间的相互作用势是产生致冷和致热的物

B(T ) ? ?2?N ? [e ?U ( r ) / kT ? 1]r 2 dr
0

?

? 定性地分析，当U(r) / kT为很小的正数时，B（T）> 0；

P? RT PB(T ) RT (1 ? ), i.e. V ? ?B V RT p 1 ?V 1 R dB ?? ( )P ? [ ? ] V ?T V P dT
V 1 dB [T? ? 1] ? [T ? B] CP CP dT

??

? 在低温区，B<0，(dB/dT)>0，故? > 0，制冷。 ? 在高温区，B足够大，使? < 0，制热。 ? 结论：低温区，分子相互吸引， B<0 ，制冷。 ? 提问：气体的节流过程利用了定义： dT ? ( ? 该定义是否应该写成?
?T ) H dP ?P ?T ?T dT ? ( ) H dP ? ( ) H dV ?P ?V

? 如果可以，温度的变化是否应该与体积的变化有关？

?提问回答： ? 节流过程可以看成是等焓过程，焓为两个热力

?H ?H dH ? ( )T dP ? ( ) P dT ? 0 ?P ?T ?T ?H ?T dT ? ?( )P ( )T dP ? ( ) H dP ?H ?P ?P 意义：任何一个特定的热力学过程，都可以用一个 偏导数表示，这些偏导数常用CP、?、?T等

?热力学函数的数学关系(实际气体)
?S ?S ?S ?S dS ? ( )V dT ? ( )T dV ? ( ) P dT ? ( )T dP ?T ?V ?T ?P
? 内能和焓都是dS的函数，可用于改变变量，得到新的关系式。 ?S ?S dU ? TdS ? PdV ? T [( )V dT ? ( ) T dV ] ? PdV ?T ?V

? CV dT ? [T (

?U ?S ) T dV ) T ? P]dV dU ? CV dT ? ( ?V ?V ?U ?S ?P ( )T ? T ( )T ? P ? T ( )V ? P ?V ?V ?T

dH ? TdS ? VdP ? T [(

? C P dT ? [T (

?S )T ?P

?S ?S ) P dT ? ( ) T dP] ? VdP ?T ?P ? V ]dP dU ? C P dT ? ( ?H ) T dP ?P

?H ?S ?V ( )T ? T ( )T ? V ? ?[T ( )P ?V ] ?P ?P ?T

?证明 ? Prove:

C P ? CV ?

VT? 2

?T

?S ?S C P ? CV ? T [( )P ? ( )V ] ?T ?T

?S ? ( S , P) ? ( S , P) ? (T , V ) ( )P ? ? ?T ? (T , P) ? (T , V ) ? (T , P)

? [(

?S ?P ?S ?P ?V )V ( )T ? ( )T ( )V ]( )T ?T ?V ?V ?T ?P
?S ?P 2 ?V ? ( )V ? ( )V ( ) T ?T ?T ?P ?S ? ( )V ? ( p? ) 2 V? T ?T

? Here,

?P ? 1 ?V ( )V ? p? , p? ? , ? T ? ? ( ) T ?T ?T V ?P

2 . 10

?1.9 The Thermodynamic Potential. Method of

Thermodynamic Functions (2.5特性函数)
? 由前面我们学过的内容可以知道： ? 1.

( ?V ?P ?V ) P , ( )V , ( ) T ?T ?T ?P

? 2.

3. 已经学过的热力学函数，其物理意义或作用是什么？
1）在确定的过程，如等焓，等内能，其作用如同偏导数； 2）用于描述特定过程的状态；

dU ? TdS ? PdV dF ? ? SdT ? PdV dH ? TdS ? VdP dG ? ? SdT ? VdP

?3)热力学函数是描述状态变化的函数
? Knowing any of the thermodynamic functions

expressed in terms of their variables, we are able in principle to find all the thermodynamic parameters of a system, including its heat capacity, and the equation of state.
? “若已知一个用热力学变量表示的均匀系统的热力学

S

? ?( ?F ?F )V , P ? ?( )T ?T ?V

? 若自由能F是用T、V的函数，通过如下变换，

?F )V ?T ?F G ? F ? PV ? F ? V ( ) T ?V ?F ?F H ? F ? TS ? PV ? F ? T ( )V ? V ( ) T ?T ?V ?U ?2F CV ? ( )V ? ?T ( 2 )V , C P ? CV ? R, ? ?T ?T U ? F ? TS ? F ? T (
?知道了热力学函数，可以导出热力学参量和其他热力学

? 例1：若压强仅仅是温度的函数：P＝P（T） ?F ?F dF ? ?SdT ? PdV S ? ?( ?T )V , P ? ?( ?V )T

F ? ? ? PdV ? ? PV , (V ? 0, F ? 0)

U ? F ?T(

?F ?P ?P )V ? F ? TV ( ) T ? V [T ( )V ? P(T )] ?T ?V ?T ?F G ? F ?V ( )T ? 0 ?V ?F ?F ?F ?P H ? F ?V ( )T ? T ( )V ? ?T ( )V ? TV ( )V ?V ?T ?T ?T
?U ?2P CV ? ( )V ? VT ( ) 2 ?T ?T

dF ? ?SdT ? PdV
S ? ?( ?F ?F )V , P ? ?( )T ?T ?V
F ? ? ? SdT ? ? ST , ( S ? 0, F ? 0)

?T ?F ?S ?S G ? F ?V ( ) T ? ? ST ? VT ( ) T ? T [V ( )T ? S ] ?V ?V ?V ?F ?F ?S H ? F ?V ( )T ? T ( )V ? TV ( )T ?V ?T ?V

?G )T ?P ?G ?G U ? G ? TS ? PV ? G ? T ?P ?T ?P F ? G ? PV ? G ? P(