Chapter 17 Heat and the First Law of Thermodynamics
Chapter 17 Heat and the First
Law of Thermodynamics
1,Heat and Energy Transfer,Internal Energy
and Specific Heat
2,The First Law of Thermodynamics
3,Applying 1st Law of Thermodynamics
4,Molar Specific Heats for Gases,and the
Equipartition of Energy
5,Adiabatic Expansion of a Gas
Chapter 17 Heat and the First Law of Thermodynamics
Heat 热量
Calorie(Cal) 卡路里
Mechanical equivalent of heat 热功当量
Internal energy 内能
Degrees of Freedom 自由度
diatomic molecules 双原子分子气体
monatomic molecules 单原子分子气体
polyatomic molecules 多原子分子气体
Quasi-Steady Process 准静态过程
Chapter 17 Heat and the First Law of Thermodynamics
Isothermal (Constant-Temperature) 等温
Isochoric (Constant-Volume) 等容
Isobaric (Constant-pressure) 等压
Adiabatic Process 绝热过程
Adiabatic Free Expansion 绝热自由膨胀
Specific heat and Molar Specific Heats
热容和摩尔热容
Chapter 17 Heat and the First Law of Thermodynamics
§ 17-1 Heat as Energy Transfer P404-405
1 Energy and heat
Heat refers to a
transfer of energy.
Joule’s Experiment
on the mechanical
equivalent of heat.
焦耳 (Joule)实验
1) Conclusion:
A given amount of work done
was always equivalent to a
particular amount of heat input.
Falling
weight
paddle
Chapter 17 Heat and the First Law of Thermodynamics
Absorbing heat,Q>0； Losing heat,Q<0
Heat is path-dependent quantity（过程量）,
2) Heat（ 热量 Q）,
Heat is the energy that is transferred
between a system and its environment because of
a temperature difference that exists between
them.（ its unit is J---Joules）
Heat and temperature are often confused.
Chapter 17 Heat and the First Law of Thermodynamics
The amount of heat necessary to raise the
temperature of 1 gram of water by 1 Celsius
degree.
3 Calorie
1 Calorie=4.186 J
In SI units,the unit for heat is the joule.
4 Mechanical equivalent of heat
Example 17-1 page 405
Chapter 17 Heat and the First Law of Thermodynamics
在研究气体的能量时，需要确定其物理模型，这个物理模型就是自由度。用自由度模型描写气体能量是有局限性的，对少原子分子气体，在常温下理论值与实验值符合得较好，但对多原子分子或在高温情况下，理论值与实验值相差较大。这得用量子物理方法进行研究。
本节我们使用自由度模型和能量均分的统计原理来研究理想气体在常温下的气体能量。
17-2&17-3 Heat as Energy Transfer ; Internal
Energy and Specific Heat (404-407)
Chapter 17 Heat and the First Law of Thermodynamics
Degrees of Freedom (自由度 )p414
1自由度,
是描写物体在空间位置所需的独立坐标数。
Degree of Freedom —– numbers of
independently describing the motion of a
molecule.
例如：物体沿一维直线运动，最少只需一个坐标，
则自由度数为 1。
Chapter 17 Heat and the First Law of Thermodynamics
轮船在海平面上行驶,要描写轮船的位置至少需要两维坐标,则自由度为 2。
飞机在天空中飞翔，要描写飞机的空间位置至少需要三维坐标,则自由度为 3。
vrti2、自由度数目平动转动振动
Chapter 17 Heat and the First Law of Thermodynamics
z
y
x
),,( zyxP
o2） 两个刚性质点
3 质点的自由度
1） 一个质点，描写它的空间位置，需要 3 个平动自由度,3?t
3?t
描写其质心位置需 3个平动自由度，
描写其绕 x,y轴转动需 2个转动自由度，绕 z轴的转动能量可不计，
2?r
总自由度数
523 rti
z
y
x
o
Chapter 17 Heat and the First Law of Thermodynamics
3） 三个或三个以上的刚性质点需 3个平动自由度和 3个转动自由度。
633 rti
3?t
3?r
平动自由度转动自由度总自由度
Chapter 17 Heat and the First Law of Thermodynamics
1） 单原子分子气体 monatomic molecules
例如：氦气 ( helium,He)等为单原子分子气体。其模型可用一个质点来代替。
y
z
x
o
平动自由度 3?t
转动自由度 0?r
303 rti
总自由度
4 气体分子的自由度 (p414)
对于理想气体在常温下，分子内各原子间的距离认为不变，只有平动自由度 translational (DF,t)、转动自由度 rotational motion (DF,r) 。
Chapter 17 Heat and the First Law of Thermodynamics
2） 双原子分子气体 diatomic molecules
例如：氢气（ H2）、氧气
（ oxygen,O2）、氮气（ N2）
等为双原子分子气体。其模型可用两个刚性质点模型来代替。
平动自由度 3?t
转动自由度 2?r
523 rti总自由度
Chapter 17 Heat and the First Law of Thermodynamics
3）多原子分子气体 polyatomic molecules
如,二氧化碳气体（ CO2）、水蒸气（ H2O）等为多原子分子气体。其模型可用多个刚性质点来代替。
平动自由度 3?t
转动自由度 3?r
633 rti总自由度
Monatomic 3 0 3
diatomic 3 2 5
polyatomic 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
Chapter 17 Heat and the First Law of Thermodynamics
由温度公式有分子平均平动动能
The theorem(principle) of the equipartition of
energy 能量均分定理 (p414)
kTm
2
3
2
1 2
kt v?
2222
3
1 vvvv
zyx
kTmmm zyx 21212121 222 vvv
由于分子运动在哪个方向都不占优势，因此，在 y、
z 方向的自由度上也都平均分配 了 kT / 2 的能量。
Chapter 17 Heat and the First Law of Thermodynamics
各平动自由度的平动动能相等；每个平动自由度上分配了一份 kT/2的能量。
平动动能 转动动能由于分子的激烈碰撞（几亿次 /秒），平动动能与平动动能，平动动能和转动动能之间不断转换；
平动动能 平动动能单原子分子平均能量 kT213
因此使得所有的自由度中没有哪个是特殊的,即各自由度的平均动能都是相等的，即：
Chapter 17 Heat and the First Law of Thermodynamics
能量均分定理：在温度为 T的平衡态下，气体分子每个自由度的平均动能都相等，而且等于 kT/2 。
Principle of equipartition of Energy:
Energy is shared equally among the active
degrees of freedom,and in particular each
active degree of freedom of a molecule has on
the average an energy equal to kT/2.
分子的平均能量
kTi
2
根据能均分定理，如果气体分子的总自由度为 i 个，
则它的平均动能就有 i份 kT/2 的能量。
Chapter 17 Heat and the First Law of Thermodynamics
单 原子分子 3 0 3
双 原子分子 3 2 5
多 原子分子 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
kTK 23?
kTK 25?
kTK 3?
Chapter 17 Heat and the First Law of Thermodynamics
Internal energy（内能）,P406
1 Internal energy
The sum total of all the energy of all the molecules
in an object.
定义：气体的内能是指它的内动能,即它所包含的所有分子的动能 （ 相对于质心参考系 ） 和分子间的相互作用势能的总和 。
对于理想气体，分子之间无势能,因此理想气体的内能就是它的所有分子的动能之和 。
Chapter 17 Heat and the First Law of Thermodynamics
一个分子的能量为 kT
i
2
1 mol气体分子的能量为,kTN
i
02 RT
i
2?
m 千 克气体的 内能 为：
RTiMmU 2?
Chapter 17 Heat and the First Law of Thermodynamics
单 原子分子 3 0 3
双 原子分子 3 2 5
多 原子分子 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
n R TU 23?
n R TU 25?
n R TU 3?
Chapter 17 Heat and the First Law of Thermodynamics
2 Discussion
1) Internal energy is a function of state,it is not
path-dependent quantity.（状态量）
RTi
M
mU
2
始末两态气体内能的变化。
内能增量 U?
12 UUU
TRi
m
MU
2
）12(2 TTRiMm
The internal energy of an ideal gas is a function
of the gas temperature only; it does not depend
on any other variables.
Chapter 17 Heat and the First Law of Thermodynamics
)( TUU?
2) 理想气体内能,表征系统状态的单值函数,理想气体的内能仅是温度的函数,
3) Distinguishing Temperature,Heat,and
Internal Energy
Temperature is a measure of the average kinetic
energy of individual molecules,Internal Energy
refer to the total energy of all the molecules in
the object,Heat,finally,refers to a transfer of
energy from one object to another because of a
difference in temperature.
Chapter 17 Heat and the First Law of Thermodynamics
Quasi-Steady Process (准静态过程 p409)
An equilibrium state of a
system can be described by
several parameters,such as
the pressure p,the volume
V,and the temperature T.
In the parameter space
(that is p-V-T phase
diagram),it is expressed by
a point.
§ 17-4 The First Law of Thermodynamics (407-408)
),,( 111 TVp
),,( 222 TVp
1V 2V
1p
2p
p
Vo
1
2
Chapter 17 Heat and the First Law of Thermodynamics
Now the system experiences a series changes
from state 1 to state 2,Each state is a new
equilibrium state.
气体活塞砂子 ),,( 111 TVp
),,( 222 TVp
1V 2V
1p
2p
p
Vo
1
2
准静态过程：从一个平衡态到另一平衡态所经过的每一中间状态均可近似当作平衡态的过程,
Chapter 17 Heat and the First Law of Thermodynamics
S
dl
dVP
由功的定义：
lpSlFW ddd
VpW dd?
21 dVV VpW
Work of Quasi-Steady Process
功是能量传递和转换的量度，它引起系统热运动状态的变化,
Chapter 17 Heat and the First Law of Thermodynamics
Discussion
1、如果 0?dV,则 0?dA
即系统体积膨胀时，系统对外界做功； If the work is done by
system,then W>0; conversely,
if the work is done on system,
then W<0.
0?dV 0?dA2、如果,则即系统体积收缩时，系统对外界做负功，即外界对系统做功。
Chapter 17 Heat and the First Law of Thermodynamics
由积分的意义可知,
功的大小等于 P~V图上过程曲线下的面积
It is the shaded area
under the p-V process curve
between initial to final state
P d VS VV 21W?
3.在 P~V图中曲线下的面积
1V
o
P
V
2V
1
2
dV
P
Chapter 17 Heat and the First Law of Thermodynamics
4,Work is path-dependent quantity (p411)
功是过程量如：从图中可看出,1?2与
1?1’?2两个过程的始末状态相同，但过程曲线不同，两条曲线下的面积不同，则作功也不同。
1V
o
P
V
2V
1
2
'1
The work done in taking a system from one state
to another depends on the initial and final states
but also on the type of process (path).
Chapter 17 Heat and the First Law of Thermodynamics
1T
2T
21 TT?
1）过程量：与过程有关；
2）等效性：改变系统热运动状态作用相同；
宏观运动 分子热运动功分子热运动分子热运动 热量
Q
3）功与热量的物理本质不同,
1卡 = 4.18 J,1 J = 0.24 卡
5 功与热量的异同
Chapter 17 Heat and the First Law of Thermodynamics
Experimentally,it is found that both work W
and heat Q are dependent on the nature of the
process,However,a surprising thing is that the
quantity Q-W is the same for all processes
between two states,
The First Law of Thermodynamics (P407)
实验证明系统从 A 状态变化到 B 状态，可以采用做功和传热的方法，不管经过什么过程，只要始末状态确定，做功和传热之和保持不变,
Chapter 17 Heat and the First Law of Thermodynamics
2
A
B
1*
*
p
Vo
02121 ABAABA QW
BABABABA QWQW 2211
2
A
B
1*
*
p
Vo
Chapter 17 Heat and the First Law of Thermodynamics
One can infer that the quantity Q-W must
represent a change in some intrinsic property of
the system,This intrinsic property is internal
energy?U.
From conservation of energy:
The change in internal energy of a closed
system will be equal to the heat added to the
system minus the work done by the system.
WQUUU if
The First Law of Thermodynamics
Chapter 17 Heat and the First Law of Thermodynamics
dWdQdU
For a differential process,it can be written as:
It is hold for any system and any process!
2
1
d
V
V
VpUQ
准静态过程定律：系统 从外界吸收的热量,一部分使系统的内能增加,另一部分使系统对外界做功,
The first law of thermodynamics is the principle
of conservation of energy for a thermodynamic
process.
Chapter 17 Heat and the First Law of Thermodynamics
The sign of W,Q,and U,(p408)
work done by the system W > 0;
Work is done on the system,W<0
Heat is added to the system Q > 0;
Heat leaves the system,Q<0
system’s energy increasing? U > 0.
system’s energy decreasing? U < 0.
Chapter 17 Heat and the First Law of Thermodynamics
+
12 EE?
系统吸热系统放热内能增加内能减少系统对外界做功外界对系统做功第一定律的符号规定
Q W
Chapter 17 Heat and the First Law of Thermodynamics
Two ways to change system’s internal energy:
Energy may be transferred into or out of a
system as either work W or heat Q.
Relationship between W and Q,1 cal =4.18 J
The transformation between work and heat is not direct,
but on the system.热 — 功转换不是直接进行的,而是间接的,内能是传递工具 。
系统吸热后，先使内能增加，再通过降低内能对外作功。
热功? 外界对系统作功，使内能增加，再通过内能降低，系统放热。
功热?
Chapter 17 Heat and the First Law of Thermodynamics
1,Specific Heats (热容 ) P407
§ 17-3 Specific Heat & § 17-6 Molar
Specific Heats for Gas (P407 &P412-414)
热容：系统和外界之间的热传递会引起系统本身温度的变化，这一温度的变化和热传递的关系用热容表示。
The amount of heat Q required to change the
temperature of a given material is proportional
to the mass m of the material and to the
temperature change,(P407)
TmcQ
T
Qmc
Chapter 17 Heat and the First Law of Thermodynamics
Unit?C?J / k g
Table 17-1 on page 407
The values of c depend on temperature,but for
temperature changes that are not too great,c can
often be considered constant.
Example 17-2 on page 407
Chapter 17 Heat and the First Law of Thermodynamics
其是 1 mol 气体升高 1 oC 所吸收的热量。
The heat require to raise 1mol of the gas by 1 oC,
2 Molar specific heat (摩尔热容 ) p413
1 mol 的物质温度升高 dT 时，如果吸收的热量为 dQ，
则 摩尔热容的定义为：
T
QC
d
d?
In analogy to this equation,the heat Q needed to raise
the temperature of n moles of gas by dT is,
T
Q
nC d
d1?
Chapter 17 Heat and the First Law of Thermodynamics
Molar specific heat at constant pressure 定体热容
Molar specific heat at constant volume 定压热容分别由定压和定体条件下物质吸收的热量决定。
3、摩尔热容的分类
Chapter 17 Heat and the First Law of Thermodynamics
因为这一部分很重要,内容又多,所以把讲过的内容先复习一遍,然后再开始第一定律的应用,
Chapter 17 Heat and the First Law of Thermodynamics
heat P404
calorie (cal) Calorie (kilocalorie) P405
mechanical equivalent of heat P405
internal energy P406
degrees of freedom P414
quasi-steady process P408
monatomic molecules P414
diatomic molecules P414
polyatomic molecules 补充
Chapter 17 Heat and the First Law of Thermodynamics
isothermal (Constant-Temperature) process P409
isochoric (Constant-Volume) process P410
isobaric (Constant-pressure) process P410
adiabatic process P409
adiabatic free expansion P412
specific heat P407
molar specific heats P413
Chapter 17 Heat and the First Law of Thermodynamics
1 Heat
Heat refers to a transfer of energy.
Three concepts
Absorbing heat,Q>0； Losing heat,Q<0
Heat is path-dependent quantity（过程量）,
Chapter 17 Heat and the First Law of Thermodynamics
2 Internal energy
RTi
M
mU
2
)( TUU?
The total of all the energy of all the molecules
in an object.
Internal energy is a function of state,it is not
path-dependent quantity.（状态量）
Chapter 17 Heat and the First Law of Thermodynamics
Internal energy of molecules
3 Temperature
Temperature is a measure of the average kinetic
energy of individual molecules.
degree of freedom internal energy
Monatomic 3
diatomic 5
polyatomic 6
n R TU 23?
n R TU 25?
n R TU 3?
Chapter 17 Heat and the First Law of Thermodynamics
Heat is path-dependent quantity.
Temperature and internal energy are path-
independent quantities.
discussion:
Chapter 17 Heat and the First Law of Thermodynamics
The First Law of Thermodynamics
S
ld
dVP由功的定义：
21 dVV VpW
1 Work of Quasi-Steady Process (P410)
Chapter 17 Heat and the First Law of Thermodynamics
Work done by an ideal
gas is the shaded area
under the PV curve
between initial to final
state,
2.在 P~V图中曲线下的面积 (P411)
1V
o
P
V
2V
1
2
Vd
P
Chapter 17 Heat and the First Law of Thermodynamics
3,Work is path-dependent quantity (p411)
1Vo
P
V
2V
1
2
'1
The work done in
taking a system from
one state to another
depends on the initial
and final states but also
on the type of process
(path).
Chapter 17 Heat and the First Law of Thermodynamics
4 The First Law of Thermodynamics (P407)
WQUUΔU if
2
1
d
V
V
VpUQ
准静态过程
The change in internal energy of a closed system,
is to equal to the heat added to the system minus
the work done by the system,
In differential form:
dWdQdU
Chapter 17 Heat and the First Law of Thermodynamics
+
12 EE?
系统吸热系统放热内能增加内能减少系统对外界做功外界对系统做功第一定律的符号规定
Q W
Chapter 17 Heat and the First Law of Thermodynamics
5 Specific Heats and Molar Specific Heats
1) Specific Heats (P407)
T
Qmc
2) Molar specific heat (P413)
T
Q
nC d
d1?
Molar specific heat at constant pressure 定体热容
Molar specific heat at constant volume 定压热容
Chapter 17 Heat and the First Law of Thermodynamics
计算各等值过程的热量、功和内能的理论基础
RT
M
mpV?（ 1） （ 理想气体的 共性）
21 dVV VpEQ
VpEQ ddd
（ 2）
解决过程中能量转换的问题
)( TUU?
（ 3） （ 理想气体的状态函数 ）
（ 4） 各等值过程的特性,
§ 17-5 Some Special Cases of
The First Law of Thermodynamics (P409-417)
Chapter 17 Heat and the First Law of Thermodynamics
1,Isothermal (Constant-Temperature) Process
(P409,P411)
The process carried out at constant temperature.
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo Vd
恒温热源
T
Movable piston
Chapter 17 Heat and the First Law of Thermodynamics
热力学第一定律
0d?U
V
RTnp?
21 dVV VpW
VpWQ T ddd
特征 常量?T
过程方程?pV 常量
TRi
M
mU
2
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo Vd
Chapter 17 Heat and the First Law of Thermodynamics
UU
1
2
V
Vn R T ln
2
1ln
p
pn R T?
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo
等温 膨胀
W
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo
W
等温 压缩
TQ TQW W
dV
V
RTnWQ V
V
T
2
1
Chapter 17 Heat and the First Law of Thermodynamics
2,Isochoric (Constant-Volume) Process (p410,P414)
The volume of the system
is held constant,no work
is done for the system.
0d,0d WV
热力学第一定律
UQ V dd?
特性 常量?V
),,( 11 TVp
),,( 22 TVp2p
1p
V
p
Vo
过程方程 常量1PT
dU > 0,system absorbs heat from its environment;< 0,then the system releases heat.
Chapter 17 Heat and the First Law of Thermodynamics
vvV T
U
nT
Q
nC )d
d(1)
d
d(1
定体摩尔热容,理想气体在等体过程中吸收的热量,使温度升高,其定体摩尔热容为,
mol n
VQd Td
n R TiU 2?
将 代入，可得,
RiC V
2
So,we may express internal energy of IG as:
TnCU V dd?TnCU V? or
Chapter 17 Heat and the First Law of Thermodynamics
TnCUQ VV ddd
1212 )( UUTTnCQ VV
1U
2U
VQ
1U
VQ
2U
),,( 11 TVp
),,( 22 TVp2p
1p
V
p
Vo
等体升压 1
2 ),,(
11 TVp
),,( 22 TVp2p
1p
V
p
Vo
等体降压
1
2
Chapter 17 Heat and the First Law of Thermodynamics
The heat absorbed by gas is used for increasing
its internal energy and doing work,气体吸收的热量一部分用于增加内能，另一部分用于对外作功。
3,Isobaric (Constant-Pressure) Process (P410,P414)
The pressure of the system is held constant.
2V
),,( 11 TVp ),,( 22 TVp
p
1V
p
Vo
1 2过程方程 常量1VT
热一律
WUQ p ddd
特 性 常量?p
)( 12 VVpW功 W
Chapter 17 Heat and the First Law of Thermodynamics
ppp T
V
n
p
T
U
nT
Q
n
C )
d
d(
d
d1)
d
d
(1
定压摩尔热容,理想气体在等压过程中吸收的热量,温度升高,其定压摩尔热容为
mo ln
pQd Td
将
n R TpVn R TiU,2
代入，
RRiC p
2
可得
Chapter 17 Heat and the First Law of Thermodynamics
)( 12 VVpW )( 12 TTnR
)( 1212 TTnCUU V
),( 12 TTnCQ pp
Chapter 17 Heat and the First Law of Thermodynamics
2V
),,( 11 TVp ),,( 22 TVp
p
1V
p
Vo
1 2
W
等压膨胀
2V
),,( 11 TVp),,( 22 TVp
p
1V
p
Vo
12
W
等压压缩
1U
2UpQ
1U
pQ
2U
W
W
Chapter 17 Heat and the First Law of Thermodynamics
RCC Vp
1) 定压摩尔热容和定体摩尔热容的关系
RRiC p
2
RiC V
2
由 和该公式给出了定压摩尔热容和定体摩尔热容的关系,
称为 迈耶公式 (Mayer Formula p414 )。 迈耶利用该公式算出了热功当量,对建立能量守恒作出了重要贡献 。
DISCUSSION
Chapter 17 Heat and the First Law of Thermodynamics
2) CP > CV 的物理意义 RCC VP
1 mol 理想气体温度升高 1 oC,对于等容过程,体积不变吸热只增加系统内能,而对于等压过程除了增加系统内能外,还要对外作功,所吸收的热量要更多一些 。
In the process done at constant volume,no work is done
since is zero,Thus according to the first law of
thermodynamics,the heat added all goes into increasing
the internal energy of the gas,In the process out at
constant pressure,work is done,and hence the heat
added,must not only increase the internal energy but lso
is used to do the work,Thus more heat must added in
this process than in the first process at constant volume.
V?
Chapter 17 Heat and the First Law of Thermodynamics
3) 摩尔热容比 p416
单原子分子 3 3R/2 5R/2 1.67
刚性双原子分子 5 5R/2 7R/2 1.40
刚性多原子分子 6 3R 4R 1.33
VC pC?i
2
2 i
C
C
V
P?RRiC
p 2 R
iC
V 2?
由 和叫摩尔热容比。
Chapter 17 Heat and the First Law of Thermodynamics
4,Adiabatic Process (绝热过程 ),P409-410,P416-417
An adiabatic process
is one in which no heat is
allowed to flow into or out
of the system.
绝热过程是系统在和外界无热量交换的条件下进行的过程。用把系统和外界隔开隔能壁（或叫绝热壁）就可以实现这一过程。
绝热的汽缸壁和活塞
Chapter 17 Heat and the First Law of Thermodynamics
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
)( 12 TTnC V
Od?Q特征
TCn VT
T
d2
1?
TnCU V dd?
21 dVV VpW
Vd
UW dd
热一律 0dd UW
Chapter 17 Heat and the First Law of Thermodynamics
)( 2211 VpVp
CC
CW
Vp
V?
1
2211
VpVpW
若已知 及 2211,,,VpVp?
)( 2211
R
Vp
R
VpCW
V
nR TpV?
从 和 可得 RCC VP
Chapter 17 Heat and the First Law of Thermodynamics
绝热过程方程的推导 P416
UWQ dd,0d
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
0?Q
QUW ddd
dTnCU V?d
Combine these two equations,we obtain:
0dd VPTnC V
We next take the differential of the ideal gas law,
nR TpV?
Chapter 17 Heat and the First Law of Thermodynamics
0ddd VP)nR PVVP(nC V
Allowing P,V,and T to vary:
TnRPVVP ddd
We solve for dT in this relation and substitute it
into the previous relation and get
or
0)( V d PCP d VRC VV
We note from that
PV CRC
So we have
Chapter 17 Heat and the First Law of Thermodynamics
0dd PVCVPC VP
0dd
V
V
P
P?
So that last equation becomes
This is integrated to become
c on s t a ntlnln VP?
This simplifies to (using the rules for addition
and multiplication of logarithms) to
c on s t a ntPV
Chapter 17 Heat and the First Law of Thermodynamics
绝热方程
TV 1?
pV
Tp 1
常量常量常量
Chapter 17 Heat and the First Law of Thermodynamics
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
W
绝 热 膨 胀
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V2V
p
Vo
W
绝 热 压 缩
1E
2E
1E
2E
W
W
Chapter 17 Heat and the First Law of Thermodynamics
Example 17-7 on page 417
Chapter 17 Heat and the First Law of Thermodynamics
5,Adiabatic Free Expansion,绝热自由膨胀 (p412)
A adiabatic process in which a gas is allowed to
expand in volume adiabatically without doing
any work.
Chapter 17 Heat and the First Law of Thermodynamics
5,Adiabatic Free Expansion,绝热自由膨胀 (p412)
定义：在绝热容器中间有一个隔板,
左半部充以理想气体并处于平衡态,
右半部为真空 。 左半部原来处在平衡态 。 现抽去隔板,气体将冲入右半部,最后可以在整个容器内达到一个新的平衡态,这种过程叫 绝热自由膨胀 。
Chapter 17 Heat and the First Law of Thermodynamics
0d?Q特征：由于过程是绝热的，
由热力学第一定律 0dd UW
由于气体是向真空中冲入，所以它对外界不做功，即
dW=0
即 气体经过自由膨胀，内能保持不变 。
The internal energy of a gas does not change in the
free expansion.
所以,0d?U
12 UU?
即
Chapter 17 Heat and the First Law of Thermodynamics
Since U depends only on T,so
对于理想气体，内能只包含分子的热运动动能，它只是温度的函数，所以经过自由膨胀，理想气体再达到平衡态时，它的温度复原。
0 T
由理想气体状态方程，对于初、末状态分别有：
,111 n R TVp? 222 n R TVp?
因为 T2=T1,V2=2V1
12 2
1 pp?得到
Chapter 17 Heat and the First Law of Thermodynamics
理想气体的各等值过程、绝热过程公式对照表过程 特征 过程方程 吸收热量 对外做功 内能增量
Isochoric
等容
Isobaric
等压
Isothermal
等温
Adiabatic
绝热
C?Tp
C?TV
C?pV
'CpV
''C1 TV?
C1 Tp
)( 12 TTCMm V?
0
)( 12 TTCMm V?
)( 12 TTCMm V?
)( 12 TTCMm V?
)( 12 TTCMm p?
)(
or)(
12
12
TTR
M
m
VVp
2
1
1
2
ln
orln
p
p
RT
M
m
V
V
RT
M
m
2
1
1
2
ln
orln
P
P
RT
M
m
V
V
RT
M
m
0
0
1
or
)(
2211
12
VPVP
TTC
M
m
V
p=C
V=C
T=C
dQ=0
Chapter 17 Heat and the First Law of Thermodynamics
Solutio
n:
Example,N2 of volume 100cm3 and pressure
1.013× 105Pa is compressed to volume 20cm3,
compute the change in internal energy?E,the
heat Q and the work done W during processes,(1)
Isothermal compress; (2) isobaric compress then
isochoric increasing pressure to same state 先等压压缩,再等体升压到同样状态,
C
V
p
O
A
B
T = C,?E = E2 – E1 =0
(1) Nitrogen,N2,diatomic gas,
and T keep constant
Chapter 17 Heat and the First Law of Thermodynamics
E=E2 – E1
0 pVp WWWWQ
J1.8)( ACAP VVpW
(2)?E=? =0
For different processes,both Q and W are different
(过程不同,Q,W 也不同 )!
Again negative W,so,release heat,too!
J
V
VVpWQ
T 3.16ln
1
2
11
W < 0,negative W,release heat!
C A
B
Reading,page 404-417
Chapter 17 Heat and the First Law of Thermodynamics
Example,Page 425,27
When a gas is taken from
a to c along the curved
path,the work done by
the gas is W=-35J and the
heat added to the gas is
Q=-63J,Along path abc,
the work done is W=-48J.
(a)What is Q for path abc?
(b) If,What is W for path cda?
bc PP 2
1?
Chapter 17 Heat and the First Law of Thermodynamics
(c) What is Q for path cda?
(d) What is?
(e) If,what is Q for path da?
ca UU?
JUU cd 5
(a)We can find the internalenergy change U
ca from theinformation
For the curved path a-c
Uc-Ua=Qac-Wac=-63-(-35)=-28
For the path a-b-c,we have
Uc-Ua=Qabc-Wabc=Qabc-Wab
-28=Qabc-(-48)
Which gives Qabc=-76J
Chapter 17 Heat and the First Law of Thermodynamics
(b)For the path c-d-a,
Work is done only during
the constant pressure
process c to d,so we have
Wcda=Pc(Vd-Vc)
=(1/2)Pb(Va-Vb)
= (1/2) Wba= -(1/2) Wab
= -(1/2)(-48)=24J
Chapter 17 Heat and the First Law of Thermodynamics
c) We use the first law of
thermodynamics for the
path c-d-a to find Qcda,
Ua-Uc= -(Uc-Ua) =Qcda-Wcda
-(-28)=Qcda-(24)
Which gives Qcda=52J
d) Ua-Uc= -(Uc-Ua)
=-(-28)=28J
Chapter 17 Heat and the First Law of Thermodynamics
e) Because there is no
work done for the path
d-a,we have,
Ua-Ud= (Ua-Uc) + (Uc-Ud)
= (Ua-Uc) - (Ud-Uc) =Qda-
Wda
28-5= Qda-0,which gives
Qda=23J
Chapter 17 Heat and the First Law of Thermodynamics
绝热线和等温线绝热线的斜率大于等温线的斜率,
Ap
BVAV
A
p
Vo
T
0?Q
V?
ap?
Tp?
B
C
常量绝热 过程曲线 (红 )的斜率等温 过程曲线 (蓝 )的斜率
0dd pVVp
0dd1 pVVpV
A
A
a V
p
V
p)
d
d(
A
A
T V
p
V
p)
d
d(
pV 常量
pV 常量
Chapter 17 Heat and the First Law of Thermodynamics
Homework,
P425,20,22,28,
P426,40
P427,47
Discuss,
P425:18,19,21,23,
P426,29,35,42,45
P427,48
Chapter 17 Heat and the First
Law of Thermodynamics
1,Heat and Energy Transfer,Internal Energy
and Specific Heat
2,The First Law of Thermodynamics
3,Applying 1st Law of Thermodynamics
4,Molar Specific Heats for Gases,and the
Equipartition of Energy
5,Adiabatic Expansion of a Gas
Chapter 17 Heat and the First Law of Thermodynamics
Heat 热量
Calorie(Cal) 卡路里
Mechanical equivalent of heat 热功当量
Internal energy 内能
Degrees of Freedom 自由度
diatomic molecules 双原子分子气体
monatomic molecules 单原子分子气体
polyatomic molecules 多原子分子气体
Quasi-Steady Process 准静态过程
Chapter 17 Heat and the First Law of Thermodynamics
Isothermal (Constant-Temperature) 等温
Isochoric (Constant-Volume) 等容
Isobaric (Constant-pressure) 等压
Adiabatic Process 绝热过程
Adiabatic Free Expansion 绝热自由膨胀
Specific heat and Molar Specific Heats
热容和摩尔热容
Chapter 17 Heat and the First Law of Thermodynamics
§ 17-1 Heat as Energy Transfer P404-405
1 Energy and heat
Heat refers to a
transfer of energy.
Joule’s Experiment
on the mechanical
equivalent of heat.
焦耳 (Joule)实验
1) Conclusion:
A given amount of work done
was always equivalent to a
particular amount of heat input.
Falling
weight
paddle
Chapter 17 Heat and the First Law of Thermodynamics
Absorbing heat,Q>0； Losing heat,Q<0
Heat is path-dependent quantity（过程量）,
2) Heat（ 热量 Q）,
Heat is the energy that is transferred
between a system and its environment because of
a temperature difference that exists between
them.（ its unit is J---Joules）
Heat and temperature are often confused.
Chapter 17 Heat and the First Law of Thermodynamics
The amount of heat necessary to raise the
temperature of 1 gram of water by 1 Celsius
degree.
3 Calorie
1 Calorie=4.186 J
In SI units,the unit for heat is the joule.
4 Mechanical equivalent of heat
Example 17-1 page 405
Chapter 17 Heat and the First Law of Thermodynamics
在研究气体的能量时，需要确定其物理模型，这个物理模型就是自由度。用自由度模型描写气体能量是有局限性的，对少原子分子气体，在常温下理论值与实验值符合得较好，但对多原子分子或在高温情况下，理论值与实验值相差较大。这得用量子物理方法进行研究。
本节我们使用自由度模型和能量均分的统计原理来研究理想气体在常温下的气体能量。
17-2&17-3 Heat as Energy Transfer ; Internal
Energy and Specific Heat (404-407)
Chapter 17 Heat and the First Law of Thermodynamics
Degrees of Freedom (自由度 )p414
1自由度,
是描写物体在空间位置所需的独立坐标数。
Degree of Freedom —– numbers of
independently describing the motion of a
molecule.
例如：物体沿一维直线运动，最少只需一个坐标，
则自由度数为 1。
Chapter 17 Heat and the First Law of Thermodynamics
轮船在海平面上行驶,要描写轮船的位置至少需要两维坐标,则自由度为 2。
飞机在天空中飞翔，要描写飞机的空间位置至少需要三维坐标,则自由度为 3。
vrti2、自由度数目平动转动振动
Chapter 17 Heat and the First Law of Thermodynamics
z
y
x
),,( zyxP
o2） 两个刚性质点
3 质点的自由度
1） 一个质点，描写它的空间位置，需要 3 个平动自由度,3?t
3?t
描写其质心位置需 3个平动自由度，
描写其绕 x,y轴转动需 2个转动自由度，绕 z轴的转动能量可不计，
2?r
总自由度数
523 rti
z
y
x
o
Chapter 17 Heat and the First Law of Thermodynamics
3） 三个或三个以上的刚性质点需 3个平动自由度和 3个转动自由度。
633 rti
3?t
3?r
平动自由度转动自由度总自由度
Chapter 17 Heat and the First Law of Thermodynamics
1） 单原子分子气体 monatomic molecules
例如：氦气 ( helium,He)等为单原子分子气体。其模型可用一个质点来代替。
y
z
x
o
平动自由度 3?t
转动自由度 0?r
303 rti
总自由度
4 气体分子的自由度 (p414)
对于理想气体在常温下，分子内各原子间的距离认为不变，只有平动自由度 translational (DF,t)、转动自由度 rotational motion (DF,r) 。
Chapter 17 Heat and the First Law of Thermodynamics
2） 双原子分子气体 diatomic molecules
例如：氢气（ H2）、氧气
（ oxygen,O2）、氮气（ N2）
等为双原子分子气体。其模型可用两个刚性质点模型来代替。
平动自由度 3?t
转动自由度 2?r
523 rti总自由度
Chapter 17 Heat and the First Law of Thermodynamics
3）多原子分子气体 polyatomic molecules
如,二氧化碳气体（ CO2）、水蒸气（ H2O）等为多原子分子气体。其模型可用多个刚性质点来代替。
平动自由度 3?t
转动自由度 3?r
633 rti总自由度
Monatomic 3 0 3
diatomic 3 2 5
polyatomic 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
Chapter 17 Heat and the First Law of Thermodynamics
由温度公式有分子平均平动动能
The theorem(principle) of the equipartition of
energy 能量均分定理 (p414)
kTm
2
3
2
1 2
kt v?
2222
3
1 vvvv
zyx
kTmmm zyx 21212121 222 vvv
由于分子运动在哪个方向都不占优势，因此，在 y、
z 方向的自由度上也都平均分配 了 kT / 2 的能量。
Chapter 17 Heat and the First Law of Thermodynamics
各平动自由度的平动动能相等；每个平动自由度上分配了一份 kT/2的能量。
平动动能 转动动能由于分子的激烈碰撞（几亿次 /秒），平动动能与平动动能，平动动能和转动动能之间不断转换；
平动动能 平动动能单原子分子平均能量 kT213
因此使得所有的自由度中没有哪个是特殊的,即各自由度的平均动能都是相等的，即：
Chapter 17 Heat and the First Law of Thermodynamics
能量均分定理：在温度为 T的平衡态下，气体分子每个自由度的平均动能都相等，而且等于 kT/2 。
Principle of equipartition of Energy:
Energy is shared equally among the active
degrees of freedom,and in particular each
active degree of freedom of a molecule has on
the average an energy equal to kT/2.
分子的平均能量
kTi
2
根据能均分定理，如果气体分子的总自由度为 i 个，
则它的平均动能就有 i份 kT/2 的能量。
Chapter 17 Heat and the First Law of Thermodynamics
单 原子分子 3 0 3
双 原子分子 3 2 5
多 原子分子 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
kTK 23?
kTK 25?
kTK 3?
Chapter 17 Heat and the First Law of Thermodynamics
Internal energy（内能）,P406
1 Internal energy
The sum total of all the energy of all the molecules
in an object.
定义：气体的内能是指它的内动能,即它所包含的所有分子的动能 （ 相对于质心参考系 ） 和分子间的相互作用势能的总和 。
对于理想气体，分子之间无势能,因此理想气体的内能就是它的所有分子的动能之和 。
Chapter 17 Heat and the First Law of Thermodynamics
一个分子的能量为 kT
i
2
1 mol气体分子的能量为,kTN
i
02 RT
i
2?
m 千 克气体的 内能 为：
RTiMmU 2?
Chapter 17 Heat and the First Law of Thermodynamics
单 原子分子 3 0 3
双 原子分子 3 2 5
多 原子分子 3 3 6
刚性 分子能量自由度
t r i分子自由度 平动 转动 总
n R TU 23?
n R TU 25?
n R TU 3?
Chapter 17 Heat and the First Law of Thermodynamics
2 Discussion
1) Internal energy is a function of state,it is not
path-dependent quantity.（状态量）
RTi
M
mU
2
始末两态气体内能的变化。
内能增量 U?
12 UUU
TRi
m
MU
2
）12(2 TTRiMm
The internal energy of an ideal gas is a function
of the gas temperature only; it does not depend
on any other variables.
Chapter 17 Heat and the First Law of Thermodynamics
)( TUU?
2) 理想气体内能,表征系统状态的单值函数,理想气体的内能仅是温度的函数,
3) Distinguishing Temperature,Heat,and
Internal Energy
Temperature is a measure of the average kinetic
energy of individual molecules,Internal Energy
refer to the total energy of all the molecules in
the object,Heat,finally,refers to a transfer of
energy from one object to another because of a
difference in temperature.
Chapter 17 Heat and the First Law of Thermodynamics
Quasi-Steady Process (准静态过程 p409)
An equilibrium state of a
system can be described by
several parameters,such as
the pressure p,the volume
V,and the temperature T.
In the parameter space
(that is p-V-T phase
diagram),it is expressed by
a point.
§ 17-4 The First Law of Thermodynamics (407-408)
),,( 111 TVp
),,( 222 TVp
1V 2V
1p
2p
p
Vo
1
2
Chapter 17 Heat and the First Law of Thermodynamics
Now the system experiences a series changes
from state 1 to state 2,Each state is a new
equilibrium state.
气体活塞砂子 ),,( 111 TVp
),,( 222 TVp
1V 2V
1p
2p
p
Vo
1
2
准静态过程：从一个平衡态到另一平衡态所经过的每一中间状态均可近似当作平衡态的过程,
Chapter 17 Heat and the First Law of Thermodynamics
S
dl
dVP
由功的定义：
lpSlFW ddd
VpW dd?
21 dVV VpW
Work of Quasi-Steady Process
功是能量传递和转换的量度，它引起系统热运动状态的变化,
Chapter 17 Heat and the First Law of Thermodynamics
Discussion
1、如果 0?dV,则 0?dA
即系统体积膨胀时，系统对外界做功； If the work is done by
system,then W>0; conversely,
if the work is done on system,
then W<0.
0?dV 0?dA2、如果,则即系统体积收缩时，系统对外界做负功，即外界对系统做功。
Chapter 17 Heat and the First Law of Thermodynamics
由积分的意义可知,
功的大小等于 P~V图上过程曲线下的面积
It is the shaded area
under the p-V process curve
between initial to final state
P d VS VV 21W?
3.在 P~V图中曲线下的面积
1V
o
P
V
2V
1
2
dV
P
Chapter 17 Heat and the First Law of Thermodynamics
4,Work is path-dependent quantity (p411)
功是过程量如：从图中可看出,1?2与
1?1’?2两个过程的始末状态相同，但过程曲线不同，两条曲线下的面积不同，则作功也不同。
1V
o
P
V
2V
1
2
'1
The work done in taking a system from one state
to another depends on the initial and final states
but also on the type of process (path).
Chapter 17 Heat and the First Law of Thermodynamics
1T
2T
21 TT?
1）过程量：与过程有关；
2）等效性：改变系统热运动状态作用相同；
宏观运动 分子热运动功分子热运动分子热运动 热量
Q
3）功与热量的物理本质不同,
1卡 = 4.18 J,1 J = 0.24 卡
5 功与热量的异同
Chapter 17 Heat and the First Law of Thermodynamics
Experimentally,it is found that both work W
and heat Q are dependent on the nature of the
process,However,a surprising thing is that the
quantity Q-W is the same for all processes
between two states,
The First Law of Thermodynamics (P407)
实验证明系统从 A 状态变化到 B 状态，可以采用做功和传热的方法，不管经过什么过程，只要始末状态确定，做功和传热之和保持不变,
Chapter 17 Heat and the First Law of Thermodynamics
2
A
B
1*
*
p
Vo
02121 ABAABA QW
BABABABA QWQW 2211
2
A
B
1*
*
p
Vo
Chapter 17 Heat and the First Law of Thermodynamics
One can infer that the quantity Q-W must
represent a change in some intrinsic property of
the system,This intrinsic property is internal
energy?U.
From conservation of energy:
The change in internal energy of a closed
system will be equal to the heat added to the
system minus the work done by the system.
WQUUU if
The First Law of Thermodynamics
Chapter 17 Heat and the First Law of Thermodynamics
dWdQdU
For a differential process,it can be written as:
It is hold for any system and any process!
2
1
d
V
V
VpUQ
准静态过程定律：系统 从外界吸收的热量,一部分使系统的内能增加,另一部分使系统对外界做功,
The first law of thermodynamics is the principle
of conservation of energy for a thermodynamic
process.
Chapter 17 Heat and the First Law of Thermodynamics
The sign of W,Q,and U,(p408)
work done by the system W > 0;
Work is done on the system,W<0
Heat is added to the system Q > 0;
Heat leaves the system,Q<0
system’s energy increasing? U > 0.
system’s energy decreasing? U < 0.
Chapter 17 Heat and the First Law of Thermodynamics
+
12 EE?
系统吸热系统放热内能增加内能减少系统对外界做功外界对系统做功第一定律的符号规定
Q W
Chapter 17 Heat and the First Law of Thermodynamics
Two ways to change system’s internal energy:
Energy may be transferred into or out of a
system as either work W or heat Q.
Relationship between W and Q,1 cal =4.18 J
The transformation between work and heat is not direct,
but on the system.热 — 功转换不是直接进行的,而是间接的,内能是传递工具 。
系统吸热后，先使内能增加，再通过降低内能对外作功。
热功? 外界对系统作功，使内能增加，再通过内能降低，系统放热。
功热?
Chapter 17 Heat and the First Law of Thermodynamics
1,Specific Heats (热容 ) P407
§ 17-3 Specific Heat & § 17-6 Molar
Specific Heats for Gas (P407 &P412-414)
热容：系统和外界之间的热传递会引起系统本身温度的变化，这一温度的变化和热传递的关系用热容表示。
The amount of heat Q required to change the
temperature of a given material is proportional
to the mass m of the material and to the
temperature change,(P407)
TmcQ
T
Qmc
Chapter 17 Heat and the First Law of Thermodynamics
Unit?C?J / k g
Table 17-1 on page 407
The values of c depend on temperature,but for
temperature changes that are not too great,c can
often be considered constant.
Example 17-2 on page 407
Chapter 17 Heat and the First Law of Thermodynamics
其是 1 mol 气体升高 1 oC 所吸收的热量。
The heat require to raise 1mol of the gas by 1 oC,
2 Molar specific heat (摩尔热容 ) p413
1 mol 的物质温度升高 dT 时，如果吸收的热量为 dQ，
则 摩尔热容的定义为：
T
QC
d
d?
In analogy to this equation,the heat Q needed to raise
the temperature of n moles of gas by dT is,
T
Q
nC d
d1?
Chapter 17 Heat and the First Law of Thermodynamics
Molar specific heat at constant pressure 定体热容
Molar specific heat at constant volume 定压热容分别由定压和定体条件下物质吸收的热量决定。
3、摩尔热容的分类
Chapter 17 Heat and the First Law of Thermodynamics
因为这一部分很重要,内容又多,所以把讲过的内容先复习一遍,然后再开始第一定律的应用,
Chapter 17 Heat and the First Law of Thermodynamics
heat P404
calorie (cal) Calorie (kilocalorie) P405
mechanical equivalent of heat P405
internal energy P406
degrees of freedom P414
quasi-steady process P408
monatomic molecules P414
diatomic molecules P414
polyatomic molecules 补充
Chapter 17 Heat and the First Law of Thermodynamics
isothermal (Constant-Temperature) process P409
isochoric (Constant-Volume) process P410
isobaric (Constant-pressure) process P410
adiabatic process P409
adiabatic free expansion P412
specific heat P407
molar specific heats P413
Chapter 17 Heat and the First Law of Thermodynamics
1 Heat
Heat refers to a transfer of energy.
Three concepts
Absorbing heat,Q>0； Losing heat,Q<0
Heat is path-dependent quantity（过程量）,
Chapter 17 Heat and the First Law of Thermodynamics
2 Internal energy
RTi
M
mU
2
)( TUU?
The total of all the energy of all the molecules
in an object.
Internal energy is a function of state,it is not
path-dependent quantity.（状态量）
Chapter 17 Heat and the First Law of Thermodynamics
Internal energy of molecules
3 Temperature
Temperature is a measure of the average kinetic
energy of individual molecules.
degree of freedom internal energy
Monatomic 3
diatomic 5
polyatomic 6
n R TU 23?
n R TU 25?
n R TU 3?
Chapter 17 Heat and the First Law of Thermodynamics
Heat is path-dependent quantity.
Temperature and internal energy are path-
independent quantities.
discussion:
Chapter 17 Heat and the First Law of Thermodynamics
The First Law of Thermodynamics
S
ld
dVP由功的定义：
21 dVV VpW
1 Work of Quasi-Steady Process (P410)
Chapter 17 Heat and the First Law of Thermodynamics
Work done by an ideal
gas is the shaded area
under the PV curve
between initial to final
state,
2.在 P~V图中曲线下的面积 (P411)
1V
o
P
V
2V
1
2
Vd
P
Chapter 17 Heat and the First Law of Thermodynamics
3,Work is path-dependent quantity (p411)
1Vo
P
V
2V
1
2
'1
The work done in
taking a system from
one state to another
depends on the initial
and final states but also
on the type of process
(path).
Chapter 17 Heat and the First Law of Thermodynamics
4 The First Law of Thermodynamics (P407)
WQUUΔU if
2
1
d
V
V
VpUQ
准静态过程
The change in internal energy of a closed system,
is to equal to the heat added to the system minus
the work done by the system,
In differential form:
dWdQdU
Chapter 17 Heat and the First Law of Thermodynamics
+
12 EE?
系统吸热系统放热内能增加内能减少系统对外界做功外界对系统做功第一定律的符号规定
Q W
Chapter 17 Heat and the First Law of Thermodynamics
5 Specific Heats and Molar Specific Heats
1) Specific Heats (P407)
T
Qmc
2) Molar specific heat (P413)
T
Q
nC d
d1?
Molar specific heat at constant pressure 定体热容
Molar specific heat at constant volume 定压热容
Chapter 17 Heat and the First Law of Thermodynamics
计算各等值过程的热量、功和内能的理论基础
RT
M
mpV?（ 1） （ 理想气体的 共性）
21 dVV VpEQ
VpEQ ddd
（ 2）
解决过程中能量转换的问题
)( TUU?
（ 3） （ 理想气体的状态函数 ）
（ 4） 各等值过程的特性,
§ 17-5 Some Special Cases of
The First Law of Thermodynamics (P409-417)
Chapter 17 Heat and the First Law of Thermodynamics
1,Isothermal (Constant-Temperature) Process
(P409,P411)
The process carried out at constant temperature.
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo Vd
恒温热源
T
Movable piston
Chapter 17 Heat and the First Law of Thermodynamics
热力学第一定律
0d?U
V
RTnp?
21 dVV VpW
VpWQ T ddd
特征 常量?T
过程方程?pV 常量
TRi
M
mU
2
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo Vd
Chapter 17 Heat and the First Law of Thermodynamics
UU
1
2
V
Vn R T ln
2
1ln
p
pn R T?
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo
等温 膨胀
W
1
2
),,( 11 TVp
),,( 22 TVp
1p
2p
1V 2V
p
Vo
W
等温 压缩
TQ TQW W
dV
V
RTnWQ V
V
T
2
1
Chapter 17 Heat and the First Law of Thermodynamics
2,Isochoric (Constant-Volume) Process (p410,P414)
The volume of the system
is held constant,no work
is done for the system.
0d,0d WV
热力学第一定律
UQ V dd?
特性 常量?V
),,( 11 TVp
),,( 22 TVp2p
1p
V
p
Vo
过程方程 常量1PT
dU > 0,system absorbs heat from its environment;< 0,then the system releases heat.
Chapter 17 Heat and the First Law of Thermodynamics
vvV T
U
nT
Q
nC )d
d(1)
d
d(1
定体摩尔热容,理想气体在等体过程中吸收的热量,使温度升高,其定体摩尔热容为,
mol n
VQd Td
n R TiU 2?
将 代入，可得,
RiC V
2
So,we may express internal energy of IG as:
TnCU V dd?TnCU V? or
Chapter 17 Heat and the First Law of Thermodynamics
TnCUQ VV ddd
1212 )( UUTTnCQ VV
1U
2U
VQ
1U
VQ
2U
),,( 11 TVp
),,( 22 TVp2p
1p
V
p
Vo
等体升压 1
2 ),,(
11 TVp
),,( 22 TVp2p
1p
V
p
Vo
等体降压
1
2
Chapter 17 Heat and the First Law of Thermodynamics
The heat absorbed by gas is used for increasing
its internal energy and doing work,气体吸收的热量一部分用于增加内能，另一部分用于对外作功。
3,Isobaric (Constant-Pressure) Process (P410,P414)
The pressure of the system is held constant.
2V
),,( 11 TVp ),,( 22 TVp
p
1V
p
Vo
1 2过程方程 常量1VT
热一律
WUQ p ddd
特 性 常量?p
)( 12 VVpW功 W
Chapter 17 Heat and the First Law of Thermodynamics
ppp T
V
n
p
T
U
nT
Q
n
C )
d
d(
d
d1)
d
d
(1
定压摩尔热容,理想气体在等压过程中吸收的热量,温度升高,其定压摩尔热容为
mo ln
pQd Td
将
n R TpVn R TiU,2
代入，
RRiC p
2
可得
Chapter 17 Heat and the First Law of Thermodynamics
)( 12 VVpW )( 12 TTnR
)( 1212 TTnCUU V
),( 12 TTnCQ pp
Chapter 17 Heat and the First Law of Thermodynamics
2V
),,( 11 TVp ),,( 22 TVp
p
1V
p
Vo
1 2
W
等压膨胀
2V
),,( 11 TVp),,( 22 TVp
p
1V
p
Vo
12
W
等压压缩
1U
2UpQ
1U
pQ
2U
W
W
Chapter 17 Heat and the First Law of Thermodynamics
RCC Vp
1) 定压摩尔热容和定体摩尔热容的关系
RRiC p
2
RiC V
2
由 和该公式给出了定压摩尔热容和定体摩尔热容的关系,
称为 迈耶公式 (Mayer Formula p414 )。 迈耶利用该公式算出了热功当量,对建立能量守恒作出了重要贡献 。
DISCUSSION
Chapter 17 Heat and the First Law of Thermodynamics
2) CP > CV 的物理意义 RCC VP
1 mol 理想气体温度升高 1 oC,对于等容过程,体积不变吸热只增加系统内能,而对于等压过程除了增加系统内能外,还要对外作功,所吸收的热量要更多一些 。
In the process done at constant volume,no work is done
since is zero,Thus according to the first law of
thermodynamics,the heat added all goes into increasing
the internal energy of the gas,In the process out at
constant pressure,work is done,and hence the heat
added,must not only increase the internal energy but lso
is used to do the work,Thus more heat must added in
this process than in the first process at constant volume.
V?
Chapter 17 Heat and the First Law of Thermodynamics
3) 摩尔热容比 p416
单原子分子 3 3R/2 5R/2 1.67
刚性双原子分子 5 5R/2 7R/2 1.40
刚性多原子分子 6 3R 4R 1.33
VC pC?i
2
2 i
C
C
V
P?RRiC
p 2 R
iC
V 2?
由 和叫摩尔热容比。
Chapter 17 Heat and the First Law of Thermodynamics
4,Adiabatic Process (绝热过程 ),P409-410,P416-417
An adiabatic process
is one in which no heat is
allowed to flow into or out
of the system.
绝热过程是系统在和外界无热量交换的条件下进行的过程。用把系统和外界隔开隔能壁（或叫绝热壁）就可以实现这一过程。
绝热的汽缸壁和活塞
Chapter 17 Heat and the First Law of Thermodynamics
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
)( 12 TTnC V
Od?Q特征
TCn VT
T
d2
1?
TnCU V dd?
21 dVV VpW
Vd
UW dd
热一律 0dd UW
Chapter 17 Heat and the First Law of Thermodynamics
)( 2211 VpVp
CC
CW
Vp
V?
1
2211
VpVpW
若已知 及 2211,,,VpVp?
)( 2211
R
Vp
R
VpCW
V
nR TpV?
从 和 可得 RCC VP
Chapter 17 Heat and the First Law of Thermodynamics
绝热过程方程的推导 P416
UWQ dd,0d
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
0?Q
QUW ddd
dTnCU V?d
Combine these two equations,we obtain:
0dd VPTnC V
We next take the differential of the ideal gas law,
nR TpV?
Chapter 17 Heat and the First Law of Thermodynamics
0ddd VP)nR PVVP(nC V
Allowing P,V,and T to vary:
TnRPVVP ddd
We solve for dT in this relation and substitute it
into the previous relation and get
or
0)( V d PCP d VRC VV
We note from that
PV CRC
So we have
Chapter 17 Heat and the First Law of Thermodynamics
0dd PVCVPC VP
0dd
V
V
P
P?
So that last equation becomes
This is integrated to become
c on s t a ntlnln VP?
This simplifies to (using the rules for addition
and multiplication of logarithms) to
c on s t a ntPV
Chapter 17 Heat and the First Law of Thermodynamics
绝热方程
TV 1?
pV
Tp 1
常量常量常量
Chapter 17 Heat and the First Law of Thermodynamics
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V 2V
p
Vo
W
绝 热 膨 胀
),,( 111 TVp
),,( 222 TVp
1
2
1p
2p
1V2V
p
Vo
W
绝 热 压 缩
1E
2E
1E
2E
W
W
Chapter 17 Heat and the First Law of Thermodynamics
Example 17-7 on page 417
Chapter 17 Heat and the First Law of Thermodynamics
5,Adiabatic Free Expansion,绝热自由膨胀 (p412)
A adiabatic process in which a gas is allowed to
expand in volume adiabatically without doing
any work.
Chapter 17 Heat and the First Law of Thermodynamics
5,Adiabatic Free Expansion,绝热自由膨胀 (p412)
定义：在绝热容器中间有一个隔板,
左半部充以理想气体并处于平衡态,
右半部为真空 。 左半部原来处在平衡态 。 现抽去隔板,气体将冲入右半部,最后可以在整个容器内达到一个新的平衡态,这种过程叫 绝热自由膨胀 。
Chapter 17 Heat and the First Law of Thermodynamics
0d?Q特征：由于过程是绝热的，
由热力学第一定律 0dd UW
由于气体是向真空中冲入，所以它对外界不做功，即
dW=0
即 气体经过自由膨胀，内能保持不变 。
The internal energy of a gas does not change in the
free expansion.
所以,0d?U
12 UU?
即
Chapter 17 Heat and the First Law of Thermodynamics
Since U depends only on T,so
对于理想气体，内能只包含分子的热运动动能，它只是温度的函数，所以经过自由膨胀，理想气体再达到平衡态时，它的温度复原。
0 T
由理想气体状态方程，对于初、末状态分别有：
,111 n R TVp? 222 n R TVp?
因为 T2=T1,V2=2V1
12 2
1 pp?得到
Chapter 17 Heat and the First Law of Thermodynamics
理想气体的各等值过程、绝热过程公式对照表过程 特征 过程方程 吸收热量 对外做功 内能增量
Isochoric
等容
Isobaric
等压
Isothermal
等温
Adiabatic
绝热
C?Tp
C?TV
C?pV
'CpV
''C1 TV?
C1 Tp
)( 12 TTCMm V?
0
)( 12 TTCMm V?
)( 12 TTCMm V?
)( 12 TTCMm V?
)( 12 TTCMm p?
)(
or)(
12
12
TTR
M
m
VVp
2
1
1
2
ln
orln
p
p
RT
M
m
V
V
RT
M
m
2
1
1
2
ln
orln
P
P
RT
M
m
V
V
RT
M
m
0
0
1
or
)(
2211
12
VPVP
TTC
M
m
V
p=C
V=C
T=C
dQ=0
Chapter 17 Heat and the First Law of Thermodynamics
Solutio
n:
Example,N2 of volume 100cm3 and pressure
1.013× 105Pa is compressed to volume 20cm3,
compute the change in internal energy?E,the
heat Q and the work done W during processes,(1)
Isothermal compress; (2) isobaric compress then
isochoric increasing pressure to same state 先等压压缩,再等体升压到同样状态,
C
V
p
O
A
B
T = C,?E = E2 – E1 =0
(1) Nitrogen,N2,diatomic gas,
and T keep constant
Chapter 17 Heat and the First Law of Thermodynamics
E=E2 – E1
0 pVp WWWWQ
J1.8)( ACAP VVpW
(2)?E=? =0
For different processes,both Q and W are different
(过程不同,Q,W 也不同 )!
Again negative W,so,release heat,too!
J
V
VVpWQ
T 3.16ln
1
2
11
W < 0,negative W,release heat!
C A
B
Reading,page 404-417
Chapter 17 Heat and the First Law of Thermodynamics
Example,Page 425,27
When a gas is taken from
a to c along the curved
path,the work done by
the gas is W=-35J and the
heat added to the gas is
Q=-63J,Along path abc,
the work done is W=-48J.
(a)What is Q for path abc?
(b) If,What is W for path cda?
bc PP 2
1?
Chapter 17 Heat and the First Law of Thermodynamics
(c) What is Q for path cda?
(d) What is?
(e) If,what is Q for path da?
ca UU?
JUU cd 5
(a)We can find the internalenergy change U
ca from theinformation
For the curved path a-c
Uc-Ua=Qac-Wac=-63-(-35)=-28
For the path a-b-c,we have
Uc-Ua=Qabc-Wabc=Qabc-Wab
-28=Qabc-(-48)
Which gives Qabc=-76J
Chapter 17 Heat and the First Law of Thermodynamics
(b)For the path c-d-a,
Work is done only during
the constant pressure
process c to d,so we have
Wcda=Pc(Vd-Vc)
=(1/2)Pb(Va-Vb)
= (1/2) Wba= -(1/2) Wab
= -(1/2)(-48)=24J
Chapter 17 Heat and the First Law of Thermodynamics
c) We use the first law of
thermodynamics for the
path c-d-a to find Qcda,
Ua-Uc= -(Uc-Ua) =Qcda-Wcda
-(-28)=Qcda-(24)
Which gives Qcda=52J
d) Ua-Uc= -(Uc-Ua)
=-(-28)=28J
Chapter 17 Heat and the First Law of Thermodynamics
e) Because there is no
work done for the path
d-a,we have,
Ua-Ud= (Ua-Uc) + (Uc-Ud)
= (Ua-Uc) - (Ud-Uc) =Qda-
Wda
28-5= Qda-0,which gives
Qda=23J
Chapter 17 Heat and the First Law of Thermodynamics
绝热线和等温线绝热线的斜率大于等温线的斜率,
Ap
BVAV
A
p
Vo
T
0?Q
V?
ap?
Tp?
B
C
常量绝热 过程曲线 (红 )的斜率等温 过程曲线 (蓝 )的斜率
0dd pVVp
0dd1 pVVpV
A
A
a V
p
V
p)
d
d(
A
A
T V
p
V
p)
d
d(
pV 常量
pV 常量
Chapter 17 Heat and the First Law of Thermodynamics
Homework,
P425,20,22,28,
P426,40
P427,47
Discuss,
P425:18,19,21,23,
P426,29,35,42,45
P427,48
